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**CHAPTER 1
**

1.1. Given the vectors **M **= −10**a***x *+ 4**a***y *− 8**a***z *and **N **= 8**a***x *+ 7**a***y *− 2**a***z*, find:
a) a unit vector in the direction of −**M **+ 2**N**.

−**M **+ 2**N **= 10**a***x *− 4**a***y *+ 8**a***z *+ 16**a***x *+ 14**a***y *− 4**a***z *= *(*26*, *10*, *4*)
*Thus

**a **= *(*26*, *10*, *4*)*|*(*26*, *10*, *4*)*| = *(*0*.*92*, *0*.*36*, *0*.*14*)
*

b) the magnitude of 5**a***x *+ **N **− 3**M**:
*(*5*, *0*, *0*)*+ *(*8*, *7*,*−2*)*− *(*−30*, *12*,*−24*) *= *(*43*,*−5*, *22*)*, and |*(*43*,*−5*, *22*)*| = 48*.*6.

c) |**M**||2**N**|*(***M **+ **N***)*:
|*(*−10*, *4*,*−8*)*||*(*16*, *14*,*−4*)*|*(*−2*, *11*,*−10*) *= *(*13*.*4*)(*21*.*6*)(*−2*, *11*,*−10*)
*= *(*−580*.*5*, *3193*,*−2902*)
*

1.2. Given three points, *A(*4*, *3*, *2*)*, *B(*−2*, *0*, *5*)*, and *C(*7*,*−2*, *1*)*:
a) Specify the vector **A **extending from the origin to the point *A*.

**A **= *(*4*, *3*, *2*) *= 4**a***x *+ 3**a***y *+ 2**a***z
*

b) Give a unit vector extending from the origin to the midpoint of line *AB*.

The vector from the origin to the midpoint is given by

**M **= *(*1*/*2*)(***A **+ **B***) *= *(*1*/*2*)(*4 − 2*, *3 + 0*, *2 + 5*) *= *(*1*, *1*.*5*, *3*.*5*)
*The unit vector will be

**m **= *(*1*, *1*.*5*, *3*.*5*)*|*(*1*, *1*.*5*, *3*.*5*)*| = *(*0*.*25*, *0*.*38*, *0*.*89*)
*

c) Calculate the length of the perimeter of triangle *ABC*:

Begin with **AB **= *(*−6*,*−3*, *3*)*, **BC **= *(*9*,*−2*,*−4*)*, **CA **= *(*3*,*−5*,*−1*)*.
Then

|**AB**| + |**BC**| + |**CA**| = 7*.*35 + 10*.*05 + 5*.*91 = 23*.*32

1.3. The vector from the origin to the point *A *is given as *(*6*,*−2*,*−4*)*, and the unit vector directed from the
origin toward point *B *is *(*2*,*−2*, *1*)/*3. If points *A *and *B *are ten units apart, find the coordinates of point
*B*.

With **A **= *(*6*,*−2*,*−4*) *and **B **= 13*B(*2*,*−2*, *1*)*, we use the fact that |**B **− **A**| = 10, or
|*(*6 − 23*B)***a***x *− *(*2 − 23*B)***a***y *− *(*4 + 13*B)***a***z*| = 10
Expanding, obtain
36 − 8*B *+ 49*B*2 + 4 − 83*B *+ 49*B*2 + 16 + 83*B *+ 19*B*2 = 100
or *B*2 − 8*B *− 44 = 0. Thus *B *= 8±

√
64−176
2 = 11*.*75 (taking positive option) and so

**B **= 2
3
*(*11*.*75*)***a***x *− 2

3
*(*11*.*75*)***a***y *+ 1

3
*(*11*.*75*)***a***z *= 7*.*83**a***x *− 7*.*83**a***y *+ 3*.*92**a***z
*

1

1.4. given points *A(*8*,*−5*, *4*) *and *B(*−2*, *3*, *2*)*, find:
a) the distance from *A *to *B*.

|**B **− **A**| = |*(*−10*, *8*,*−2*)*| = 12*.*96

b) a unit vector directed from *A *towards *B*. This is found through

**a***AB *= **B **− **A**|**B **− **A**| = *(*−0*.*77*, *0*.*62*,*−0*.*15*)
*

c) a unit vector directed from the origin to the midpoint of the line *AB*.

**a**0*M *= *(***A **+ **B***)/*2|*(***A **+ **B***)/*2| =
*(*3*,*−1*, *3*)*√

19
= *(*0*.*69*,*−0*.*23*, *0*.*69*)
*

d) the coordinates of the point on the line connecting *A *to *B *at which the line intersects the plane *z *= 3.
Note that the midpoint, *(*3*,*−1*, *3*)*, as determined from part *c *happens to have *z *coordinate of 3. This
is the point we are looking for.

1.5. A vector field is specified as **G **= 24*xy***a***x *+ 12*(x*2 + 2*)***a***y *+ 18*z*2**a***z*. Given two points, *P(*1*, *2*,*−1*) *and
*Q(*−2*, *1*, *3*)*, find:

a) **G **at *P *: **G***(*1*, *2*,*−1*) *= *(*48*, *36*, *18*)
*b) a unit vector in the direction of **G **at *Q*: **G***(*−2*, *1*, *3*) *= *(*−48*, *72*, *162*)*, so

**a***G *= *(*−48*, *72*, *162*)*|*(*−48*, *72*, *162*)*| = *(*−0*.*26*, *0*.*39*, *0*.*88*)
*

c) a unit vector directed from *Q *toward *P *:

**a***QP *= **P **− **Q**|**P **− **Q**| =
*(*3*,*−1*, *4*)*√

26
= *(*0*.*59*, *0*.*20*,*−0*.*78*)
*

d) the equation of the surface on which |**G**| = 60: We write 60 = |*(*24*xy, *12*(x*2 + 2*), *18*z*2*)*|, or
10 = |*(*4*xy, *2*x*2 + 4*, *3*z*2*)*|, so the equation is

100 = 16*x*2*y*2 + 4*x*4 + 16*x*2 + 16 + 9*z*4

2

1.6. For the **G **field in Problem 1.5, make sketches of *Gx *, *Gy *, *Gz *and |**G**| along the line *y *= 1, *z *= 1, for
0 ≤ *x *≤ 2. We find **G***(x, *1*, *1*) *= *(*24*x, *12*x*2 + 24*, *18*)*, from which *Gx *= 24*x*, *Gy *= 12*x*2 + 24,
*Gz *= 18, and |**G**| = 6

√
4*x*4 + 32*x*2 + 25. Plots are shown below.

1.7. Given the vector field **E **= 4*zy*2 cos 2*x***a***x *+ 2*zy *sin 2*x***a***y *+ *y*2 sin 2*x***a***z *for the region |*x*|, |*y*|, and |*z*| less
than 2, find:

a) the surfaces on which *Ey *= 0. With *Ey *= 2*zy *sin 2*x *= 0, the surfaces are 1) the plane *z *= 0, with
|*x*| *< *2, |*y*| *< *2; 2) the plane *y *= 0, with |*x*| *< *2, |*z*| *< *2; 3) the plane *x *= 0, with |*y*| *< *2, |*z*| *< *2;
4) the plane *x *= *π/*2, with |*y*| *< *2, |*z*| *< *2.

b) the region in which *Ey *= *Ez*: This occurs when 2*zy *sin 2*x *= *y*2 sin 2*x*, or on the plane 2*z *= *y*, with
|*x*| *< *2, |*y*| *< *2, |*z*| *< *1.

c) the region in which **E **= 0: We would have *Ex *= *Ey *= *Ez *= 0, or *zy*2 cos 2*x *= *zy *sin 2*x *=
*y*2 sin 2*x *= 0. This condition is met on the plane *y *= 0, with |*x*| *< *2, |*z*| *< *2.

1.8. Two vector fields are **F **= −10**a***x*+20*x(y*−1*)***a***y *and **G **= 2*x*2*y***a***x*−4**a***y*+*z***a***z*. For the point *P(*2*, *3*,*−4*)*,
find:

a) |**F**|: **F **at *(*2*, *3*,*−4*) *= *(*−10*, *80*, *0*)*, so |**F**| = 80*.*6.
b) |**G**|: **G **at *(*2*, *3*,*−4*) *= *(*24*,*−4*,*−4*)*, so |**G**| = 24*.*7.
c) a unit vector in the direction of **F **− **G**: **F **− **G **= *(*−10*, *80*, *0*)*− *(*24*,*−4*,*−4*) *= *(*−34*, *84*, *4*)*. So

**a **= **F **− **G**|**F **− **G**| =
*(*−34*, *84*, *4*)
*

90*.*7
= *(*−0*.*37*, *0*.*92*, *0*.*04*)
*

d) a unit vector in the direction of **F **+ **G**: **F **+ **G **= *(*−10*, *80*, *0*)*+ *(*24*,*−4*,*−4*) *= *(*14*, *76*,*−4*)*. So

**a **= **F **+ **G**|**F **+ **G**| =
*(*14*, *76*,*−4*)
*

77*.*4
= *(*0*.*18*, *0*.*98*,*−0*.*05*)
*

3

1.9. A field is given as

**G **= 25
*(x*2 + *y*2*) (x***a***x *+ *y***a***y)
*

Find:
a) a unit vector in the direction of **G **at *P(*3*, *4*,*−2*)*: Have **G***p *= 25*/(*9 + 16*)*× *(*3*, *4*, *0*) *= 3**a***x *+ 4**a***y *,

and |**G***p*| = 5. Thus **a***G *= *(*0*.*6*, *0*.*8*, *0*)*.
b) the angle between **G **and **a***x *at *P *: The angle is found through **a***G *· **a***x *= cos *θ *. So cos *θ *=

*(*0*.*6*, *0*.*8*, *0*) *· *(*1*, *0*, *0*) *= 0*.*6. Thus *θ *= 53◦.
c) the value of the following double integral on the plane *y *= 7:

∫ 4 0

∫ 2 0

**G **· **a***ydzdx
*

∫ 4 0

∫ 2 0

25

*x*2 + *y*2 *(x***a***x *+ *y***a***y) *· **a***ydzdx *=
∫ 4

0

∫ 2 0

25

*x*2 + 49 × 7 *dzdx *=
∫ 4

0

350

*x*2 + 49*dx
*

= 350 × 1 7

[ tan−1

( 4

7

) − 0

] = 26

1.10. Use the definition of the dot product to find the interior angles at *A *and *B *of the triangle defined by the
three points *A(*1*, *3*,*−2*)*, *B(*−2*, *4*, *5*)*, and *C(*0*,*−2*, *1*)*:

a) Use **R***AB *= *(*−3*, *1*, *7*) *and **R***AC *= *(*−1*,*−5*, *3*) *to form **R***AB *· **R***AC *= |**R***AB *||**R***AC *| cos *θA*. Obtain
3 + 5 + 21 = √59√35 cos *θA*. Solve to find *θA *= 65*.*3◦.

b) Use **R***BA *= *(*3*,*−1*,*−7*) *and **R***BC *= *(*2*,*−6*,*−4*) *to form **R***BA *· **R***BC *= |**R***BA*||**R***BC *| cos *θB *. Obtain
6 + 6 + 28 = √59√56 cos *θB *. Solve to find *θB *= 45*.*9◦.

1.11. Given the points *M(*0*.*1*,*−0*.*2*,*−0*.*1*)*, *N(*−0*.*2*, *0*.*1*, *0*.*3*)*, and *P(*0*.*4*, *0*, *0*.*1*)*, find:
a) the vector **R***MN *: **R***MN *= *(*−0*.*2*, *0*.*1*, *0*.*3*)*− *(*0*.*1*,*−0*.*2*,*−0*.*1*) *= *(*−0*.*3*, *0*.*3*, *0*.*4*)*.
b) the dot product **R***MN *· **R***MP *: **R***MP *= *(*0*.*4*, *0*, *0*.*1*) *− *(*0*.*1*,*−0*.*2*,*−0*.*1*) *= *(*0*.*3*, *0*.*2*, *0*.*2*)*. **R***MN *·

**R***MP *= *(*−0*.*3*, *0*.*3*, *0*.*4*) *· *(*0*.*3*, *0*.*2*, *0*.*2*) *= −0*.*09 + 0*.*06 + 0*.*08 = 0*.*05.
c) the scalar projection of **R***MN *on **R***MP *:

**R***MN *· **a***RMP *= *(*−0*.*3*, *0*.*3*, *0*.*4*) *· *(*0*.*3*, *0*.*2*, *0*.*2*)*√
0*.*09 + 0*.*04 + 0*.*04 =

0*.*05√
0*.*17

= 0*.*12

d) the angle between **R***MN *and **R***MP *:

*θM *= cos−1
(

**R***MN *· **R***MP
*|**R***MN *||**R***MP *|

) = cos−1

(
0*.*05√

0*.*34
√

0*.*17

) = 78◦

4

1.12. Given points *A(*10*, *12*,*−6*)*, *B(*16*, *8*,*−2*)*, *C(*8*, *1*,*−4*)*, and *D(*−2*,*−5*, *8*)*, determine:
a) the vector projection of **R***AB *+ **R***BC *on **R***AD*: **R***AB *+ **R***BC *= **R***AC *= *(*8*, *1*, *4*) *− *(*10*, *12*,*−6*) *=

*(*−2*,*−11*, *10*) *Then **R***AD *= *(*−2*,*−5*, *8*) *− *(*10*, *12*,*−6*) *= *(*−12*,*−17*, *14*)*. So the projection will
be:

*(***R***AC *· **a***RAD)***a***RAD *=
[
*(*−2*,*−11*, *10*) *· *(*−12*,*−17*, *14*)*√

629

]
*(*−12*,*−17*, *14*)*√

629
= *(*−6*.*7*,*−9*.*5*, *7*.*8*)
*

b) the vector projection of **R***AB *+**R***BC *on **R***DC *: **R***DC *= *(*8*,*−1*, *4*)*− *(*−2*,*−5*, *8*) *= *(*10*, *6*,*−4*)*. The
projection is:

*(***R***AC *· **a***RDC)***a***RDC *=
[
*(*−2*,*−11*, *10*) *· *(*10*, *6*,*−4*)*√

152

]
*(*10*, *6*,*−4*)*√

152
= *(*−8*.*3*,*−5*.*0*, *3*.*3*)
*

c) the angle between **R***DA *and **R***DC *: Use **R***DA *= −**R***AD *= *(*12*, *17*,*−14*) *and **R***DC *= *(*10*, *6*,*−4*)*.
The angle is found through the dot product of the associated unit vectors, or:

*θD *= cos−1*(***a***RDA *· **a***RDC) *= cos−1
(
*(*12*, *17*,*−14*) *· *(*10*, *6*,*−4*)*√

629 √

152

) = 26◦

1.13. a) Find the vector component of **F **= *(*10*,*−6*, *5*) *that is parallel to **G **= *(*0*.*1*, *0*.*2*, *0*.*3*)*:

**F**||*G *= **F **· **G**|**G**|2 **G **=
*(*10*,*−6*, *5*) *· *(*0*.*1*, *0*.*2*, *0*.*3*)
*

0*.*01 + 0*.*04 + 0*.*09 *(*0*.*1*, *0*.*2*, *0*.*3*) *= *(*0*.*93*, *1*.*86*, *2*.*79*)
*

b) Find the vector component of **F **that is perpendicular to **G**:

**F***pG *= **F **− **F**||*G *= *(*10*,*−6*, *5*)*− *(*0*.*93*, *1*.*86*, *2*.*79*) *= *(*9*.*07*,*−7*.*86*, *2*.*21*)
*

c) Find the vector component of **G **that is perpendicular to **F**:

**G***pF *= **G**−**G**||*F *= **G**− **G **· **F**|**F**|2 **F **= *(*0*.*1*, *0*.*2*, *0*.*3*)*−
1*.*3

100 + 36 + 25 *(*10*,*−6*, *5*) *= *(*0*.*02*, *0*.*25*, *0*.*26*)
*

1.14. The four vertices of a regular tetrahedron are located at *O(*0*, *0*, *0*)*, *A(*0*, *1*, *0*)*, *B(*0*.*5
√

3*, *0*.*5*, *0*)*, and
*C(
*

√
3*/*6*, *0*.*5*,
*

√
2*/*3*)*.

a) Find a unit vector perpendicular (outward) to the face *ABC*: First find

**R***BA *× **R***BC *= [*(*0*, *1*, *0*)*− *(*0*.*5
√

3*, *0*.*5*, *0*)*] × [*(
*√

3*/*6*, *0*.*5*,
*√

2*/*3*)*− *(*0*.*5
√

3*, *0*.*5*, *0*)*]

= *(*−0*.*5
√

3*, *0*.*5*, *0*)*× *(*−
√

3*/*3*, *0*,
*√

2*/*3*) *= *(*0*.*41*, *0*.*71*, *0*.*29*)
*The required unit vector will then be:

**R***BA *× **R***BC
*|**R***BA *× **R***BC *| = *(*0*.*47*, *0*.*82*, *0*.*33*)
*

b) Find the area of the face *ABC*:

Area = 1
2
|**R***BA *× **R***BC *| = 0*.*43

5

1.15. Three vectors extending from the origin are given as **r**1 = *(*7*, *3*,*−2*)*, **r**2 = *(*−2*, *7*,*−3*)*, and **r**3 = *(*0*, *2*, *3*)*.
Find:

a) a unit vector perpendicular to both **r**1 and **r**2:

**a***p*12 = **r**1 × **r**2|**r**1 × **r**2| =
*(*5*, *25*, *55*)
*

60*.*6
= *(*0*.*08*, *0*.*41*, *0*.*91*)
*

b) a unit vector perpendicular to the vectors **r**1 − **r**2 and **r**2 − **r**3: **r**1 − **r**2 = *(*9*,*−4*, *1*) *and **r**2 − **r**3 =
*(*−2*, *5*,*−6*)*. So **r**1 − **r**2 × **r**2 − **r**3 = *(*19*, *52*, *32*)*. Then

**a***p *= *(*19*, *52*, *32*)*|*(*19*, *52*, *32*)*| =
*(*19*, *52*, *32*)
*

63*.*95
= *(*0*.*30*, *0*.*81*, *0*.*50*)
*

c) the area of the triangle defined by **r**1 and **r**2:

Area = 1
2
|**r**1 × **r**2| = 30*.*3

d) the area of the triangle defined by the heads of **r**1, **r**2, and **r**3:

Area = 1
2
|*(***r**2 − **r**1*)*× *(***r**2 − **r**3*)*| = 1

2
|*(*−9*, *4*,*−1*)*× *(*−2*, *5*,*−6*)*| = 32*.*0

1.16. Describe the surfaces defined by the equations:

a) **r **· **a***x *= 2, where **r **= *(x, y, z)*: This will be the plane *x *= 2.
b) |**r **× **a***x *| = 2: **r **× **a***x *= *(*0*, z,*−*y)*, and |**r **× **a***x *| =

√
*z*2 + *y*2 = 2. This is the equation of a cylinder,

centered on the *x *axis, and of radius 2.

1.17. Point *A(*−4*, *2*, *5*) *and the two vectors, **R***AM *= *(*20*, *18*,*−10*) *and **R***AN *= *(*−10*, *8*, *15*)*, define a triangle.
a) Find a unit vector perpendicular to the triangle: Use

**a***p *= **R***AM *× **R***AN*|**R***AM *× **R***AN *| =
*(*350*,*−200*, *340*)
*

527*.*35
= *(*0*.*664*,*−0*.*379*, *0*.*645*)
*

The vector in the opposite direction to this one is also a valid answer.

b) Find a unit vector in the plane of the triangle and perpendicular to **R***AN *:

**a***AN *= *(*−10*, *8*, *15*)*√
389

= *(*−0*.*507*, *0*.*406*, *0*.*761*)
*

Then

**a***pAN *= **a***p *× **a***AN *= *(*0*.*664*,*−0*.*379*, *0*.*645*)*× *(*−0*.*507*, *0*.*406*, *0*.*761*) *= *(*−0*.*550*,*−0*.*832*, *0*.*077*)
*The vector in the opposite direction to this one is also a valid answer.

c) Find a unit vector in the plane of the triangle that bisects the interior angle at *A*: A non-unit vector
in the required direction is *(*1*/*2*)(***a***AM *+ **a***AN)*, where

**a***AM *= *(*20*, *18*,*−10*)*|*(*20*, *18*,*−10*)*| = *(*0*.*697*, *0*.*627*,*−0*.*348*)
*

6

1.17c. (continued) Now

1

2
*(***a***AM *+ **a***AN) *= 1

2
[*(*0*.*697*, *0*.*627*,*−0*.*348*)*+ *(*−0*.*507*, *0*.*406*, *0*.*761*)*] = *(*0*.*095*, *0*.*516*, *0*.*207*)
*

Finally,

**a***bis *= *(*0*.*095*, *0*.*516*, *0*.*207*)*|*(*0*.*095*, *0*.*516*, *0*.*207*)*| = *(*0*.*168*, *0*.*915*, *0*.*367*)
*

1.18. Given points *A(ρ *= 5*, φ *= 70◦*, z *= −3*) *and *B(ρ *= 2*, φ *= −30◦*, z *= 1*)*, find:
a) unit vector in cartesian coordinates at *A *toward *B*: *A(*5 cos 70◦*, *5 sin 70◦*,*−3*) *= *A(*1*.*71*, *4*.*70*,*−3*)*, In

the same manner, *B(*1*.*73*,*−1*, *1*)*. So **R***AB *= *(*1*.*73*,*−1*, *1*) *− *(*1*.*71*, *4*.*70*,*−3*) *= *(*0*.*02*,*−5*.*70*, *4*) *and
therefore

**a***AB *= *(*0*.*02*,*−5*.*70*, *4*)*|*(*0*.*02*,*−5*.*70*, *4*)*| = *(*0*.*003*,*−0*.*82*, *0*.*57*)
*

b) a vector in cylindrical coordinates at *A *directed toward *B*: **a***AB *· **a***ρ *= 0*.*003 cos 70◦ − 0*.*82 sin 70◦ =
−0*.*77. **a***AB *· **a***φ *= −0*.*003 sin 70◦ − 0*.*82 cos 70◦ = −0*.*28. Thus

**a***AB *= −0*.*77**a***ρ *− 0*.*28**a***φ *+ 0*.*57**a***z
*

.

c) a unit vector in cylindrical coordinates at *B *directed toward *A*:
Use **a***BA *= *(*−0*, *003*, *0*.*82*,*−0*.*57*)*. Then **a***BA *·**a***ρ *= −0*.*003 cos*(*−30◦*)*+0*.*82 sin*(*−30◦*) *= −0*.*43, and
**a***BA *· **a***φ *= 0*.*003 sin*(*−30◦*)*+ 0*.*82 cos*(*−30◦*) *= 0*.*71. Finally,

**a***BA *= −0*.*43**a***ρ *+ 0*.*71**a***φ *− 0*.*57**a***z
*

1.19 a) Express the field **D **= *(x*2 + *y*2*)*−1*(x***a***x *+ *y***a***y) *in cylindrical components and cylindrical variables:
Have *x *= *ρ *cos*φ*, *y *= *ρ *sin *φ*, and *x*2 + *y*2 = *ρ*2. Therefore

**D **= 1
*ρ
(*cos*φ***a***x *+ sin *φ***a***y)
*

Then

*Dρ *= **D **· **a***ρ *= 1
*ρ
*

[
cos*φ(***a***x *· **a***ρ)*+ sin *φ(***a***y *· **a***ρ)
*

] = 1
*ρ
*

[
cos2 *φ *+ sin2 *φ
*

] = 1

*ρ
*

and

*Dφ *= **D **· **a***φ *= 1
*ρ
*

[
cos*φ(***a***x *· **a***φ)*+ sin *φ(***a***y *· **a***φ)
*

] = 1
*ρ
*

[cos*φ(*− sin *φ)*+ sin *φ *cos*φ*] = 0

Therefore

**D **= 1
*ρ
*

**a***ρ
*

7

1.19b. Evaluate **D **at the point where *ρ *= 2, *φ *= 0*.*2*π *, and *z *= 5, expressing the result in cylindrical and
cartesian coordinates: At the given point, and in cylindrical coordinates, **D **= 0*.*5**a***ρ *. To express this in
cartesian, we use

**D **= 0*.*5*(***a***ρ *· **a***x)***a***x *+ 0*.*5*(***a***ρ *· **a***y)***a***y *= 0*.*5 cos 36◦**a***x *+ 0*.*5 sin 36◦**a***y *= 0*.*41**a***x *+ 0*.*29**a***y
*

1.20. Express in cartesian components:
a) the vector at *A(ρ *= 4*, φ *= 40◦*, z *= −2*) *that extends to *B(ρ *= 5*, φ *= −110◦*, z *= 2*)*: We

have *A(*4 cos 40◦*, *4 sin 40◦*,*−2*) *= *A(*3*.*06*, *2*.*57*,*−2*)*, and *B(*5 cos*(*−110◦*), *5 sin*(*−110◦*), *2*) *=
*B(*−1*.*71*,*−4*.*70*, *2*) *in cartesian. Thus **R***AB *= *(*−4*.*77*,*−7*.*30*, *4*)*.

b) a unit vector at *B *directed toward *A*: Have **R***BA *= *(*4*.*77*, *7*.*30*,*−4*)*, and so

**a***BA *= *(*4*.*77*, *7*.*30*,*−4*)*|*(*4*.*77*, *7*.*30*,*−4*)*| = *(*0*.*50*, *0*.*76*,*−0*.*42*)
*

c) a unit vector at *B *directed toward the origin: Have **r***B *= *(*−1*.*71*,*−4*.*70*, *2*)*, and so −**r***B *=
*(*1*.*71*, *4*.*70*,*−2*)*. Thus

**a **= *(*1*.*71*, *4*.*70*,*−2*)*|*(*1*.*71*, *4*.*70*,*−2*)*| = *(*0*.*32*, *0*.*87*,*−0*.*37*)
*

1.21. Express in cylindrical components:

a) the vector from *C(*3*, *2*,*−7*) *to *D(*−1*,*−4*, *2*)*:
*C(*3*, *2*,*−7*) *→ *C(ρ *= 3*.*61*, φ *= 33*.*7◦*, z *= −7*) *and
*D(*−1*,*−4*, *2*) *→ *D(ρ *= 4*.*12*, φ *= −104*.*0◦*, z *= 2*)*.
Now **R***CD *= *(*−4*,*−6*, *9*) *and *Rρ *= **R***CD *· **a***ρ *= −4 cos*(*33*.*7*) *− 6 sin*(*33*.*7*) *= −6*.*66. Then
*Rφ *= **R***CD *· **a***φ *= 4 sin*(*33*.*7*)*− 6 cos*(*33*.*7*) *= −2*.*77. So **R***CD *= −6*.*66**a***ρ *− 2*.*77**a***φ *+ 9**a***z
*

b) a unit vector at *D *directed toward *C*:
**R***CD *= *(*4*, *6*,*−9*) *and *Rρ *= **R***DC *· **a***ρ *= 4 cos*(*−104*.*0*) *+ 6 sin*(*−104*.*0*) *= −6*.*79. Then *Rφ *=
**R***DC *· **a***φ *= 4[− sin*(*−104*.*0*)*] + 6 cos*(*−104*.*0*) *= 2*.*43. So **R***DC *= −6*.*79**a***ρ *+ 2*.*43**a***φ *− 9**a***z
*Thus **a***DC *= −0*.*59**a***ρ *+ 0*.*21**a***φ *− 0*.*78**a***z
*

c) a unit vector at *D *directed toward the origin: Start with **r***D *= *(*−1*,*−4*, *2*)*, and so the vector toward
the origin will be −**r***D *= *(*1*, *4*,*−2*)*. Thus in cartesian the unit vector is **a **= *(*0*.*22*, *0*.*87*,*−0*.*44*)*.
Convert to cylindrical:
*aρ *= *(*0*.*22*, *0*.*87*,*−0*.*44*) *· **a***ρ *= 0*.*22 cos*(*−104*.*0*)*+ 0*.*87 sin*(*−104*.*0*) *= −0*.*90, and
*aφ *= *(*0*.*22*, *0*.*87*,*−0*.*44*) *· **a***φ *= 0*.*22[− sin*(*−104*.*0*)*] + 0*.*87 cos*(*−104*.*0*) *= 0, so that finally,
**a **= −0*.*90**a***ρ *− 0*.*44**a***z*.

1.22. A field is given in cylindrical coordinates as

**F **=
[

40

*ρ*2 + 1 + 3*(*cos*φ *+ sin *φ)
*]

**a***ρ *+ 3*(*cos*φ *− sin *φ)***a***φ *− 2**a***z
*

where the magnitude of **F **is found to be:

|**F**| =
√

**F **· **F **=
[

1600

*(ρ*2 + 1*)*2 +
240

*ρ*2 + 1 *(*cos*φ *+ sin *φ)*+ 22
]1*/*2

8

Sketch |**F**|:
a) vs. *φ *with *ρ *= 3: in this case the above simplifies to

|**F***(ρ *= 3*)*| = |*Fa*| = [38 + 24*(*cos*φ *+ sin *φ)*]1*/*2

b) vs. *ρ *with *φ *= 0, in which:

|**F***(φ *= 0*)*| = |*Fb*| =
[

1600

*(ρ*2 + 1*)*2 +
240

*ρ*2 + 1 + 22
]1*/*2

c) vs. *ρ *with *φ *= 45◦, in which

|**F***(φ *= 45◦*)*| = |*Fc*| =
[

1600

*(ρ*2 + 1*)*2 +
240

√ 2

*ρ*2 + 1 + 22
]1*/*2

9

1.23. The surfaces *ρ *= 3, *ρ *= 5, *φ *= 100◦, *φ *= 130◦, *z *= 3, and *z *= 4*.*5 define a closed surface.
a) Find the enclosed volume:

Vol =
∫ 4*.*5

3

∫ 130◦ 100◦

∫ 5 3

*ρ dρ dφ dz *= 6*.*28

NOTE: The limits on the *φ *integration must be converted to radians (as was done here, but not shown).

b) Find the total area of the enclosing surface:

Area = 2 ∫ 130◦

100◦

∫ 5 3

*ρ dρ dφ *+
∫ 4*.*5

3

∫ 130◦ 100◦

3 *dφ dz
*

+
∫ 4*.*5

3

∫ 130◦ 100◦

5 *dφ dz *+ 2
∫ 4*.*5

3

∫ 5 3

*dρ dz *= 20*.*7

c) Find the total length of the twelve edges of the surfaces:

Length = 4 × 1*.*5 + 4 × 2 + 2 ×
[

30◦

360◦
× 2*π *× 3 + 30

◦

360◦
× 2*π *× 5

]
= 22*.*4

d) Find the length of the longest straight line that lies entirely within the volume: This will be between
the points A(*ρ *= 3, *φ *= 100◦, *z *= 3) and B(*ρ *= 5, *φ *= 130◦, *z *= 4*.*5). Performing point
transformations to cartesian coordinates, these become A(*x *= −0*.*52, *y *= 2*.*95, *z *= 3) and B(*x *=
−3*.*21, *y *= 3*.*83, *z *= 4*.*5). Taking A and B as vectors directed from the origin, the requested length
is

Length = |**B **− **A**| = |*(*−2*.*69*, *0*.*88*, *1*.*5*)*| = 3*.*21

1.24. At point *P(*−3*, *4*, *5*)*, express the vector that extends from *P *to *Q(*2*, *0*,*−1*) *in:
a) rectangular coordinates.

**R***PQ *= **Q **− **P **= 5**a***x *− 4**a***y *− 6**a***z
*Then |**R***PQ*| =

√
25 + 16 + 36 = 8*.*8

b) cylindrical coordinates. At *P *, *ρ *= 5, *φ *= tan−1*(*4*/*− 3*) *= −53*.*1◦, and *z *= 5. Now,
**R***PQ *· **a***ρ *= *(*5**a***x *− 4**a***y *− 6**a***z) *· **a***ρ *= 5 cos*φ *− 4 sin *φ *= 6*.*20

**R***PQ *· **a***φ *= *(*5**a***x *− 4**a***y *− 6**a***z) *· **a***φ *= −5 sin *φ *− 4 cos*φ *= 1*.*60
Thus

**R***PQ *= 6*.*20**a***ρ *+ 1*.*60**a***φ *− 6**a***z
*and |**R***PQ*| =

√
6*.*202 + 1*.*602 + 62 = 8*.*8

c) spherical coordinates. At *P *, *r *= √9 + 16 + 25 = √50 = 7*.*07, *θ *= cos−1*(*5*/*7*.*07*) *= 45◦, and
*φ *= tan−1*(*4*/*− 3*) *= −53*.*1◦.

**R***PQ *· **a***r *= *(*5**a***x *− 4**a***y *− 6**a***z) *· **a***r *= 5 sin *θ *cos*φ *− 4 sin *θ *sin *φ *− 6 cos *θ *= 0*.*14
**R***PQ *· **a***θ *= *(*5**a***x *− 4**a***y *− 6**a***z) *· **a***θ *= 5 cos *θ *cos*φ *− 4 cos *θ *sin *φ *− *(*−6*) *sin *θ *= 8*.*62

**R***PQ *· **a***φ *= *(*5**a***x *− 4**a***y *− 6**a***z) *· **a***φ *= −5 sin *φ *− 4 cos*φ *= 1*.*60

10

1.24. (continued)

Thus
**R***PQ *= 0*.*14**a***r *+ 8*.*62**a***θ *+ 1*.*60**a***φ
*

and |**R***PQ*| =
√

0*.*142 + 8*.*622 + 1*.*602 = 8*.*8
d) Show that each of these vectors has the same magnitude. Each does, as shown above.

1.25. Given point *P(r *= 0*.*8*, θ *= 30◦*, φ *= 45◦*)*, and

**E **= 1
*r*2

(
cos*φ ***a***r *+ sin *φ
*

sin *θ
***a***φ
*

)

a) Find **E **at *P *: **E **= 1*.*10**a***ρ *+ 2*.*21**a***φ *.
b) Find |**E**| at *P *: |**E**| = √1*.*102 + 2*.*212 = 2*.*47.
c) Find a unit vector in the direction of **E **at *P *:

**a***E *= **E**|**E**| = 0*.*45**a***r *+ 0*.*89**a***φ
*

1.26. a) Determine an expression for **a***y *in spherical coordinates at *P(r *= 4*, θ *= 0*.*2*π, φ *= 0*.*8*π)*: Use
**a***y *· **a***r *= sin *θ *sin *φ *= 0*.*35, **a***y *· **a***θ *= cos *θ *sin *φ *= 0*.*48, and **a***y *· **a***φ *= cos*φ *= −0*.*81 to obtain

**a***y *= 0*.*35**a***r *+ 0*.*48**a***θ *− 0*.*81**a***φ
*

b) Express **a***r *in cartesian components at *P *: Find *x *= *r *sin *θ *cos*φ *= −1*.*90, *y *= *r *sin *θ *sin *φ *= 1*.*38,
and *z *= *r *cos *θ *= −3*.*24. Then use **a***r *· **a***x *= sin *θ *cos*φ *= −0*.*48, **a***r *· **a***y *= sin *θ *sin *φ *= 0*.*35, and
**a***r *· **a***z *= cos *θ *= 0*.*81 to obtain

**a***r *= −0*.*48**a***x *+ 0*.*35**a***y *+ 0*.*81**a***z
*

1.27. The surfaces *r *= 2 and 4, *θ *= 30◦ and 50◦, and *φ *= 20◦ and 60◦ identify a closed surface.
a) Find the enclosed volume: This will be

Vol = ∫ 60◦

20◦

∫ 50◦ 30◦

∫ 4 2

*r*2 sin *θdrdθdφ *= 2*.*91

where degrees have been converted to radians. b) Find the total area of the enclosing surface:

Area = ∫ 60◦

20◦

∫ 50◦ 30◦

*(*42 + 22*) *sin *θdθdφ *+
∫ 4

2

∫ 60◦ 20◦

*r(*sin 30◦ + sin 50◦*)drdφ
*

+ 2 ∫ 50◦

30◦

∫ 4 2

*rdrdθ *= 12*.*61

c) Find the total length of the twelve edges of the surface:

Length = 4 ∫ 4

2
*dr *+ 2

∫ 50◦ 30◦

*(*4 + 2*)dθ *+
∫ 60◦

20◦
*(*4 sin 50◦ + 4 sin 30◦ + 2 sin 50◦ + 2 sin 30◦*)dφ
*

= 17*.*49

11

1.27. (continued)

d) Find the length of the longest straight line that lies entirely within the surface: This will be from
*A(r *= 2*, θ *= 50◦*, φ *= 20◦*) *to *B(r *= 4*, θ *= 30◦*, φ *= 60◦*) *or

*A(x *= 2 sin 50◦ cos 20◦*, y *= 2 sin 50◦ sin 20◦*, z *= 2 cos 50◦*)
*

to
*B(x *= 4 sin 30◦ cos 60◦*, y *= 4 sin 30◦ sin 60◦*, z *= 4 cos 30◦*)
*

or finally *A(*1*.*44*, *0*.*52*, *1*.*29*) *to *B(*1*.*00*, *1*.*73*, *3*.*46*)*. Thus **B **− **A **= *(*−0*.*44*, *1*.*21*, *2*.*18*) *and

Length = |**B **− **A**| = 2*.*53

1.28. a) Determine the cartesian components of the vector from *A(r *= 5*, θ *= 110◦*, φ *= 200◦*) *to *B(r *=
7*, θ *= 30◦*, φ *= 70◦*)*: First transform the points to cartesian: *xA *= 5 sin 110◦ cos 200◦ = −4*.*42,
*yA *= 5 sin 110◦ sin 200◦ = −1*.*61, and *zA *= 5 cos 110◦ = −1*.*71; *xB *= 7 sin 30◦ cos 70◦ = 1*.*20,
*yB *= 7 sin 30◦ sin 70◦ = 3*.*29, and *zB *= 7 cos 30◦ = 6*.*06. Now

**R***AB *= **B **− **A **= 5*.*62**ax **+ 4*.*90**a***y *+ 7*.*77**a***z
*

b) Find the spherical components of the vector at *P(*2*,*−3*, *4*) *extending to *Q(*−3*, *2*, *5*)*: First, **R***PQ *=
**Q **− **P **= *(*−5*, *5*, *1*)*. Then at *P *, *r *= √4 + 9 + 16 = 5*.*39, *θ *= cos−1*(*4*/*√29*) *= 42*.*0◦, and *φ *=
tan−1*(*−3*/*2*) *= −56*.*3◦. Now

**R***PQ *· **a***r *= −5 sin*(*42◦*) *cos*(*−56*.*3◦*)*+ 5 sin*(*42◦*) *sin*(*−56*.*3◦*)*+ 1 cos*(*42◦*) *= −3*.*90

**R***PQ *· **a***θ *= −5 cos*(*42◦*) *cos*(*−56*.*3◦*)*+ 5 cos*(*42◦*) *sin*(*−56*.*3◦*)*− 1 sin*(*42◦*) *= −5*.*82
**R***PQ *· **a***φ *= −*(*−5*) *sin*(*−56*.*3◦*)*+ 5 cos*(*−56*.*3◦*) *= −1*.*39

So finally,
**R***PQ *= −3*.*90**a***r *− 5*.*82**a***θ *− 1*.*39**a***φ
*

c) If **D **= 5**a***r *− 3**a***θ *+ 4**a***φ *, find **D **· **a***ρ *at *M(*1*, *2*, *3*)*: First convert **a***ρ *to cartesian coordinates at the
specified point. Use **a***ρ *= *(***a***ρ *· **a***x)***a***x *+ *(***a***ρ *· **a***y)***a***y *. At *A(*1*, *2*, *3*)*, *ρ *=

√
5, *φ *= tan−1*(*2*) *= 63*.*4◦,

*r *= √14, and *θ *= cos−1*(*3*/*√14*) *= 36*.*7◦. So **a***ρ *= cos*(*63*.*4◦*)***a***x *+ sin*(*63*.*4◦*)***a***y *= 0*.*45**a***x *+ 0*.*89**a***y *.
Then

*(*5**a***r *− 3**a***θ *+ 4**a***φ) *· *(*0*.*45**a***x *+ 0*.*89**a***y) *=
5*(*0*.*45*) *sin *θ *cos*φ *+ 5*(*0*.*89*) *sin *θ *sin *φ *− 3*(*0*.*45*) *cos *θ *cos*φ
*− 3*(*0*.*89*) *cos *θ *sin *φ *+ 4*(*0*.*45*)(*− sin *φ) *+ 4*(*0*.*89*) *cos*φ *= 0*.*59

1.29. Express the unit vector **a***x *in spherical components at the point:
a) *r *= 2, *θ *= 1 rad, *φ *= 0*.*8 rad: Use

**a***x *= *(***a***x *· **a***r )***a***r *+ *(***a***x *· **a***θ )***a***θ *+ *(***a***x *· **a***φ)***a***φ *=
sin*(*1*) *cos*(*0*.*8*)***a***r *+ cos*(*1*) *cos*(*0*.*8*)***a***θ *+ *(*− sin*(*0*.*8*))***a***φ *= 0*.*59**a***r *+ 0*.*38**a***θ *− 0*.*72**a***φ
*

12

1.29 (continued) Express the unit vector **a***x *in spherical components at the point:

b) *x *= 3, *y *= 2, *z *= −1: First, transform the point to spherical coordinates. Have *r *= √14,
*θ *= cos−1*(*−1*/*√14*) *= 105*.*5◦, and *φ *= tan−1*(*2*/*3*) *= 33*.*7◦. Then

**a***x *= sin*(*105*.*5◦*) *cos*(*33*.*7◦*)***a***r *+ cos*(*105*.*5◦*) *cos*(*33*.*7◦*)***a***θ *+ *(*− sin*(*33*.*7◦*))***a***φ
*= 0*.*80**a***r *− 0*.*22**a***θ *− 0*.*55**a***φ
*

c) *ρ *= 2*.*5, *φ *= 0*.*7 rad, *z *= 1*.*5: Again, convert the point to spherical coordinates. *r *=
√
*ρ*2 + *z*2 =√

8*.*5, *θ *= cos−1*(z/r) *= cos−1*(*1*.*5*/*√8*.*5*) *= 59*.*0◦, and *φ *= 0*.*7 rad = 40*.*1◦. Now

**a***x *= sin*(*59◦*) *cos*(*40*.*1◦*)***a***r *+ cos*(*59◦*) *cos*(*40*.*1◦*)***a***θ *+ *(*− sin*(*40*.*1◦*))***a***φ
*= 0*.*66**a***r *+ 0*.*39**a***θ *− 0*.*64**a***φ
*

1.30. Given *A(r *= 20*, θ *= 30◦*, φ *= 45◦*) *and *B(r *= 30*, θ *= 115◦*, φ *= 160◦*)*, find:
a) |**R***AB *|: First convert *A *and *B *to cartesian: Have *xA *= 20 sin*(*30◦*) *cos*(*45◦*) *= 7*.*07, *yA *=

20 sin*(*30◦*) *sin*(*45◦*) *= 7*.*07, and *zA *= 20 cos*(*30◦*) *= 17*.*3. *xB *= 30 sin*(*115◦*) *cos*(*160◦*) *= −25*.*6,
*yB *= 30 sin*(*115◦*) *sin*(*160◦*) *= 9*.*3, and *zB *= 30 cos*(*115◦*) *= −12*.*7. Now **R***AB *= **R***B *− **R***A *=
*(*−32*.*6*, *2*.*2*,*−30*.*0*)*, and so |**R***AB *| = 44*.*4.

b) |**R***AC *|, given *C(r *= 20*, θ *= 90◦*, φ *= 45◦*)*. Again, converting *C *to cartesian, obtain *xC *=
20 sin*(*90◦*) *cos*(*45◦*) *= 14*.*14, *yC *= 20 sin*(*90◦*) *sin*(*45◦*) *= 14*.*14, and *zC *= 20 cos*(*90◦*) *= 0. So
**R***AC *= **R***C *− **R***A *= *(*7*.*07*, *7*.*07*,*−17*.*3*)*, and |**R***AC *| = 20*.*0.

c) the distance from *A *to *C *on a great circle path: Note that *A *and *C *share the same *r *and *φ *coordinates;
thus moving from *A *to *C *involves only a change in *θ *of 60◦. The requested arc length is then

distance = 20 × [

60

(
2*π
*

360

)]
= 20*.*9

13

**CHAPTER 2
**

2.1. Four 10nC positive charges are located in the *z *= 0 plane at the corners of a square 8cm on a side.
A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the
magnitude of the total force on this fifth charge for * *= 0:
Arrange the charges in the *xy *plane at locations (4,4), (4,-4), (-4,4), and (-4,-4). Then the fifth charge
will be on the *z *axis at location *z *= 4√2, which puts it at 8cm distance from the other four. By
symmetry, the force on the fifth charge will be *z*-directed, and will be four times the *z *component of
force produced by each of the four other charges.

*F *= 4√
2
× *q
*

2

4*π*0*d*2
= 4√

2
× *(*10

−8*)*2

4*π(*8*.*85 × 10−12*)(*0*.*08*)*2 = 4*.*0 × 10
−4 N

2.2. A charge *Q*1 = 0*.*1 *µ*C is located at the origin, while *Q*2 = 0*.*2 *µ*C is at *A(*0*.*8*,*−0*.*6*, *0*)*. Find the
locus of points in the *z *= 0 plane at which the *x *component of the force on a third positive charge is
zero.

To solve this problem, the *z *coordinate of the third charge is immaterial, so we can place it in the
*xy *plane at coordinates *(x, y, *0*)*. We take its magnitude to be *Q*3. The vector directed from the first
charge to the third is **R**13 = *x***a***x *+ *y***a***y *; the vector directed from the second charge to the third is
**R**23 = *(x *− 0*.*8*)***a***x *+ *(y *+ 0*.*6*)***a***y *. The force on the third charge is now

**F**3 = *Q*3
4*π*0

[
*Q*1**R**13
|**R**13|3 +

*Q*2**R**23
|**R**23|3

]

= *Q*3 × 10
−6

4*π*0

[
0*.*1*(x***a***x *+ *y***a***y)
(x*2 + *y*2*)*1*.*5 +

0*.*2[*(x *− 0*.*8*)***a***x *+ *(y *+ 0*.*6*)***a***y*]
[*(x *− 0*.*8*)*2 + *(y *+ 0*.*6*)*2]1*.*5

]

We desire the *x *component to be zero. Thus,

0 = [

0*.*1*x***a***x
(x*2 + *y*2*)*1*.*5 +

0*.*2*(x *− 0*.*8*)***a***x
*[*(x *− 0*.*8*)*2 + *(y *+ 0*.*6*)*2]1*.*5

]

or
*x*[*(x *− 0*.*8*)*2 + *(y *+ 0*.*6*)*2]1*.*5 = 2*(*0*.*8 − *x)(x*2 + *y*2*)*1*.*5

2.3. Point charges of 50nC each are located at *A(*1*, *0*, *0*)*, *B(*−1*, *0*, *0*)*, *C(*0*, *1*, *0*)*, and *D(*0*,*−1*, *0*) *in free
space. Find the total force on the charge at *A*.

The force will be:

**F **= *(*50 × 10
−9*)*2

4*π*0

[
**R***CA
*

|**R***CA*|3 +
**R***DA
*

|**R***DA*|3 +
**R***BA
*

|**R***BA*|3
]

where **R***CA *= **a***x *− **a***y *, **R***DA *= **a***x *+ **a***y *, and **R***BA *= 2**a***x *. The magnitudes are |**R***CA*| = |**R***DA*| =
√

2,
and |**R***BA*| = 2. Substituting these leads to

**F **= *(*50 × 10
−9*)*2

4*π*0

[ 1

2 √

2 + 1

2 √

2 + 2

8

]
**a***x *= 21*.*5**a***x µ*N

where distances are in meters.

14

2.4. Let *Q*1 = 8 *µ*C be located at *P*1*(*2*, *5*, *8*) *while *Q*2 = −5 *µ*C is at *P*2*(*6*, *15*, *8*)*. Let * *= 0.
a) Find **F**2, the force on *Q*2: This force will be

**F**2 = *Q*1*Q*2
4*π*0

**R**12
|**R**12|3 =

*(*8 × 10−6*)(*−5 × 10−6*)
*4*π*0

*(*4**a***x *+ 10**a***y)
(*116*)*1*.*5

= *(*−1*.*15**a***x *− 2*.*88**a***y)*mN

b) Find the coordinates of *P*3 if a charge *Q*3 experiences a total force **F**3 = 0 at *P*3: This force in
general will be:

**F**3 = *Q*3
4*π*0

[
*Q*1**R**13
|**R**13|3 +

*Q*2**R**23
|**R**23|3

]
where **R**13 = *(x *− 2*)***a***x *+ *(y *− 5*)***a***y *and **R**23 = *(x *− 6*)***a***x *+ *(y *− 15*)***a***y *. Note, however, that
all three charges must lie in a straight line, and the location of *Q*3 will be along the vector **R**12
extended past *Q*2. The slope of this vector is *(*15 − 5*)/(*6 − 2*) *= 2*.*5. Therefore, we look for *P*3
at coordinates *(x, *2*.*5*x, *8*)*. With this restriction, the force becomes:

**F**3 = *Q*3
4*π*0

[
8[*(x *− 2*)***a***x *+ 2*.*5*(x *− 2*)***a***y*]

[*(x *− 2*)*2 + *(*2*.*5*)*2*(x *− 2*)*2]1*.*5 −
5[*(x *− 6*)***a***x *+ 2*.*5*(x *− 6*)***a***y*]

[*(x *− 6*)*2 + *(*2*.*5*)*2*(x *− 6*)*2]1*.*5
]

where we require the term in large brackets to be zero. This leads to

8*(x *− 2*)*[*((*2*.*5*)*2 + 1*)(x *− 6*)*2]1*.*5 − 5*(x *− 6*)*[*((*2*.*5*)*2 + 1*)(x *− 2*)*2]1*.*5 = 0
which reduces to

8*(x *− 6*)*2 − 5*(x *− 2*)*2 = 0
or

*x *= 6
√

8 − 2√5√
8 −√5 = 21*.*1

The coordinates of *P*3 are thus *P*3*(*21*.*1*, *52*.*8*, *8*)
*

2.5. Let a point charge *Q*125 nC be located at *P*1*(*4*,*−2*, *7*) *and a charge *Q*2 = 60 nC be at *P*2*(*−3*, *4*,*−2*)*.
a) If * *= 0, find **E **at *P*3*(*1*, *2*, *3*)*: This field will be

**E **= 10
−9

4*π*0

[
25**R**13
|**R**13|3 +

60**R**23
|**R**23|3

]

where **R**13 = −3**a***x*+4**a***y*−4**a***z *and **R**23 = 4**a***x*−2**a***y*+5**a***z*. Also, |**R**13| =
√

41 and |**R**23| =
√

45. So

**E **= 10
−9

4*π*0

[
25 × *(*−3**a***x *+ 4**a***y *− 4**a***z)
*

*(*41*)*1*.*5
+ 60 × *(*4**a***x *− 2**a***y *+ 5**a***z)
*

*(*45*)*1*.*5

]
= 4*.*58**a***x *− 0*.*15**a***y *+ 5*.*51**a***z
*

b) At what point on the *y *axis is *Ex *= 0? *P*3 is now at *(*0*, y, *0*)*, so **R**13 = −4**a***x *+ *(y *+ 2*)***a***y *− 7**a***z
*and **R**23 = 3**a***x *+ *(y *− 4*)***a***y *+ 2**a***z*. Also, |**R**13| =

√
65 + *(y *+ 2*)*2 and |**R**23| =

√
13 + *(y *− 4*)*2.

Now the *x *component of **E **at the new *P*3 will be:

*Ex *= 10
−9

4*π*0

[
25 × *(*−4*)
*

[65 + *(y *+ 2*)*2]1*.*5 +
60 × 3

[13 + *(y *− 4*)*2]1*.*5
]

To obtain *Ex *= 0, we require the expression in the large brackets to be zero. This expression
simplifies to the following quadratic:

0*.*48*y*2 + 13*.*92*y *+ 73*.*10 = 0
which yields the two values: *y *= −6*.*89*,*−22*.*11

15

2.6. Point charges of 120 nC are located at *A(*0*, *0*, *1*) *and *B(*0*, *0*,*−1*) *in free space.
a) Find **E **at *P(*0*.*5*, *0*, *0*)*: This will be

**E***P *= 120 × 10
−9

4*π*0

[
**R***AP
*

|**R***AP *|3 +
**R***BP
*

|**R***BP *|3
]

where **R***AP *= 0*.*5**a***x *− **a***z *and **R***BP *= 0*.*5**a***x *+ **a***z*. Also, |**R***AP *| = |**R***BP *| =
√

1*.*25. Thus:

**E***P *= 120 × 10
−9**a***x
*

4*π(*1*.*25*)*1*.*50
= 772 V*/*m

b) What single charge at the origin would provide the identical field strength? We require

*Q*0

4*π*0*(*0*.*5*)*2
= 772

from which we find *Q*0 = 21*.*5 nC.

2.7. A 2 *µ*C point charge is located at *A(*4*, *3*, *5*) *in free space. Find *Eρ *, *Eφ *, and *Ez *at *P(*8*, *12*, *2*)*. Have

**E***P *= 2 × 10
−6

4*π*0

**R***AP
*|**R***AP *|3 =

2 × 10−6
4*π*0

[
4**a***x *+ 9**a***y *− 3**a***z
*

*(*106*)*1*.*5

]
= 65*.*9**a***x *+ 148*.*3**a***y *− 49*.*4**a***z
*

Then, at point *P *, *ρ *= √82 + 122 = 14*.*4, *φ *= tan−1*(*12*/*8*) *= 56*.*3◦, and *z *= *z*. Now,
*Eρ *= **E***p *· **a***ρ *= 65*.*9*(***a***x *· **a***ρ)*+ 148*.*3*(***a***y *· **a***ρ) *= 65*.*9 cos*(*56*.*3◦*)*+ 148*.*3 sin*(*56*.*3◦*) *= 159*.*7

and

*Eφ *= **E***p *· **a***φ *= 65*.*9*(***a***x *· **a***φ)*+ 148*.*3*(***a***y *· **a***φ) *= −65*.*9 sin*(*56*.*3◦*)*+ 148*.*3 cos*(*56*.*3◦*) *= 27*.*4
Finally, *Ez *= −49*.*4

2.8. Given point charges of −1 *µ*C at *P*1*(*0*, *0*, *0*.*5*) *and *P*2*(*0*, *0*,*−0*.*5*)*, and a charge of 2 *µ*C at the origin,
find **E **at *P(*0*, *2*, *1*) *in spherical components, assuming * *= 0.
The field will take the general form:

**E***P *= 10
−6

4*π*0

[
− **R**1|**R**1|3 +

2**R**2
|**R**2|3 −

**R**3
|**R**3|3

]

where **R**1, **R**2, **R**3 are the vectors to*P *from each of the charges in their original listed order. Specifically,
**R**1 = *(*0*, *2*, *0*.*5*)*, **R**2 = *(*0*, *2*, *1*)*, and **R**3 = *(*0*, *2*, *1*.*5*)*. The magnitudes are |**R**1| = 2*.*06, |**R**2| = 2*.*24,
and |**R**3| = 2*.*50. Thus

**E***P *= 10
−6

4*π*0

[−*(*0*, *2*, *0*.*5*)
(*2*.*06*)*3

+ 2*(*0*, *2*, *1*)
(*2*.*24*)*3

− *(*0*, *2*, *1*.*5*)
(*2*.*50*)*3

]
= 89*.*9**a***y *+ 179*.*8**a***z
*

Now, at *P *, *r *= √5, *θ *= cos−1*(*1*/*√5*) *= 63*.*4◦, and *φ *= 90◦. So
*Er *= **E***P *· **a***r *= 89*.*9*(***a***y *· **a***r )*+ 179*.*8*(***a***z *· **a***r ) *= 89*.*9 sin *θ *sin *φ *+ 179*.*8 cos *θ *= 160*.*9

*Eθ *= **E***P *· **a***θ *= 89*.*9*(***a***y *· **a***θ )*+ 179*.*8*(***a***z *· **a***θ ) *= 89*.*9 cos *θ *sin *φ *+ 179*.*8*(*− sin *θ) *= −120*.*5
*Eφ *= **E***P *· **a***φ *= 89*.*9*(***a***y *· **a***φ)*+ 179*.*8*(***a***z *· **a***φ) *= 89*.*9 cos*φ *= 0

16

2.9. A 100 nC point charge is located at *A(*−1*, *1*, *3*) *in free space.
a) Find the locus of all points *P(x, y, z) *at which *Ex *= 500 V/m: The total field at *P *will be:

**E***P *= 100 × 10
−9

4*π*0

**R***AP
*|**R***AP *|3

where **R***AP *= *(x *+ 1*)***a***x *+ *(y *− 1*)***a***y *+ *(z *− 3*)***a***z*, and where |**R***AP *| = [*(x *+ 1*)*2 + *(y *− 1*)*2 +
*(z*− 3*)*2]1*/*2. The *x *component of the field will be

*Ex *= 100 × 10
−9

4*π*0

[
*(x *+ 1*)
*

[*(x *+ 1*)*2 + *(y *− 1*)*2 + *(z*− 3*)*2]1*.*5
]
= 500 V*/*m

And so our condition becomes:

*(x *+ 1*) *= 0*.*56 [*(x *+ 1*)*2 + *(y *− 1*)*2 + *(z*− 3*)*2]1*.*5

b) Find *y*1 if *P(*−2*, y*1*, *3*) *lies on that locus: At point *P *, the condition of part *a *becomes

3*.*19 =
[
1 + *(y*1 − 1*)*2

]3
from which *(y*1 − 1*)*2 = 0*.*47, or *y*1 = 1*.*69 or 0*.*31

2.10. Charges of 20 and -20 nC are located at *(*3*, *0*, *0*) *and *(*−3*, *0*, *0*)*, respectively. Let * *= 0.
Determine |**E**| at *P(*0*, y, *0*)*: The field will be

**E***P *= 20 × 10
−9

4*π*0

[
**R**1

|**R**1|3 −
**R**2

|**R**2|3
]

where **R**1, the vector from the positive charge to point *P *is *(*−3*, y, *0*)*, and **R**2, the vector from
the negative charge to point *P *, is *(*3*, y, *0*)*. The magnitudes of these vectors are |**R**1| = |**R**2| =√

9 + *y*2. Substituting these into the expression for **E***P *produces

**E***P *= 20 × 10
−9

4*π*0

[ −6**a***x
(*9 + *y*2*)*1*.*5

]

from which

|**E***P *| = 1079
*(*9 + *y*2*)*1*.*5 V*/*m

2.11. A charge *Q*0 located at the origin in free space produces a field for which *Ez *= 1 kV/m at point
*P(*−2*, *1*,*−1*)*.

a) Find *Q*0: The field at *P *will be

**E***P *= *Q*0
4*π*0

[−2**a***x *+ **a***y *− **a***z
*61*.*5

]

Since the *z *component is of value 1 kV/m, we find *Q*0 = −4*π*061*.*5 × 103 = −1*.*63 *µ*C.

17

2.11. (continued)

b) Find **E **at *M(*1*, *6*, *5*) *in cartesian coordinates: This field will be:

**E***M *= −1*.*63 × 10
−6

4*π*0

[
**a***x *+ 6**a***y *+ 5**a***z
*[1 + 36 + 25]1*.*5

]

or **E***M *= −30*.*11**a***x *− 180*.*63**a***y *− 150*.*53**a***z*.
c) Find **E **at *M(*1*, *6*, *5*) *in cylindrical coordinates: At *M *, *ρ *= √1 + 36 = 6*.*08, *φ *= tan−1*(*6*/*1*) *=

80*.*54◦, and *z *= 5. Now

*Eρ *= **E***M *· **a***ρ *= −30*.*11 cos*φ *− 180*.*63 sin *φ *= −183*.*12

*Eφ *= **E***M *· **a***φ *= −30*.*11*(*− sin *φ)*− 180*.*63 cos*φ *= 0 *(*as expected*)
*so that **E***M *= −183*.*12**a***ρ *− 150*.*53**a***z*.

d) Find **E **at *M(*1*, *6*, *5*) *in spherical coordinates: At *M *, *r *= √1 + 36 + 25 = 7*.*87, *φ *= 80*.*54◦ (as
before), and *θ *= cos−1*(*5*/*7*.*87*) *= 50*.*58◦. Now, since the charge is at the origin, we expect to
obtain only a radial component of **E***M *. This will be:

*Er *= **E***M *· **a***r *= −30*.*11 sin *θ *cos*φ *− 180*.*63 sin *θ *sin *φ *− 150*.*53 cos *θ *= −237*.*1

2.12. The volume charge density *ρv *= *ρ*0*e*−|*x*|−|*y*|−|*z*| exists over all free space. Calculate the total charge
present: This will be 8 times the integral of *ρv *over the first octant, or

*Q *= 8
∫ ∞

0

∫ ∞ 0

∫ ∞ 0

*ρ*0*e
*−*x*−*y*−*z dx dy dz *= 8*ρ*0

2.13. A uniform volume charge density of 0.2 *µ*C*/*m3 (note typo in book) is present throughout the spherical
shell extending from *r *= 3 cm to *r *= 5 cm. If *ρv *= 0 elsewhere:

a) find the total charge present throughout the shell: This will be

*Q *=
∫ 2*π
*

0

∫ *π
*0

∫ *.*05
*.*03

0*.*2 *r*2 sin *θ dr dθ dφ *=
[

4*π(*0*.*2*)
r*3

3

]*.*05
*.*03

= 8*.*21 × 10−5 *µ*C = 82*.*1 pC

b) find *r*1 if half the total charge is located in the region 3 cm *< r < r*1: If the integral over *r *in part
*a *is taken to *r*1, we would obtain[

4*π(*0*.*2*)
r*3

3

]*r*1
*.*03

= 4*.*105 × 10−5

Thus

*r*1 =
[

3 × 4*.*105 × 10−5
0*.*2 × 4*π *+ *(.*03*)
*

3

]1*/*3
= 4*.*24 cm

18

2.14. Let

*ρv *= 5*e*−0*.*1*ρ (π *− |*φ*|*) *1
*z*2 + 10 *µ*C*/*m

3

in the region 0 ≤ *ρ *≤ 10, −*π < φ < π *, all *z*, and *ρv *= 0 elsewhere.
a) Determine the total charge present: This will be the integral of *ρv *over the region where it exists;
specifically,

*Q *=
∫ ∞
−∞

∫ *π
*−*π
*

∫ 10 0

5*e*−0*.*1*ρ (π *− |*φ*|*) *1
*z*2 + 10*ρ dρ dφ dz
*

which becomes

*Q *= 5
[
*e*−0*.*1*ρ
*

*(*0*.*1*)*2
*(*−0*.*1 − 1*)
*

]10 0

∫ ∞ −∞

2
∫ *π
*

0
*(π *− *φ) *1

*z*2 + 10*dφ dz
*

or

*Q *= 5 × 26*.*4
∫ ∞
−∞

*π*2
1

*z*2 + 10 *dz
*

Finally,

*Q *= 5 × 26*.*4 × *π*2
[

1√ 10

tan−1 (

*z*√
10

)]∞ −∞

= 5*(*26*.*4*)π
*3

√ 10

= 1*.*29 × 103 *µ*C = 1*.*29 mC

b) Calculate the charge within the region 0 ≤ *ρ *≤ 4, −*π/*2 *< φ < π/*2, −10 *< z < *10: With the
limits thus changed, the integral for the charge becomes:

*Q*′ =
∫ 10
−10

2
∫ *π/*2

0

∫ 4 0

5*e*−0*.*1*ρ (π *− *φ) *1
*z*2 + 10*ρ dρ dφ dz
*

Following the same evaulation procedure as in part *a*, we obtain *Q*′ = 0*.*182 mC.

2.15. A spherical volume having a 2 *µ*m radius contains a uniform volume charge density of 1015 C*/*m3.

a) What total charge is enclosed in the spherical volume?
This will be *Q *= *(*4*/*3*)π(*2 × 10−6*)*3 × 1015 = 3*.*35 × 10−2 C.

b) Now assume that a large region contains one of these little spheres at every corner of a cubical grid 3mm on a side, and that there is no charge between spheres. What is the average volume charge density throughout this large region? Each cube will contain the equivalent of one little sphere. Neglecting the little sphere volume, the average density becomes

*ρv,avg *= 3*.*35 × 10
−2

*(*0*.*003*)*3
= 1*.*24 × 106 C*/*m3

2.16. The region in which 4 *< r < *5, 0 *< θ < *25◦, and 0*.*9*π < φ < *1*.*1*π *contains the volume charge
density of *ρv *= 10*(r *− 4*)(r *− 5*) *sin *θ *sin*(φ/*2*)*. Outside the region, *ρv *= 0. Find the charge within
the region: The integral that gives the charge will be

*Q *= 10
∫ 1*.*1*π
.*9*π
*

∫ 25◦ 0

∫ 5 4

*(r *− 4*)(r *− 5*) *sin *θ *sin*(φ/*2*) r*2 sin *θ dr dθ dφ
*

19

2.16. (continued) Carrying out the integral, we obtain

*Q *= 10
[
*r*5

5
− 9 *r
*

4

4
+ 20 *r
*

3

3

]5 4

[ 1

2
*θ *− 1

4
sin*(*2*θ)
*

]25◦ 0

[ −2 cos

(
*θ
*

2

)]1*.*1*π
.*9*π
*

= 10*(*−3*.*39*)(.*0266*)(.*626*) *= 0*.*57 C

2.17. A uniform line charge of 16 nC/m is located along the line defined by *y *= −2, *z *= 5. If * *= 0:
a) Find **E **at *P(*1*, *2*, *3*)*: This will be

**E***P *= *ρl
*2*π*0

**R***P
*|**R***P *|2

where **R***P *= *(*1*, *2*, *3*)*− *(*1*,*−2*, *5*) *= *(*0*, *4*,*−2*)*, and |**R***P *|2 = 20. So

**E***P *= 16 × 10
−9

2*π*0

[
4**a***y *− 2**a***z
*

20

]
= 57*.*5**a***y *− 28*.*8**a***z *V*/*m

b) Find **E **at that point in the *z *= 0 plane where the direction of **E **is given by *(*1*/*3*)***a***y *− *(*2*/*3*)***a***z*:
With *z *= 0, the general field will be

**E***z*=0 = *ρl
*2*π*0

[
*(y *+ 2*)***a***y *− 5**a***z
(y *+ 2*)*2 + 25

]

We require |*Ez*| = −|2*Ey *|, so 2*(y *+ 2*) *= 5. Thus *y *= 1*/*2, and the field becomes:

**E***z*=0 = *ρl
*2*π*0

[
2*.*5**a***y *− 5**a***z
(*2*.*5*)*2 + 25

]
= 23**a***y *− 46**a***z
*

2.18. Uniform line charges of 0*.*4 *µ*C/m and −0*.*4 *µ*C/m are located in the *x *= 0 plane at *y *= −0*.*6 and
*y *= 0*.*6 m respectively. Let * *= 0.

a) Find **E **at *P(x, *0*, z)*: In general, we have

**E***P *= *ρl
*2*π*0

[
**R**+*P
*|**R**+*P *| −

**R**−*P
*|**R**−*P *|

]

where **R**+*P *and **R**−*P *are, respectively, the vectors directed from the positive and negative line
charges to the point *P *, and these are normal to the *z *axis. We thus have **R**+*P *= *(x, *0*, z) *−
*(*0*,*−*.*6*, z) *= *(x, .*6*, *0*)*, and **R**−*P *= *(x, *0*, z)*− *(*0*, .*6*, z) *= *(x,*−*.*6*, *0*)*. So

**E***P *= *ρl
*2*π*0

[
*x***a***x *+ 0*.*6**a***y
x*2 + *(*0*.*6*)*2 −

*x***a***x *− 0*.*6**a***y
x*2 + *(*0*.*6*)*2

]
= 0*.*4 × 10

−6

2*π*0

[
1*.*2**a***y
*

*x*2 + 0*.*36
]
= 8*.*63**a***y
*

*x*2 + 0*.*36 kV*/*m

20

2.18. (continued)
b) Find **E **at *Q(*2*, *3*, *4*)*: This field will in general be:

**E***Q *= *ρl
*2*π*0

[
**R**+*Q
*|**R**+*Q*| −

**R**−*Q
*|**R**−*Q*|

]

where **R**+*Q *= *(*2*, *3*, *4*)*−*(*0*,*−*.*6*, *4*) *= *(*2*, *3*.*6*, *0*)*, and **R**−*Q *= *(*2*, *3*, *4*)*−*(*0*, .*6*, *4*) *= *(*2*, *2*.*4*, *0*)*.
Thus

**E***Q *= *ρl
*2*π*0

[
2**a***x *+ 3*.*6**a***y
*22 + *(*3*.*6*)*2 −

2**a***x *+ 2*.*4**a***y
*22 + *(*2*.*4*)*2

]
= −625*.*8**a***x *− 241*.*6**a***y *V*/*m

2.19. A uniform line charge of 2 *µ*C/m is located on the *z *axis. Find **E **in cartesian coordinates at *P(*1*, *2*, *3*)
*if the charge extends from

a) −∞ *< z < *∞: With the infinite line, we know that the field will have only a radial component
in cylindrical coordinates (or *x *and *y *components in cartesian). The field from an infinite line on
the z axis is generally **E **= [*ρl/(*2*π*0*ρ)*]**a***ρ *. Therefore, at point *P *:

**E***P *= *ρl
*2*π*0

**R***zP
*|**R***zP *|2 =

*(*2 × 10−6*)
*2*π*0

**a***x *+ 2**a***y
*5

= 7*.*2**a***x *+ 14*.*4**a***y *kV*/*m

where **R***zP *is the vector that extends from the line charge to point *P *, and is perpendicular to the *z
*axis; i.e., **R***zP *= *(*1*, *2*, *3*)*− *(*0*, *0*, *3*) *= *(*1*, *2*, *0*)*.

b) −4 ≤ *z *≤ 4: Here we use the general relation

**E***P *=
∫

*ρldz
*

4*π*0

**r **− **r**′
|**r **− **r**′|3

where **r **= **a***x *+ 2**a***y *+ 3**a***z *and **r**′ = *z***a***z*. So the integral becomes

**E***P *= *(*2 × 10
−6*)
*

4*π*0

∫ 4 −4

**a***x *+ 2**a***y *+ *(*3 − *z)***a***z
*[5 + *(*3 − *z)*2]1*.*5 *dz
*

Using integral tables, we obtain:

**E***P *= 3597
[
*(***a***x *+ 2**a***y)(z*− 3*)*+ 5**a***z
*

*(z*2 − 6*z*+ 14*)
*]4
−4

V*/*m = 4*.*9**a***x *+ 9*.*8**a***y *+ 4*.*9**a***z *kV*/*m

The student is invited to verify that when evaluating the above expression over the limits −∞ *<
z < *∞, the *z *component vanishes and the *x *and *y *components become those found in part *a*.

2.20. Uniform line charges of 120 nC/m lie along the entire extent of the three coordinate axes. Assuming
free space conditions, find **E **at *P(*−3*, *2*,*−1*)*: Since all line charges are infinitely-long, we can write:

**E***P *= *ρl
*2*π*0

[
**R***xP
*

|**R***xP *|2 +
**R***yP
*

|**R***yP *|2 +
**R***zP
*

|**R***zP *|2
]

where **R***xP *, **R***yP *, and **R***zP *are the normal vectors from each of the three axes that terminate on point
*P *. Specifically, **R***xP *= *(*−3*, *2*,*−1*) *− *(*−3*, *0*, *0*) *= *(*0*, *2*,*−1*)*, **R***yP *= *(*−3*, *2*,*−1*) *− *(*0*, *2*, *0*) *=
*(*−3*, *0*,*−1*)*, and **R***zP *= *(*−3*, *2*,*−1*)*− *(*0*, *0*,*−1*) *= *(*−3*, *2*, *0*)*. Substituting these into the expression
for **E***P *gives

**E***P *= *ρl
*2*π*0

[
2**a***y *− **a***z
*

5
+ −3**a***x *− **a***z
*

10
+ −3**a***x *+ 2**a***y
*

13

]
= −1*.*15**a***x *+ 1*.*20**a***y *− 0*.*65**a***z *kV*/*m

21

2.21. Two identical uniform line charges with *ρl *= 75 nC/m are located in free space at *x *= 0, *y *= ±0*.*4 m.
What force per unit length does each line charge exert on the other? The charges are parallel to the *z
*axis and are separated by 0*.*8 m. Thus the field from the charge at *y *= −0*.*4 evaluated at the location
of the charge at *y *= +0*.*4 will be **E **= [*ρl/(*2*π*0*(*0*.*8*))*]**a***y *. The force on a differential length of the
line at the positive *y *location is *d***F **= *dq***E **= *ρldz***E**. Thus the force per unit length acting on the line
at postive *y *arising from the charge at negative *y *is

**F **=
∫ 1

0

*ρ*2*l dz
*

2*π*0*(*0*.*8*)
***a***y *= 1*.*26 × 10−4 **a***y *N*/*m = 126 **a***y µ*N*/*m

The force on the line at negative *y *is of course the same, but with −**a***y *.

2.22. A uniform surface charge density of 5 nC*/*m2 is present in the region *x *= 0, −2 *< y < *2, and all *z*. If
* *= 0, find **E **at:

a) *PA(*3*, *0*, *0*)*: We use the superposition integral:

**E **=
∫ ∫

*ρsda
*

4*π*0

**r **− **r**′
|**r **− **r**′|3

where **r **= 3**a***x *and **r**′ = *y***a***y *+ *z***a***z*. The integral becomes:

**E***PA *= *ρs
*4*π*0

∫ ∞ −∞

∫ 2 −2

3**a***x *− *y***a***y *− *z***a***z
*[9 + *y*2 + *z*2]1*.*5 *dy dz
*

Since the integration limits are symmetric about the origin, and since the *y *and *z *components of
the integrand exhibit odd parity (change sign when crossing the origin, but otherwise symmetric),
these will integrate to zero, leaving only the *x *component. This is evident just from the symmetry
of the problem. Performing the *z *integration first on the *x *component, we obtain (using tables):

*Ex,PA *= 3*ρs
*4*π*0

∫ 2 −2

*dy
*

*(*9 + *y*2*)
*

[
*z*√

9 + *y*2 + *z*2

]∞ −∞

= 3*ρs
*2*π*0

∫ 2 −2

*dy
*

*(*9 + *y*2*)
*

= 3*ρs
*2*π*0

( 1

3

) tan−1

(*y
*3

) ∣∣∣2−2 = 106 V*/*m
The student is encouraged to verify that if the *y *limits were −∞ to ∞, the result would be that of
the infinite charged plane, or *Ex *= *ρs/(*20*)*.

b) *PB(*0*, *3*, *0*)*: In this case, **r **= 3**a***y *, and symmetry indicates that only a *y *component will exist.
The integral becomes

*Ey,PB *= *ρs
*4*π*0

∫ ∞ −∞

∫ 2 −2

*(*3 − *y) dy dz
*[*(z*2 + 9*)*− 6*y *+ *y*2]1*.*5 =

*ρs
*

2*π*0

∫ 2 −2

*(*3 − *y) dy
(*3 − *y)*2

= − *ρs
*2*π*0

ln*(*3 − *y)
*∣∣∣2−2 = 145 V*/*m

22

2.23. Given the surface charge density, *ρs *= 2*µ*C*/*m2, in the region *ρ < *0*.*2 m, *z *= 0, and is zero elsewhere,
find **E **at:

a) *PA(ρ *= 0*, z *= 0*.*5*)*: First, we recognize from symmetry that only a *z *component of **E **will be
present. Considering a general point *z *on the *z *axis, we have **r **= *z***a***z*. Then, with **r**′ = *ρ***a***ρ *, we
obtain **r **− **r**′ = *z***a***z *− *ρ***a***ρ *. The superposition integral for the *z *component of **E **will be:

*Ez,PA *=
*ρs
*

4*π*0

∫ 2*π
*0

∫ 0*.*2
0

*z ρ dρ dφ
*

*(ρ*2 + *z*2*)*1*.*5 = −
2*πρs
*4*π*0

*z
*

[ 1√

*z*2 + *ρ*2

]0*.*2
0

= *ρs
*20

*z
*

[
1√
*z*2

− 1√
*z*2 + 0*.*4

]

With *z *= 0*.*5 m, the above evaluates as *Ez,PA *= 8*.*1 kV*/*m.
b) With *z *at −0*.*5 m, we evaluate the expression for *Ez *to obtain *Ez,PB *= −8*.*1 kV*/*m.

2.24. Surface charge density is positioned in free space as follows: 20 nC*/*m2 at *x *= −3, −30 nC*/*m2 at
*y *= 4, and 40 nC*/*m2 at *z *= 2. Find the magnitude of **E **at the three points, *(*4*, *3*,*−2*)*, *(*−2*, *5*,*−1*)*,
and *(*0*, *0*, *0*)*. Since all three sheets are infinite, the field magnitude associated with each one will be
*ρs/(*20*)*, which is position-independent. For this reason, the *net *field magnitude will be the same
everywhere, whereas the field direction will depend on which side of a given sheet one is positioned.
We take the first point, for example, and find

**E***A *= 20 × 10
−9

20
**a***x *+ 30 × 10

−9

20
**a***y *− 40 × 10

−9

20
**a***z *= 1130**a***x *+ 1695**a***y *− 2260**a***z *V*/*m

The magnitude of **E***A *is thus 3*.*04 kV*/*m. This will be the magnitude at the other two points as well.

2.25. Find **E **at the origin if the following charge distributions are present in free space: point charge, 12 nC
at *P(*2*, *0*, *6*)*; uniform line charge density, 3nC*/*m at *x *= −2, *y *= 3; uniform surface charge density,
0*.*2 nC*/*m2 at *x *= 2. The sum of the fields at the origin from each charge in order is:

**E **=
[
*(*12 × 10−9*)
*

4*π*0

*(*−2**a***x *− 6**a***z)
(*4 + 36*)*1*.*5

]
+
[
*(*3 × 10−9*)
*

2*π*0

*(*2**a***x *− 3**a***y)
(*4 + 9*)
*

]
−
[
*(*0*.*2 × 10−9*)***a***x
*

20

]
= −3*.*9**a***x *− 12*.*4**a***y *− 2*.*5**a***z *V*/*m

2.26. A uniform line charge density of 5 nC*/*m is at *y *= 0, *z *= 2 m in free space, while −5 nC*/*m is located
at *y *= 0, *z *= −2 m. A uniform surface charge density of 0*.*3 nC*/*m2 is at *y *= 0*.*2 m, and −0*.*3 nC*/*m2
is at *y *= −0*.*2 m. Find |**E**| at the origin: Since each pair consists of equal and opposite charges, the
effect at the origin is to double the field produce by one of each type. Taking the sum of the fields at
the origin from the surface and line charges, respectively, we find:

**E***(*0*, *0*, *0*) *= −2 × 0*.*3 × 10
−9

20
**a***y *− 2 × 5 × 10

−9

2*π*0*(*2*)
***a***z *= −33*.*9**a***y *− 89*.*9**a***z
*

so that |**E**| = 96*.*1 V*/*m.

23

2.27. Given the electric field **E **= *(*4*x *− 2*y)***a***x *− *(*2*x *+ 4*y)***a***y *, find:
a) the equation of the streamline that passes through the point *P(*2*, *3*,*−4*)*: We write

*dy
*

*dx
*= *Ey
*

*Ex
*= −*(*2*x *+ 4*y)
*

*(*4*x *− 2*y)
*Thus

2*(x dy *+ *y dx) *= *y dy *− *x dx
*or

2 *d(xy) *= 1
2
*d(y*2*)*− 1

2
*d(x*2*)
*

So

*C*1 + 2*xy *= 1
2
*y*2 − 1

2
*x*2

or
*y*2 − *x*2 = 4*xy *+ *C*2

Evaluating at *P(*2*, *3*,*−4*)*, obtain:

9 − 4 = 24 + *C*2*, *or *C*2 = −19

Finally, at *P *, the requested equation is

*y*2 − *x*2 = 4*xy *− 19

b) a unit vector specifying the direction of **E **at *Q(*3*,*−2*, *5*)*: Have **E***Q *= [4*(*3*)*+ 2*(*2*)*]**a***x *− [2*(*3*)*−
4*(*2*)*]**a***y *= 16**a***x *+ 2**a***y *. Then |**E**| =

√
162 + 4 = 16*.*12 So

**a***Q *= 16**a***x *+ 2**a***y
*16*.*12

= 0*.*99**a***x *+ 0*.*12**a***y
*

2.28. Let **E **= 5*x*3 **a***x *− 15*x*2*y ***a***y *, and find:
a) the equation of the streamline that passes through *P(*4*, *2*, *1*)*: Write

*dy
*

*dx
*= *Ey
*

*Ex
*= −15*x
*

2*y
*

5*x*3
= −3*y
*

*x
*

So
*dy
*

*y
*= −3 *dx
*

*x
*⇒ ln *y *= −3 ln *x *+ ln*C
*

Thus

*y *= *e*−3 ln *xe*ln*C *= *C
x*3

At *P *, have 2 = *C/(*4*)*3 ⇒ *C *= 128. Finally, at *P *,

*y *= 128
*x*3

24

2.28. (continued)
b) a unit vector **a***E *specifying the direction of **E **at *Q(*3*,*−2*, *5*)*: At *Q*, **E***Q *= 135**a***x *+ 270**a***y *, and

|**E***Q*| = 301*.*9. Thus **a***E *= 0*.*45**a***x *+ 0*.*89**a***y *.
c) a unit vector **a***N *= *(l, m, *0*) *that is perpendicular to **a***E *at*Q*: Since this vector is to have no *z *compo-

nent, we can find it through **a***N *= ±*(***a***E*×**a***z)*. Performing this, we find **a***N *= ±*(*0*.*89**a***x *− 0*.*45**a***y)*.

2.29. If **E **= 20*e*−5*y *(cos 5*x***a***x *− sin 5*x***a***y*), find:
a) |**E**| at *P(π/*6*, *0*.*1*, *2*)*: Substituting this point, we obtain **E***P *= −10*.*6**a***x *− 6*.*1**a***y *, and so |**E***P *| =

12*.*2.

b) a unit vector in the direction of **E***P *: The unit vector associated with **E **is just
(
cos 5*x***a***x *− sin 5*x***a***y
*

) ,

which evaluated at *P *becomes **a***E *= −0*.*87**a***x *− 0*.*50**a***y *.
c) the equation of the direction line passing through *P *: Use

*dy
*

*dx
*= − sin 5*x
*

cos 5*x
*= − tan 5*x *⇒ *dy *= − tan 5*x dx
*

Thus *y *= 15 ln cos 5*x *+ *C*. Evaluating at *P *, we find *C *= 0*.*13, and so

*y *= 1
5

ln cos 5*x *+ 0*.*13

2.30. Given the electric field intensity **E **= 400*y***a***x *+ 400*x***a***y *V/m, find:
a) the equation of the streamline passing through the point *A(*2*, *1*,*−2*)*: Write:

*dy
*

*dx
*= *Ey
*

*Ex
*= *x
*

*y
*⇒ *x dx *= *y dy
*

Thus *x*2 = *y*2 + *C*. Evaluating at *A *yields *C *= 3, so the equation becomes

*x*2

3
− *y
*

2

3 = 1

b) the equation of the surface on which |**E**| = 800 V/m: Have |**E**| = 400
√
*x*2 + *y*2 = 800. Thus

*x*2 + *y*2 = 4, or we have a circular-cylindrical surface, centered on the *z *axis, and of radius 2.
c) A sketch of the part *a *equation would yield a parabola, centered at the origin, whose axis is the

positive *x *axis, and for which the slopes of the asymptotes are ±1.
d) A sketch of the trace produced by the intersection of the surface of part *b *with the *z *= 0 plane

would yield a circle centered at the origin, of radius 2.

25

2.31. In cylindrical coordinates with **E***(ρ, φ) *= *Eρ(ρ, φ)***a***ρ*+*Eφ(ρ, φ)***a***φ *, the differential equation describ-
ing the direction lines is *Eρ/Eφ *= *dρ/(ρdφ) *in any constant-*z *plane. Derive the equation of the line
passing through the point *P(ρ *= 4*, φ *= 10◦*, z *= 2*) *in the field **E **= 2*ρ*2 cos 3*φ***a***ρ *+ 2*ρ*2 sin 3*φ***a***φ *:
Using the given information, we write

*Eρ
*

*Eφ
*= *dρ
*

*ρdφ
*= cot 3*φ
*

Thus
*dρ
*

*ρ
*= cot 3*φ dφ *⇒ ln *ρ *= 1

3
ln sin 3*φ *+ ln*C
*

or *ρ *= *C(*sin 3*φ)*1*/*3. Evaluate this at *P *to obtain *C *= 7*.*14. Finally,

*ρ*3 = 364 sin 3*φ
*

26

**CHAPTER 3
**

3.1. An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged (honorably) by touching them to ground. An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a dime are glued to the thread so that they are not touching each other. The penny is given a charge of +5 nC, and the nickel and dime are discharged. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured. The outside of the can is again touched momentarily to ground. The device is carefully disassembled with insulating gloves and tools.

a) What charges are found on each of the five metallic pieces? All coins were insulated during the entire procedure, so they will retain their original charges: Penny: +5 nC; nickel: 0; dime: 0. The penny’s charge will have induced an equal and opposite negative charge (-5 nC) on the inside wall of the can and lid. This left a charge layer of +5 nC on the outside surface which was neutralized by the ground connection. Therefore, the can retained a net charge of −5 nC after disassembly.

b) If the penny had been given a charge of +5 nC, the dime a charge of −2 nC, and the nickel a charge of −1 nC, what would the final charge arrangement have been? Again, since the coins are insulated, they retain their original charges. The charge induced on the inside wall of the can and lid is equal to negative the sum of the coin charges, or −2 nC. This is the charge that the can/lid contraption retains after grounding and disassembly.

3.2. A point charge of 12 nC is located at the origin. four uniform line charges are located in the *x *= 0
plane as follows: 80 nC/m at *y *= −1 and −5 m, −50 nC/m at *y *= −2 and −4 m.

a) Find **D **at *P(*0*,*−3*, *2*)*: Note that this point lies in the center of a symmetric arrangement of line
charges, whose fields will all cancel at that point. Thus **D **arise from the point charge alone, and
will be

**D **= 12 × 10
−9*(*−3**a***y *+ 2**a***z)
*

4*π(*32 + 22*)*1*.*5 = −6*.*11 × 10
−11**a***y *+ 4*.*07 × 10−11**a***z *C*/*m2

= −61*.*1**a***y *+ 40*.*7**a***z *pC*/*m2

b) How much electric flux crosses the plane *y *= −3 and in what direction? The plane intercepts all
flux that enters the −*y *half-space, or exactly half the total flux of 12 nC. The answer is thus 6 nC
and in the −**a***y *direction.

c) How much electric flux leaves the surface of a sphere, 4m in radius, centered at *C(*0*,*−3*, *0*)*? This
sphere encloses the point charge, so its flux of 12 nC is included. The line charge contributions
are most easily found by translating the whole assembly (sphere and line charges) such that the
sphere is centered at the origin, with line charges now at *y *= ±1 and ±2. The flux from the line
charges will equal the total line charge that lies within the sphere. The length of each of the inner
two line charges (at *y *= ±1) will be

*h*1 = 2*r *cos *θ*1 = 2*(*4*) *cos
[

sin−1 (

1

4

)]
= 1*.*94 m

That of each of the outer two line charges (at *y *= ±2) will be

*h*2 = 2*r *cos *θ*2 = 2*(*4*) *cos
[

sin−1 (

2

4

)]
= 1*.*73 m

27

3.2c. (continued) The total charge enclosed in the sphere (and the outward flux from it) is now

*Ql *+*Qp *= 2*(*1*.*94*)(*−50 × 10−9*)*+ 2*(*1*.*73*)(*80 × 10−9*)*+ 12 × 10−9 = 348 nC

3.3. The cylindrical surface *ρ *= 8 cm contains the surface charge density, *ρs *= 5*e*−20|*z*| nC*/*m2.
a) What is the total amount of charge present? We integrate over the surface to find:

*Q *= 2
∫ ∞

0

∫ 2*π
*0

5*e*−20*z(.*08*)dφ dz *nC = 20*π(.*08*)
*(−1

20

)
*e*−20*z
*

∣∣∣∣∣ ∞

0

= 0*.*25 nC

b) How much flux leaves the surface *ρ *= 8 cm, 1 cm *< z < *5cm, 30◦ *< φ < *90◦? We just integrate
the charge density on that surface to find the flux that leaves it.

* *= *Q*′ =
∫ *.*05
*.*01

∫ 90◦ 30◦

5*e*−20*z(.*08*) dφ dz *nC =
(

90 − 30 360

)
2*π(*5*)(.*08*)
*

(−1 20

)
*e*−20*z
*

∣∣∣∣∣
*.*05

*.*01

= 9*.*45 × 10−3 nC = 9*.*45 pC

3.4. The cylindrical surfaces *ρ *= 1*, *2*, *and 3 cm carry uniform surface charge densities of 20, −8, and 5
nC*/*m2, respectively.

a) How much electric flux passes through the closed surface *ρ *= 5 cm, 0 *< z < *1 m? Since the
densities are uniform, the flux will be

* *= 2*π(aρs*1 + *bρs*2 + *cρs*3*)(*1 m*) *= 2*π *[*(.*01*)(*20*)*− *(.*02*)(*8*)*+ *(.*03*)(*5*)*] × 10−9 = 1*.*2 nC

b) Find **D **at *P(*1 cm*, *2 cm*, *3 cm*)*: This point lies at radius
√

5 cm, and is thus inside the outermost
charge layer. This layer, being of uniform density, will not contribute to **D **at *P *. We know that in
cylindrical coordinates, the layers at 1 and 2 cm will produce the flux density:

**D **= *Dρ***a***ρ *= *aρs*1 + *bρs*2
*ρ
*

**a***ρ
*

or

*Dρ *= *(.*01*)(*20*)*+ *(.*02*)(*−8*)*√
*.*05

= 1*.*8 nC*/*m2

At *P *, *φ *= tan−1*(*2*/*1*) *= 63*.*4◦. Thus *Dx *= 1*.*8 cos*φ *= 0*.*8 and *Dy *= 1*.*8 sin *φ *= 1*.*6. Finally,

**D***P *= *(*0*.*8**a***x *+ 1*.*6**a***y) *nC*/*m2

28

3.5. Let **D **= 4*xy***a***x *+ 2*(x*2 + *z*2*)***a***y *+ 4*yz***a***z *C*/*m2 and evaluate surface integrals to find the total charge
enclosed in the rectangular parallelepiped 0 *< x < *2, 0 *< y < *3, 0 *< z < *5 m: Of the 6 surfaces to
consider, only 2 will contribute to the net outward flux. Why? First consider the planes at *y *= 0 and 3.
The *y *component of **D **will penetrate those surfaces, but will be inward at *y *= 0 and outward at *y *= 3,
while having the same magnitude in both cases. These fluxes will thus cancel. At the *x *= 0 plane,
*Dx *= 0 and at the *z *= 0 plane, *Dz *= 0, so there will be no flux contributions from these surfaces.
This leaves the 2 remaining surfaces at *x *= 2 and *z *= 5. The net outward flux becomes:

* *=
∫ 5

0

∫ 3 0

**D
**∣∣
*x*=2 · **a***x dy dz*+

∫ 3 0

∫ 2 0

**D
**∣∣
*z*=5 · **a***z dx dy
*

= 5 ∫ 3

0
4*(*2*)y dy *+ 2

∫ 3 0

4*(*5*)y dy *= 360 C

3.6. Two uniform line charges, each 20 nC/m, are located at *y *= 1, *z *= ±1 m. Find the total flux leaving a
sphere of radius 2 m if it is centered at

a) *A(*3*, *1*, *0*)*: The result will be the same if we move the sphere to the origin and the line charges to
*(*0*, *0*,*±1*)*. The length of the line charge within the sphere is given by *l *= 4 sin[cos−1*(*1*/*2*)*] =
3*.*46. With two line charges, symmetrically arranged, the total charge enclosed is given by *Q *=
2*(*3*.*46*)(*20 nC*/*m*) *= 139 nC

b) *B(*3*, *2*, *0*)*: In this case the result will be the same if we move the sphere to the origin and keep
the charges where they were. The length of the line joining the origin to the midpoint of the line
charge (in the *yz *plane) is *l*1 =

√ 2. The length of the line joining the origin to either endpoint

of the line charge is then just the sphere radius, or 2. The half-angle subtended at the origin by
the line charge is then *ψ *= cos−1*(*√2*/*2*) *= 45◦. The length of each line charge in the sphere
is then *l*2 = 2 × 2 sin*ψ *= 2

√ 2. The total charge enclosed (with two line charges) is now

*Q*′ = 2*(*2√2*)(*20 nC*/*m*) *= 113 nC

3.7. Volume charge density is located in free space as *ρv *= 2*e*−1000*r *nC*/*m3 for 0 *< r < *1 mm, and *ρv *= 0
elsewhere.

a) Find the total charge enclosed by the spherical surface *r *= 1 mm: To find the charge we integrate:

*Q *=
∫ 2*π
*

0

∫ *π
*0

∫ *.*001
0

2*e*−1000*r r*2 sin *θ dr dθ dφ
*

Integration over the angles gives a factor of 4*π *. The radial integration we evaluate using tables;
we obtain

*Q *= 8*π
*[−*r*2*e*−1000*r
*

1000

∣∣∣*.*001
0

+ 2 1000

*e*−1000*r
*

*(*1000*)*2
*(*−1000*r *− 1*)
*

∣∣∣*.*001
0

]
= 4*.*0 × 10−9 nC

b) By using Gauss’s law, calculate the value of *Dr *on the surface *r *= 1 mm: The gaussian surface
is a spherical shell of radius 1 mm. The enclosed charge is the result of part *a*. We thus write
4*πr*2*Dr *= *Q*, or

*Dr *= *Q
*4*πr*2

= 4*.*0 × 10
−9

4*π(.*001*)*2
= 3*.*2 × 10−4 nC*/*m2

29

3.8. Uniform line charges of 5 nC/m ar located in free space at *x *= 1, *z *= 1, and at *y *= 1, *z *= 0.
a) Obtain an expression for **D **in cartesian coordinates at *P(*0*, *0*, z)*. In general, we have

**D***(z) *= *ρs
*2*π
*

[
**r**1 − **r**′1

|**r**1 − **r**′1|2
+ **r**2 − **r
**

′ 2

|**r**2 − **r**′2|2
]

where **r**1 = **r**2 = *z***a***z*, **r**′1 = **a***y *, and **r**′2 = **a***x *+ **a***z*. Thus

**D***(z) *= *ρs
*2*π
*

[
[*z***a***z *− **a***y*]
[1 + *z*2] +

[*(z*− 1*)***a***z *− **a***x*]
[1 + *(z*− 1*)*2]

]

= *ρs
*2*π
*

[ −**a***x
*[1 + *(z*− 1*)*2] −

**a***y
*[1 + *z*2] +

(
*(z*− 1*)
*

[1 + *(z*− 1*)*2] +
*z
*

[1 + *z*2]
)

**a***z
*

]

b) Plot |**D**| vs. *z *at *P *, −3 *< z < *10: Using part *a*, we find the magnitude of **D **to be

|**D**| = *ρs
*2*π
*

[ 1

[1 + *(z*− 1*)*2]2 +
1

[1 + *z*2]2 +
(

*(z*− 1*)
*[1 + *(z*− 1*)*2] +

*z
*

[1 + *z*2]
)2]1*/*2

A plot of this over the specified range is shown in Prob3.8.pdf.

3.9. A uniform volume charge density of 80*µ*C*/*m3 is present throughout the region 8 mm *< r < *10 mm.
Let *ρv *= 0 for 0 *< r < *8 mm.

a) Find the total charge inside the spherical surface *r *= 10 mm: This will be

*Q *=
∫ 2*π
*

0

∫ *π
*0

∫ *.*010
*.*008

*(*80 × 10−6*)r*2 sin *θ dr dθ dφ *= 4*π *× *(*80 × 10−6*) r
*3

3

∣∣∣*.*010
*.*008

= 1*.*64 × 10−10 C = 164 pC

b) Find *Dr *at *r *= 10 mm: Using a spherical gaussian surface at *r *= 10, Gauss’ law is written as
4*πr*2*Dr *= *Q *= 164 × 10−12, or

*Dr(*10 mm*) *= 164 × 10
−12

4*π(.*01*)*2
= 1*.*30 × 10−7 C*/*m2 = 130 nC*/*m2

c) If there is no charge for *r > *10 mm, find *Dr *at *r *= 20 mm: This will be the same computation
as in part *b*, except the gaussian surface now lies at 20 mm. Thus

*Dr(*20 mm*) *= 164 × 10
−12

4*π(.*02*)*2
= 3*.*25 × 10−8 C*/*m2 = 32*.*5 nC*/*m2

3.10. Let *ρs *= 8*µ*C*/*m2 in the region where *x *= 0 and −4 *< z < *4 m, and let *ρs *= 0 elsewhere. Find **D **at
*P(x, *0*, z)*, where *x > *0: The sheet charge can be thought of as an assembly of infinitely-long parallel
strips that lie parallel to the *y *axis in the *yz *plane, and where each is of thickness *dz*. The field from
each strip is that of an infinite line charge, and so we can construct the field at *P *from a single strip as:

*d***D***P *= *ρs dz
*′

2*π
*

**r **− **r**′
|**r **− **r**′|2

30

3.10 (continued) where **r **= *x***a***x *+ *z***a***z *and **r**′ = *z*′**a***z *We distinguish between the fixed coordinate of *P *, *z*,
and the variable coordinate, *z*′, that determines the location of each charge strip. To find the net field at
*P *, we sum the contributions of each strip by integrating over *z*′:

**D***P *=
∫ 4
−4

8 × 10−6 *dz*′ *(x***a***x *+ *(z*− *z*′*)***a***z)
*2*π *[*x*2 + *(z*− *z*′*)*2]

We can re-arrange this to determine the integral forms:

**D***P *= 8 × 10
−6

2*π
*

[
*(x***a***x *+ *z***a***z)
*

∫ 4 −4

*dz*′

*(x*2 + *z*2*)*− 2*zz*′ + *(z*′*)*2 − **a***z
*∫ 4
−4

*z*′ *dz*′

*(x*2 + *z*2*)*− 2*zz*′ + *(z*′*)*2
]

Using integral tables, we find

**D***P *= 4 × 10
−6

*π
*

[
*(x***a***x *+ *z***a***z)*1

*x
*tan−1

(
2*z*′ − 2*z
*

2*x
*

)

− [

1

2
ln*(x*2 + *z*2 − 2*zz*′ + *(z*′*)*2*)*+ 2*z
*

2

1

*x
*tan−1

(
2*z*′ − 2*z
*

2*x
*

)]
**a***z
*

]4 −4

which evaluates as

**D***P *= 4 × 10
−6

*π
*

{[ tan−1

(
*z*+ 4
*x
*

) − tan−1

(
*z*− 4
*x
*

)]
**a***x *+ 1

2 ln

[
*x*2 + *(z*+ 4*)*2
*x*2 + *(z*− 4*)*2

]
**a***z
*

}
C*/*m2

The student is invited to verify that for very small *x *or for a very large sheet (allowing *z*′ to approach
infinity), the above expression reduces to the expected form, **D***P *= *ρs/*2. Note also that the expression
is valid for all *x *(positive or negative values).

3.11. In cylindrical coordinates, let *ρv *= 0 for *ρ < *1 mm, *ρv *= 2 sin*(*2000*πρ) *nC*/*m3 for 1 mm *< ρ <
*1*.*5 mm, and *ρv *= 0 for *ρ > *1*.*5 mm. Find **D **everywhere: Since the charge varies only with radius,
and is in the form of a cylinder, symmetry tells us that the flux density will be radially-directed and will
be constant over a cylindrical surface of a fixed radius. Gauss’ law applied to such a surface of unit
length in *z *gives:

a) for *ρ < *1 mm, *Dρ *= 0, since no charge is enclosed by a cylindrical surface whose radius lies
within this range.

b) for 1 mm *< ρ < *1*.*5 mm, we have

2*πρDρ *= 2*π
*∫ *ρ
.*001

2 × 10−9 sin*(*2000*πρ*′*)ρ*′ *dρ*′

= 4*π *× 10−9
[

1

*(*2000*π)*2
sin*(*2000*πρ)*− *ρ
*

2000*π
*cos*(*2000*πρ)
*

]*ρ
.*001

or finally,

*Dρ *= 10
−15

2*π*2*ρ
*

[
sin*(*2000*πρ)*+ 2*π
*

[
1 − 103*ρ *cos*(*2000*πρ)
*

] ]
C*/*m2 *(*1 mm *< ρ < *1*.*5 mm*)
*

31

3.11. (continued)
c) for *ρ > *1*.*5 mm, the gaussian cylinder now lies at radius *ρ outside *the charge distribution, so

the integral that evaluates the enclosed charge now includes the entire charge distribution. To
accomplish this, we change the upper limit of the integral of part *b *from *ρ *to 1*.*5 mm, finally
obtaining:

*Dρ *= 2*.*5 × 10
−15

*πρ
*C*/*m2 *(ρ > *1*.*5 mm*)
*

3.12. A nonuniform volume charge density, *ρv *= 120*r *C*/*m3, lies within the spherical surface *r *= 1 m, and
*ρv *= 0 everywhere else.

a) Find *Dr *everywhere. For *r < *1 m, we apply Gauss’ law to a spherical surface of radius *r *within
this range to find

4*πr*2*Dr *= 4*π
*∫ *r
*

0
120*r *′*(r *′*)*2 *dr *′ = 120*πr*4

Thus *Dr *= *(*30*r*2*) *for *r < *1 m. For *r > *1 m, the gaussian surface lies outside the charge
distribution. The set up is the same, except the upper limit of the above integral is 1 instead of *r *.
This results in *Dr *= *(*30*/r*2*) *for *r > *1 m.

b) What surface charge density, *ρs*2, should be on the surface *r *= 2 such that *Dr,r*=2− = 2*Dr,r*=2+?
At *r *= 2−, we have *Dr,r*=2− = 30*/*22 = 15*/*2, from part *a*. The flux density in the region *r > *2
arising from a surface charge at *r *= 2 is found from Gauss’ law through

4*πr*2*Drs *= 4*π(*2*)*2*ρs*2 ⇒ *Drs *= 4*ρs*2
*r*2

The total flux density in the region *r > *2 arising from the two distributions is

*DrT *= 30
*r*2

+ 4*ρs*2
*r*2

Our requirement that *Dr,r*=2− = 2*Dr,r*=2+ becomes

30

22 = 2

( 30

22
+ *ρs*2

)
⇒ *ρs*2 = −15

4
C*/*m2

c) Make a sketch of *Dr *vs. *r *for 0 *< r < *5 m with both distributions present. With both charges,
*Dr(r < *1*) *= 30*r*2, *Dr(*1 *< r < *2*) *= 30*/r*2, and *Dr(r > *2*) *= 15*/r*2. These are plotted on the
next page.

32

.

3.13. Spherical surfaces at *r *= 2*, *4*, *and 6 m carry uniform surface charge densities of 20 nC*/*m2,−4 nC*/*m2,
and *ρs*0, respectively.

a) Find **D **at *r *= 1*, *3 and 5 m: Noting that the charges are spherically-symmetric, we ascertain that
**D **will be radially-directed and will vary only with radius. Thus, we apply Gauss’ law to spherical
shells in the following regions: *r < *2: Here, no charge is enclosed, and so *Dr *= 0.

2 *< r < *4 : 4*πr*2*Dr *= 4*π(*2*)*2*(*20 × 10−9*) *⇒ *Dr *= 80 × 10
−9

*r*2
C*/*m2

So *Dr(r *= 3*) *= 8*.*9 × 10−9 C*/*m2.

4 *< r < *6 : 4*πr*2*Dr *= 4*π(*2*)*2*(*20 × 10−9*)*+ 4*π(*4*)*2*(*−4 × 10−9*) *⇒ *Dr *= 16 × 10
−9

*r*2

So *Dr(r *= 5*) *= 6*.*4 × 10−10 C*/*m2.
b) Determine *ρs*0 such that **D **= 0 at *r *= 7 m. Since fields will decrease as 1*/r*2, the question could

be re-phrased to ask for *ρs*0 such that **D **= 0 at *all *points where *r > *6 m. In this region, the total
field will be

*Dr(r > *6*) *= 16 × 10
−9

*r*2
+ *ρs*0*(*6*)
*

2

*r*2

Requiring this to be zero, we find *ρs*0 = −*(*4*/*9*)*× 10−9 C*/*m2.

3.14. If *ρv *= 5 nC*/*m3 for 0 *< ρ < *1 mm and no other charges are present:
a) find *Dρ *for *ρ < *1 mm: Applying Gauss’ law to a cylindrical surface of unit length in *z*, and of

radius *ρ < *1 mm, we find

2*πρDρ *= *πρ*2*(*5 × 10−9*) *⇒ *Dρ *= 2*.*5 *ρ *× 10−9 C*/*m2

33

3.14b. find *Dρ *for *ρ > *1 mm: The Gaussian cylinder now lies outside the charge, so

2*πρDρ *= *π(.*001*)*2*(*5 × 10−9*) *⇒ *Dρ *= 2*.*5 × 10
−15

*ρ
*C*/*m2

c) What line charge *ρL *at *ρ *= 0 would give the same result for part *b*? The line charge field will be

*Dr *= *ρL
*2*πρ
*

= 2*.*5 × 10
−15

*ρ
(*part *b)
*

Thus *ρL *= 5*π *× 10−15 C*/*m. In all answers, *ρ *is expressed in meters.

3.15. Volume charge density is located as follows: *ρv *= 0 for *ρ < *1 mm and for *ρ > *2 mm, *ρv *= 4*ρ µ*C*/*m3
for 1 *< ρ < *2 mm.

a) Calculate the total charge in the region 0 *< ρ < ρ*1, 0 *< z < L*, where 1 *< ρ*1 *< *2 mm: We find

*Q *=
∫ *L
*

0

∫ 2*π
*0

∫ *ρ*1
*.*001

4*ρ ρ dρ dφ dz *= 8*πL
*3

[*ρ*31 − 10−9] *µ*C

where *ρ*1 is in meters.

b) Use Gauss’ law to determine *Dρ *at *ρ *= *ρ*1: Gauss’ law states that 2*πρ*1*LDρ *= *Q*, where *Q *is
the result of part *a*. Thus

*Dρ(ρ*1*) *=
4*(ρ*31 − 10−9*)
*

3*ρ*1
*µ*C*/*m2

where *ρ*1 is in meters.

c) Evaluate *Dρ *at *ρ *= 0*.*8 mm*, *1*.*6 mm*, *and 2*.*4 mm: At *ρ *= 0*.*8 mm, no charge is enclosed by a
cylindrical gaussian surface of that radius, so *Dρ(*0*.*8mm*) *= 0. At *ρ *= 1*.*6 mm, we evaluate the
part *b *result at *ρ*1 = 1*.*6 to obtain:

*Dρ(*1*.*6mm*) *= 4[*(.*0016*)
*3 − *(.*0010*)*3]

3*(.*0016*)
*= 3*.*6 × 10−6 *µ*C*/*m2

At *ρ *= 2*.*4, we evaluate the charge integral of part *a *from .001 to .002, and Gauss’ law is written
as

2*πρLDρ *= 8*πL
*3

[*(.*002*)*2 − *(.*001*)*2] *µ*C
from which *Dρ(*2*.*4mm*) *= 3*.*9 × 10−6 *µ*C*/*m2.

3.16. Given the electric flux density, **D **= 2*xy ***a***x *+ *x*2 **a***y *+ 6*z*3 **a***z *C*/*m2:
a) use Gauss’ law to evaluate the total charge enclosed in the volume 0 *< x, y, z < a*: We call the

surfaces at *x *= *a *and *x *= 0 the front and back surfaces respectively, those at *y *= *a *and *y *= 0
the right and left surfaces, and those at *z *= *a *and *z *= 0 the top and bottom surfaces. To evaluate
the total charge, we integrate **D **· **n **over all six surfaces and sum the results:

* *= *Q *=
∮

**D **· **n ***da *=
∫ *a
*

0

∫ *a
*0

2*ay dy dz*︸ ︷︷ ︸
front

+
∫ *a
*

0

∫ *a
*0

−2*(*0*)y dy dz*︸ ︷︷ ︸
back

+
∫ *a
*

0

∫ *a
*0

−*x*2 *dx dz*︸ ︷︷ ︸
left

+
∫ *a
*

0

∫ *a
*0

*x*2 *dx dz*︸ ︷︷ ︸
right

+
∫ *a
*

0

∫ *a
*0

−6*(*0*)*3 *dx dy*︸ ︷︷ ︸
bottom

+
∫ *a
*

0

∫ *a
*0

6*a*3 *dx dy*︸ ︷︷ ︸
top

34

3.16a. (continued) Noting that the back and bottom integrals are zero, and that the left and right integrals
cancel, we evaluate the remaining two (front and top) to obtain *Q *= 6*a*5 + *a*4.

b) use Eq. (8) to find an approximate value for the above charge. Evaluate the derivatives at
*P(a/*2*, a/*2*, a/*2*)*: In this application, Eq. (8) states that *Q
*

*.*= *(*∇ · **D**∣∣
*P
)v*. We find ∇ · **D **=

2*x*+18*z*2, which when evaluated at *P *becomes ∇ ·**D**∣∣
*P
*= *a*+4*.*5*a*2. Thus *Q .*= *(a*+4*.*5*a*2*)a*3 =

4*.*5*a*5 + *a*4

c) Show that the results of parts *a *and *b *agree in the limit as *a *→ 0. In this limit, both expressions
reduce to *Q *= *a*4, and so they agree.

3.17. A cube is defined by 1 *< x, y, z < *1*.*2. If **D **= 2*x*2*y***a***x *+ 3*x*2*y*2**a***y *C*/*m2:
a) apply Gauss’ law to find the total flux leaving the closed surface of the cube. We call the surfaces

at *x *= 1*.*2 and *x *= 1 the front and back surfaces respectively, those at *y *= 1*.*2 and *y *= 1 the
right and left surfaces, and those at *z *= 1*.*2 and *z *= 1 the top and bottom surfaces. To evaluate
the total charge, we integrate **D **· **n **over all six surfaces and sum the results. We note that there
is no *z *component of **D**, so there will be no outward flux contributions from the top and bottom
surfaces. The fluxes through the remaining four are

* *= *Q *=
∮

**D **· **n ***da *=
∫ 1*.*2

1

∫ 1*.*2
1

2*(*1*.*2*)*2*y dy dz*︸ ︷︷ ︸
front

+
∫ 1*.*2

1

∫ 1*.*2
1

−2*(*1*)*2*y dy dz*︸ ︷︷ ︸
back

+
∫ 1*.*2

1

∫ 1*.*2
1

−3*x*2*(*1*)*2 *dx dz*︸ ︷︷ ︸
left

+
∫ 1*.*2

1

∫ 1*.*2
1

3*x*2*(*1*.*2*)*2 *dx dz*︸ ︷︷ ︸
right

= 0*.*1028 C

b) evaluate ∇ · **D **at the center of the cube: This is

∇ · **D **=
[
4*xy *+ 6*x*2*y
*

]
*(*1*.*1*,*1*.*1*)
*

= 4*(*1*.*1*)*2 + 6*(*1*.*1*)*3 = 12*.*83

c) Estimate the total charge enclosed within the cube by using Eq. (8): This is

*Q
.*= ∇ · **D**∣∣center ×*v *= 12*.*83 × *(*0*.*2*)*3 = 0*.*1026 Close!

3.18. Let a vector field by given by **G **= 5*x*4*y*4*z*4 **a***y *. Evaluate both sides of Eq. (8) for this **G **field and the
volume defined by *x *= 3 and 3.1, *y *= 1 and 1.1, and *z *= 2 and 2.1. Evaluate the partial derivatives at
the center of the volume. First find

∇ · **G **= *∂Gy
∂y
*

= 20*x*4*y*3*z*4

The center of the cube is located at (3.05,1.05,2.05), and the volume is *v *= *(*0*.*1*)*3 = 0*.*001. Eq. (8)
then becomes

*
.*= 20*(*3*.*05*)*4*(*1*.*05*)*3*(*2*.*05*)*4*(*0*.*001*) *= 35*.*4

35

3.19. A spherical surface of radius 3 mm is centered at *P(*4*, *1*, *5*) *in free space. Let **D **= *x***a***x *C*/*m2. Use the
results of Sec. 3.4 to estimate the net electric flux leaving the spherical surface: We use *
*

*.*= ∇ ·**D***v*,
where in this case ∇ · **D **= *(∂/∂x)x *= 1 C*/*m3. Thus

*
.*= 4

3
*π(.*003*)*3*(*1*) *= 1*.*13 × 10−7 C = 113 nC

3.20. A cube of volume *a*3 has its faces parallel to the cartesian coordinate surfaces. It is centered at
*P(*3*,*−2*, *4*)*. Given the field **D **= 2*x*3**a***x *C*/*m2:

a) calculate div **D **at *P *: In the present case, this will be

∇ · **D **= *∂Dx
∂x
*

= *dDx
dx
*

= 54 C*/*m3

b) evaluate the fraction in the rightmost side of Eq. (13) for *a *= 1 m, 0.1 m, and 1 mm: With the
field having only an *x *component, flux will pentrate only the two surfaces at *x *= 3 ± *a/*2, each
of which has surface area *a*2. The cube volume is *v *= *a*3. The equation reads:

∮
**D **· *d***S
***v
*

= 1
*a*3

[ 2 (

3 + *a
*2

)3
*a*2 − 2

(
3 − *a
*

2

)3
*a*2
]
= 2

*a
*

[
*(*3 + *a
*

2
*)*3 − *(*3 − *a
*

2
*)*3
]

evaluating the above formula at *a *= 1 m, .1 m, and 1 mm, yields respectively

54*.*50*, *54*.*01*, *and 54*.*00 C*/*m3*,
*

thus demonstrating the approach to the exact value as *v *gets smaller.

3.21. Calculate the divergence of **D **at the point specified if
a) **D **= *(*1*/z*2*) *[10*xyz ***a***x *+ 5*x*2*z ***a***y *+ *(*2*z*3 − 5*x*2*y) ***a***z*] at *P(*−2*, *3*, *5*)*: We find

∇ · **D **=
[

10*y
*

*z
*+ 0 + 2 + 10*x
*

2*y
*

*z*3

]
*(*−2*,*3*,*5*)
*

= 8*.*96

b) **D **= 5*z*2**a***ρ *+ 10*ρz ***a***z *at *P(*3*,*−45◦*, *5*)*: In cylindrical coordinates, we have

∇ · **D **= 1
*ρ
*

*∂
*

*∂ρ
(ρDρ)*+ 1

*ρ
*

*∂Dφ
*

*∂φ
*+ *∂Dz
*

*∂z
*=
[

5*z*2

*ρ
*+ 10*ρ
*

]
*(*3*,*−45◦*,*5*)
*

= 71*.*67

c) **D **= 2*r *sin *θ *sin *φ ***a***r *+ *r *cos *θ *sin *φ ***a***θ *+ *r *cos*φ ***a***φ *at *P(*3*, *45◦*,*−45◦*)*: In spherical coordinates,
we have

∇ · **D **= 1
*r*2

*∂
*

*∂r
(r*2*Dr)*+ 1

*r *sin *θ
*

*∂
*

*∂θ
(*sin *θDθ)*+ 1

*r *sin *θ
*

*∂Dφ
*

*∂φ
*

= [

6 sin *θ *sin *φ *+ cos 2*θ *sin *φ
*sin *θ
*

− sin *φ
*sin *θ
*

]
*(*3*,*45◦*,*−45◦*)
*

= −2

36

3.22. Let **D **= 8*ρ *sin *φ ***a***ρ *+ 4*ρ *cos*φ ***a***φ *C*/*m2.
a) Find div **D**: Using the divergence formula for cylindrical coordinates (see problem 3.21), we find

∇ · **D **= 12 sin *φ*.
b) Find the volume charge density at *P(*2*.*6*, *38◦*,*−6*.*1*)*: Since *ρv *= ∇ · **D**, we evaluate the result of

part *a *at this point to find *ρvP *= 12 sin 38◦ = 7*.*39 C*/*m3.
c) How much charge is located inside the region defined by 0 *< ρ < *1*.*8, 20◦ *< φ < *70◦,

2*.*4 *< z < *3*.*1? We use

*Q *=
∫
*vol
*

*ρvdv *=
∫ 3*.*1

2*.*4

∫ 70◦ 20◦

∫ 1*.*8
0

12 sin *φρ dρ dφ dz *= −*(*3*.*1 − 2*.*4*)*12 cos*φ
*∣∣∣70◦
20◦

*ρ*2

2

∣∣∣1*.*8
0

= 8*.*13 C

3.23. a) A point charge *Q *lies at the origin. Show that div **D **is zero everywhere except at the origin. For
a point charge at the origin we know that **D **= *Q/(*4*πr*2*) ***a***r *. Using the formula for divergence in
spherical coordinates (see problem 3.21 solution), we find in this case that

∇ · **D **= 1
*r*2

*d
*

*dr
*

(
*r*2

*Q
*

4*πr*2

) = 0

The above is true provided *r > *0. When *r *= 0, we have a singularity in **D**, so its divergence is not
defined.

b) Replace the point charge with a uniform volume charge density *ρv*0 for 0 *< r < a*. Relate *ρv*0
to *Q *and *a *so that the total charge is the same. Find div **D **everywhere: To achieve the same net
charge, we require that *(*4*/*3*)πa*3*ρv*0 = *Q*, so *ρv*0 = 3*Q/(*4*πa*3*) *C*/*m3. Gauss’ law tells us that
inside the charged sphere

4*πr*2*Dr *= 4
3
*πr*3*ρv*0 = *Qr
*

3

*a*3

Thus

*Dr *= *Qr
*4*πa*3

C*/*m2 and ∇ · **D **= 1
*r*2

*d
*

*dr
*

(
*Qr*3

4*πa*3

)
= 3*Q
*

4*πa*3

as expected. Outside the charged sphere, **D **= *Q/(*4*πr*2*) ***a***r *as before, and the divergence is zero.

3.24. Inside the cylindrical shell, 3 *< ρ < *4 m, the electric flux density is given as

**D **= 5*(ρ *− 3*)*3 **a***ρ *C*/*m2

a) What is the volume charge density at *ρ *= 4 m? In this case we have

*ρv *= ∇ · **D **= 1
*ρ
*

*d
*

*dρ
(ρDρ) *= 1

*ρ
*

*d
*

*dρ
*[5*ρ(ρ *− 3*)*3] = 5*(ρ *− 3*)
*

2

*ρ
(*4*ρ *− 3*) *C*/*m3

Evaluating this at *ρ *= 4 m, we find *ρv(*4*) *= 16*.*25 C*/*m3

b) What is the electric flux density at *ρ *= 4 m? We evaluate the given **D **at this point to find
**D***(*4*) *= 5 **a***ρ *C*/*m2

37

3.24c. How much electric flux leaves the closed surface 3 *< ρ < *4, 0 *< φ < *2*π *, −2*.*5 *< z < *2*.*5? We note
that **D **has only a radial component, and so flux would leave only through the cylinder sides. Also, **D
**does not vary with *φ *or *z*, so the flux is found by a simple product of the side area and the flux density.
We further note that **D **= 0 at *ρ *= 3, so only the outer side (at *ρ *= 4) will contribute. We use the result
of part *b*, and write the flux as

* *= [2*.*5 − *(*−2*.*5*)*]2*π(*4*)(*5*) *= 200*π *C

d) How much charge is contained within the volume used in part *c*? By Gauss’ law, this will be the
same as the net outward flux through that volume, or again, 200*π *C.

3.25. Within the spherical shell, 3 *< r < *4 m, the electric flux density is given as

**D **= 5*(r *− 3*)*3 **a***r *C*/*m2

a) What is the volume charge density at *r *= 4? In this case we have

*ρv *= ∇ · **D **= 1
*r*2

*d
*

*dr
(r*2*Dr) *= 5

*r
(r *− 3*)*2*(*5*r *− 6*) *C*/*m3

which we evaluate at *r *= 4 to find *ρv(r *= 4*) *= 17*.*50 C*/*m3.
b) What is the electric flux density at *r *= 4? Substitute *r *= 4 into the given expression to

find **D***(*4*) *= 5 **a***r *C*/*m2

c) How much electric flux leaves the sphere *r *= 4? Using the result of part *b*, this will be * *=
4*π(*4*)*2*(*5*) *= 320*π *C

d) How much charge is contained within the sphere, *r *= 4? From Gauss’ law, this will be the same
as the outward flux, or again, *Q *= 320*π *C.

3.26. Given the field

**D **= 5 sin *θ *cos*φ
r
*

**a***r *C*/*m2*,
*

find: a) the volume charge density: Use

*ρv *= ∇ · **D **= 1
*r*2

*d
*

*dr
(r*2*Dr) *= 5 sin *θ *cos*φ
*

*r*2
C*/*m3

b) the total charge contained within the region *r < *2 m: To find this, we integrate over the volume:

*Q *=
∫ 2*π
*

0

∫ *π
*0

∫ 2 0

5 sin *θ *cos*φ
*

*r*2
*r*2 sin *θ dr dθ dφ
*

Before plunging into this one notice that the *φ *integration is of cos*φ *from zero to 2*π *. This yields
a zero result, and so the total enclosed charge is *Q *= 0.

c) the value of **D **at the surface *r *= 2: Substituting *r *= 2 into the given field produces

**D***(r *= 2*) *= 5
2

sin *θ *cos*φ ***a***r *C*/*m2

38

3.26d. the total electric flux leaving the surface *r *= 2 Since the total enclosed charge is zero (from part *b*), the
net outward flux is also zero, from Gauss’ law.

3.27. Let **D **= 5*.*00*r*2**a***r *mC*/*m2 for *r *≤ 0*.*08 m and **D **= 0*.*205 **a***r/r*2 *µ*C*/*m2 for *r *≥ 0*.*08 m (note error in
problem statement).

a) Find *ρv *for *r *= 0*.*06 m: This radius lies within the first region, and so

*ρv *= ∇ · **D **= 1
*r*2

*d
*

*dr
(r*2*Dr) *= 1

*r*2

*d
*

*dr
(*5*.*00*r*4*) *= 20*r *mC*/*m3

which when evaluated at *r *= 0*.*06 yields *ρv(r *= *.*06*) *= 1*.*20 mC*/*m3.
b) Find *ρv *for *r *= 0*.*1 m: This is in the region where the second field expression is valid. The 1*/r*2

dependence of this field yields a zero divergence (shown in Problem 3.23), and so the volume charge density is zero at 0.1 m.

c) What surface charge density could be located at *r *= 0*.*08 m to cause **D **= 0 for *r > *0*.*08 m? The
total surface charge should be equal and opposite to the total volume charge. The latter is

*Q *=
∫ 2*π
*

0

∫ *π
*0

∫ *.*08
0

20*r(*mC*/*m3*) r*2 sin *θ dr dθ dφ *= 2*.*57 × 10−3 mC = 2*.*57*µ*C

So now

*ρs *= −
[

2*.*57

4*π(.*08*)*2

]
= −32*µ*C*/*m2

3.28. The electric flux density is given as **D **= 20*ρ*3 **a***ρ *C*/*m2 for *ρ < *100*µ*m, and *k ***a***ρ/ρ *for *ρ > *100*µ*m.
a) Find *k *so that **D **is continuous at *ρ *= 100*µ*m: We require

20 × 10−12 = *k
*10−4

⇒ *k *= 2 × 10−15 C*/*m

b) Find and sketch *ρv *as a function of *ρ*: In cylindrical coordinates, with only a radial component of **D**,
we use

*ρv *= ∇ · **D **= 1
*ρ
*

*∂
*

*∂ρ
(ρDρ) *= 1

*ρ
*

*∂
*

*∂ρ
(*20*ρ*4*) *= 80*ρ*2 C*/*m3 *(ρ < *100*µ*m*)
*

For *ρ > *100*µ*m, we obtain

*ρv *= 1
*ρ
*

*∂
*

*∂ρ
(ρ
*

*k
*

*ρ
) *= 0

The sketch of *ρv *vs. *ρ *would be a parabola, starting at the origin, reaching a maximum value of
8 × 10−7 C*/*m3 at *ρ *= 100 *µ*m. The plot is zero at larger radii.

3.29. In the region of free space that includes the volume 2 *< x, y, z < *3,

**D **= 2
*z*2

*(yz ***a***x *+ *xz ***a***y *− 2*xy ***a***z) *C*/*m2

a) Evaluate the volume integral side of the divergence theorem for the volume defined above: In
cartesian, we find ∇ · **D **= 8*xy/z*3. The volume integral side is now∫

*vol
*

∇ · **D ***dv *=
∫ 3

2

∫ 3 2

∫ 3 2

8*xy
*

*z*3
*dxdydz *= *(*9 − 4*)(*9 − 4*)
*

( 1

4 − 1

9

)
= 3*.*47 C

39

3.29b. Evaluate the surface integral side for the corresponding closed surface: We call the surfaces at *x *= 3
and *x *= 2 the front and back surfaces respectively, those at *y *= 3 and *y *= 2 the right and left surfaces,
and those at *z *= 3 and *z *= 2 the top and bottom surfaces. To evaluate the surface integral side, we
integrate **D **· **n **over all six surfaces and sum the results. Note that since the *x *component of **D **does not
vary with *x*, the outward fluxes from the front and back surfaces will cancel each other. The same is
true for the left and right surfaces, since *Dy *does not vary with *y*. This leaves only the top and bottom
surfaces, where the fluxes are:∮

**D **· *d***S **=
∫ 3

2

∫ 3 2

−4*xy
*32

*dxdy*︸ ︷︷ ︸
top

− ∫ 3

2

∫ 3 2

−4*xy
*22

*dxdy*︸ ︷︷ ︸
bottom

= *(*9 − 4*)(*9 − 4*)
*(

1

4 − 1

9

)
= 3*.*47 C

3.30. If **D **= 15*ρ*2 sin 2*φ ***a***ρ *+ 10*ρ*2 cos 2*φ ***a***φ *C*/*m2, evaluate both sides of the divergence theorem for the
region 1 *< ρ < *2 m, 1 *< φ < *2 rad, 1 *< z < *2 m: Taking the surface integral side first, the six sides
over which the flux must be evaluated are only four, since there is no *z *component of **D**. We are left
with the sides at *φ *= 1 and *φ *= 2 rad (left and right sides, respectively), and those at *ρ *= 1 and *ρ *= 2
(back and front sides). We evaluate∮

**D **· *d***S **=
∫ 2

1

∫ 2 1

15*(*2*)*2 sin*(*2*φ) (*2*)dφdz*︸ ︷︷ ︸
front

− ∫ 2

1

∫ 2 1

15*(*1*)*2 sin*(*2*φ) (*1*)dφdz*︸ ︷︷ ︸
back

− ∫ 2

1

∫ 2 1

10*ρ*2 cos*(*2*) dρdz*︸ ︷︷ ︸
left

+ ∫ 2

1

∫ 2 1

10*ρ*2 cos*(*4*) dρdz*︸ ︷︷ ︸
right

= 6*.*93 C

For the volume integral side, we first evaluate the divergence of **D**, which is

∇ · **D **= 1
*ρ
*

*∂
*

*∂ρ
(*15*ρ*3 sin 2*φ)*+ 1

*ρ
*

*∂
*

*∂φ
(*10*ρ*2 cos 2*φ) *= 25*ρ *sin 2*φ
*

Next ∫
*vol
*

∇ · **D ***dv *=
∫ 2

1

∫ 2 1

∫ 2 1

25*ρ *sin*(*2*φ) ρdρ dφ dz *= 25
3
*ρ*3
∣∣∣2
1

[− cos*(*2*φ)
*2

]2
1
= 6*.*93 C

3.31. Given the flux density

**D **= 16
*r
*

cos*(*2*θ) ***a***θ *C*/*m2*,
*

use two different methods to find the total charge within the region 1 *< r < *2 m, 1 *< θ < *2 rad,
1 *< φ < *2 rad: We use the divergence theorem and first evaluate the surface integral side. We are
evaluating the net outward flux through a curvilinear “cube”, whose boundaries are defined by the
specified ranges. The flux contributions will be only through the surfaces of constant *θ *, however, since
**D **has only a *θ *component. On a constant-theta surface, the differential area is *da *= *r *sin *θdrdφ*,
where *θ *is fixed at the surface location. Our flux integral becomes∮

**D **· *d***S **= −
∫ 2

1

∫ 2 1

16

*r
*cos*(*2*) r *sin*(*1*) drdφ*︸ ︷︷ ︸

*θ*=1

+ ∫ 2

1

∫ 2 1

16

*r
*cos*(*4*) r *sin*(*2*) drdφ*︸ ︷︷ ︸

*θ*=2
= −16 [cos*(*2*) *sin*(*1*)*− cos*(*4*) *sin*(*2*)*] = −3*.*91 C

40

3.31. (continued) We next evaluate the volume integral side of the divergence theorem, where in this case,

∇ · **D **= 1
*r *sin *θ
*

*d
*

*dθ
(*sin *θ Dθ) *= 1

*r *sin *θ
*

*d
*

*dθ
*

[ 16

*r
*cos 2*θ *sin *θ
*

] = 16

*r*2

[
cos 2*θ *cos *θ
*

sin *θ
*− 2 sin 2*θ
*

]

We now evaluate:

∫
*vol
*

∇ · **D ***dv *=
∫ 2

1

∫ 2 1

∫ 2 1

16

*r*2

[
cos 2*θ *cos *θ
*

sin *θ
*− 2 sin 2*θ
*

]
*r*2 sin *θ drdθdφ
*

The integral simplifies to

∫ 2 1

∫ 2 1

∫ 2 1

16[cos 2*θ *cos *θ *− 2 sin 2*θ *sin *θ *] *drdθdφ *= 8
∫ 2

1
[3 cos 3*θ *− cos *θ *] *dθ *= −3*.*91 C

3.32. If **D **= 2*r ***a***r *C*/*m2, find the total electric flux leaving the surface of the cube, 0 *< x, y, z < *0*.*4: This
is where the divergence theorem really saves you time! First find

∇ · **D **= 1
*r*2

*d
*

*dr
(r*2 × 2*r) *= 6

Then the net outward flux will be ∫
*vol
*

∇ · **D ***dv *= 6*(*0*.*4*)*3 = 0*.*38 C

41

**CHAPTER 4
**

4.1. The value of **E **at *P(ρ *= 2*, φ *= 40◦*, z *= 3*) *is given as **E **= 100**a***ρ *−200**a***φ*+300**a***z *V/m. Determine
the incremental work required to move a 20*µ*C charge a distance of 6 *µ*m:

a) in the direction of **a***ρ *: The incremental work is given by *dW *= −*q ***E **· *d***L**, where in this case,
*d***L **= *dρ ***a***ρ *= 6 × 10−6 **a***ρ *. Thus

*dW *= −*(*20 × 10−6 C*)(*100 V*/*m*)(*6 × 10−6 m*) *= −12 × 10−9 J = −12 nJ

b) in the direction of **a***φ *: In this case *d***L **= 2 *dφ ***a***φ *= 6 × 10−6 **a***φ *, and so

*dW *= −*(*20 × 10−6*)(*−200*)(*6 × 10−6*) *= 2*.*4 × 10−8 J = 24 nJ

c) in the direction of **a***z*: Here, *d***L **= *dz ***a***z *= 6 × 10−6 **a***z*, and so

*dW *= −*(*20 × 10−6*)(*300*)(*6 × 10−6*) *= −3*.*6 × 10−8 J = −36 nJ

d) in the direction of **E**: Here, *d***L **= 6 × 10−6 **a***E *, where

**a***E *= 100**a***ρ *− 200**a***φ *+ 300**a***z
*[1002 + 2002 + 3002]1*/*2 = 0*.*267 **a***ρ *− 0*.*535 **a***φ *+ 0*.*802 **a***z
*

Thus

*dW *= −*(*20 × 10−6*)*[100**a***ρ *− 200**a***φ *+ 300**a***z*] · [0*.*267 **a***ρ *− 0*.*535 **a***φ *+ 0*.*802 **a***z*]*(*6 × 10−6*)
*= −44*.*9 nJ

e) In the direction of **G **= 2 **a***x *− 3 **a***y *+ 4 **a***z*: In this case, *d***L **= 6 × 10−6 **a***G*, where

**a***G *= 2**a***x *− 3**a***y *+ 4**a***z
*[22 + 32 + 42]1*/*2 = 0*.*371 **a***x *− 0*.*557 **a***y *+ 0*.*743 **a***z
*

So now

*dW *= −*(*20 × 10−6*)*[100**a***ρ *− 200**a***φ *+ 300**a***z*] · [0*.*371 **a***x *− 0*.*557 **a***y *+ 0*.*743 **a***z*]*(*6 × 10−6*)
*= −*(*20 × 10−6*) *[37*.*1*(***a***ρ *· **a***x)*− 55*.*7*(***a***ρ *· **a***y)*− 74*.*2*(***a***φ *· **a***x)*+ 111*.*4*(***a***φ *· **a***y)
*+ 222*.*9] *(*6 × 10−6*)
*

where, at *P *, *(***a***ρ *· **a***x) *= *(***a***φ *· **a***y) *= cos*(*40◦*) *= 0*.*766, *(***a***ρ *· **a***y) *= sin*(*40◦*) *= 0*.*643, and
*(***a***φ *· **a***x) *= − sin*(*40◦*) *= −0*.*643. Substituting these results in

*dW *= −*(*20 × 10−6*)*[28*.*4 − 35*.*8 + 47*.*7 + 85*.*3 + 222*.*9]*(*6 × 10−6*) *= −41*.*8 nJ

42

4.2. Let **E **= 400**a***x *− 300**a***y *+ 500**a***z *in the neighborhood of point *P(*6*, *2*,*−3*)*. Find the incremental work
done in moving a 4-C charge a distance of 1 mm in the direction specified by:

a) **a***x *+ **a***y *+ **a***z*: We write

*dW *= −*q***E **· *d***L **= −4*(*400**a***x *− 300**a***y *+ 500**a***z) *· *(***a***x *+ **a***y *+ **a***z)*√
3

*(*10−3*)
*

= − *(*4 × 10
−3*)*√

3
*(*400 − 300 + 500*) *= −1*.*39 J

b) −2**a***x *+ 3**a***y *− **a***z*: The computation is similar to that of part *a*, but we change the direction:

*dW *= −*q***E **· *d***L **= −4*(*400**a***x *− 300**a***y *+ 500**a***z) *· *(*−2**a***x *+ 3**a***y *− **a***z)*√
14

*(*10−3*)
*

= − *(*4 × 10
−3*)*√

14
*(*−800 − 900 − 500*) *= 2*.*35 J

4.3. If **E **= 120 **a***ρ *V*/*m, find the incremental amount of work done in moving a 50*µ*m charge a distance
of 2 mm from:

a) *P(*1*, *2*, *3*) *toward *Q(*2*, *1*, *4*)*: The vector along this direction will be *Q *− *P *= *(*1*,*−1*, *1*) *from
which **a***PQ *= [**a***x *− **a***y *+ **a***z*]*/
*

√ 3. We now write

*dW *= −*q***E **· *d***L **= −*(*50 × 10−6*)
*[

120**a***ρ *· *(***a***x *− **a***y *+ **a***z*√
3

]
*(*2 × 10−3*)
*

= −*(*50 × 10−6*)(*120*) *[*(***a***ρ *· **a***x)*− *(***a***ρ *· **a***y)*] 1√
3
*(*2 × 10−3*)
*

At *P *, *φ *= tan−1*(*2*/*1*) *= 63*.*4◦. Thus *(***a***ρ *· **a***x) *= cos*(*63*.*4*) *= 0*.*447 and *(***a***ρ *· **a***y) *= sin*(*63*.*4*) *=
0*.*894. Substituting these, we obtain *dW *= 3*.*1*µ*J.

b) *Q(*2*, *1*, *4*) *toward *P(*1*, *2*, *3*)*: A little thought is in order here: Note that the field has only a radial
component and does not depend on *φ *or *z*. Note also that *P *and *Q *are at the same radius (

√ 5)

from the *z *axis, but have different *φ *and *z *coordinates. We could just as well position the two
points at the same *z *location and the problem would not change. If this were so, then moving
along a straight line between *P *and *Q *would thus involve moving along a chord of a circle whose
radius is

√ 5. Halfway along this line is a point of symmetry in the field (make a sketch to see

this). This means that when starting from either point, the initial force will be the same. Thus the
answer is *dW *= 3*.*1*µ*J as in part *a*. This is also found by going through the same procedure as
in part *a*, but with the direction (roles of *P *and *Q*) reversed.

4.4. Find the amount of energy required to move a 6-C charge from the origin to *P(*3*, *1*,*−1*) *in the field
**E **= 2*x***a***x *− 3*y*2**a***y *+ 4**a***z *V/m along the straight-line path *x *= −3*z*, *y *= *x *+ 2*z*: We set up the
computation as follows, and find the the result *does not depend on the path*.

*W *= −*q
*∫

**E **· *d***L **= −6
∫
*(*2*x***a***x *− 3*y*2**a***y *+ 4**a***z) *· *(dx***a***x *+ *dy***a***y *+ *dz***a***z)
*

= −6 ∫ 3

0
2*xdx *+ 6

∫ 1 0

3*y*2*dy *− 6
∫ −1

0
4*dz *= −24 J

43

4.5. Compute the value of
∫ *P
A
*

**G **· *d***L **for **G **= 2*y***a***x *with *A(*1*,*−1*, *2*) *and *P(*2*, *1*, *2*) *using the path:
a) straight-line segments *A(*1*,*−1*, *2*) *to *B(*1*, *1*, *2*) *to *P(*2*, *1*, *2*)*: In general we would have

∫ *P
A
*

**G **· *d***L **=
∫ *P
A
*

2*y dx
*

The change in *x *occurs when moving between *B *and *P *, during which *y *= 1. Thus
∫ *P
A
*

**G **· *d***L **=
∫ *P
B
*

2*y dx *=
∫ 2

1
2*(*1*)dx *= 2

b) straight-line segments*A(*1*,*−1*, *2*) *to*C(*2*,*−1*, *2*) *to*P(*2*, *1*, *2*)*: In this case the change in *x *occurs
when moving from *A *to *C*, during which *y *= −1. Thus

∫ *P
A
*

**G **· *d***L **=
∫ *C
A
*

2*y dx *=
∫ 2

1
2*(*−1*)dx *= −2

4.6. Let **G **= 4*x***a***x*+2*z***a***y*+2*y***a***z*. Given an initial point*P(*2*, *1*, *1*) *and a final point*Q(*4*, *3*, *1*)*, find
∫

**G **·*d***L
**using the path: a) straight line: *y *= *x *− 1, *z *= 1; b) parabola: 6*y *= *x*2 + 2, *z *= 1:
With **G **as given, the line integral will be

∫
**G **· *d***L **=

∫ 4 2

4*x dx *+
∫ 3

1
2*z dy *+

∫ 1 1

2*y dz
*

Clearly, we are going nowhere in *z*, so the last integral is zero. With *z *= 1, the first two evaluate as
∫

**G **· *d***L **= 2*x*2
∣∣∣4
2
+ 2*y
*

∣∣∣3 1 = 28

The paths specified in parts *a *and *b *did not play a role, meaning that the integral between the specified
points is path-independent.

4.7. Repeat Problem 4.6 for **G **= 3*xy*3**a***x *+ 2*z***a***y *. Now things are different in that the path does matter:
a) straight line: *y *= *x *− 1, *z *= 1: We obtain:

∫
**G **· *d***L **=

∫ 4 2

3*xy*2 *dx *+
∫ 3

1
2*z dy *=

∫ 4 2

3*x(x *− 1*)*2 *dx *+
∫ 3

1
2*(*1*) dy *= 90

b) parabola: 6*y *= *x*2 + 2, *z *= 1: We obtain:
∫

**G **· *d***L **=
∫ 4

2
3*xy*2 *dx *+

∫ 3 1

2*z dy *=
∫ 4

2

1

12
*x(x*2 + 2*)*2 *dx *+

∫ 3 1

2*(*1*) dy *= 82

44

4.8. A point charge*Q*1 is located at the origin in free space. Find the work done in carrying a charge*Q*2 from:
(a) *B(rB, θB, φB) *to *C(rA, θB, φB) *with *θ *and *φ *held constant; (b) *C(rA, θB, φB) *to *D(rA, θA, φB)
*with *r *and *φ *held constant; (c) *D(rA, θA, φB) *to *A(rA, θA, φA) *with *r *and *θ *held constant: The general
expression for the work done in this instance is

*W *= −*Q*2
∫

**E **· *d***L **= −*Q*2
∫

*Q*1

4*π*0*r*2
**a***r *· *(dr***a***r *+ *rdθ***a***θ *+ *r *sin *θdφ***a***φ) *= −*Q*1*Q*2

4*π*0

∫
*dr
*

*r*2

We see that only changes in *r *will produce non-zero results. Thus for part *a *we have

*W *= −*Q*1*Q*2
4*π*0

∫ *rA
rB
*

*dr
*

*r*2
= *Q*1*Q*2

4*π*0

[ 1

*rA
*− 1

*rB
*

] J

The answers to parts *b *and *c *(involving paths over which *r *is held constant) are both 0.

4.9. A uniform surface charge density of 20 nC*/*m2 is present on the spherical surface *r *= 0*.*6 cm in free
space.

a) Find the absolute potential at *P(r *= 1 cm*, θ *= 25◦*, φ *= 50◦*)*: Since the charge density
is uniform and is spherically-symmetric, the angular coordinates do not matter. The potential
function for *r > *0*.*6 cm will be that of a point charge of *Q *= 4*πa*2*ρs *, or

*V (r) *= 4*π(*0*.*6 × 10
−2*)*2*(*20 × 10−9*)
*

4*π*0*r
*= 0*.*081

*r
*V with *r *in meters

At *r *= 1 cm, this becomes *V (r *= 1 cm*) *= 8*.*14 V
b) Find *VAB *given points *A(r *= 2 cm*, θ *= 30◦*, φ *= 60◦*) *and *B(r *= 3 cm*, θ *= 45◦*, φ *= 90◦*)*:

Again, the angles do not matter because of the spherical symmetry. We use the part *a *result to
obtain

*VAB *= *VA *− *VB *= 0*.*081
[

1

0*.*02
− 1

0*.*03

]
= 1*.*36 V

4.10. Given a surface charge density of 8 nC*/*m2 on the plane *x *= 2, a line charge density of 30 nC/m on
the line *x *= 1, *y *= 2, and a 1-*µC *point charge at *P(*−1*,*−1*, *2*)*, find *VAB *for points *A(*3*, *4*, *0*) *and
*B(*4*, *0*, *1*)*: We need to find a potential function for the combined charges. That for the point charge we
know to be

*Vp(r) *= *Q
*4*π*0*r
*

Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have

*Vl(ρ) *= −
∫

*ρl
*

2*π*0*ρ
dρ *+ *C*1 = − *ρl
*

2*π*0
ln*(ρ)*+ *C*1

For the sheet charge, we have

*Vs(x) *= −
∫

*ρs
*

20
*dx *+ *C*2 = − *ρs
*

20
*x *+ *C*2

45

4.10. (continued) The total potential function will be the sum of the three. Combining the integration con- stants, we obtain:

*V *= *Q
*4*π*0*r
*

− *ρl
*2*π*0

ln*(ρ)*− *ρs
*20

*x *+ *C
*

The terms in this expression are not referenced to a common origin, since the charges are at different
positions. The parameters *r *, *ρ*, and *x *are *scalar distances *from the charges, and will be treated
as such here. For point *A *we have *rA *=

√
*(*3 − *(*−1*))*2 + *(*4 − *(*−1*))*2 + *(*−2*)*2 = √45, *ρA *=√

*(*3 − 1*)*2 + *(*4 − 2*)*2 = √8, and its distance from the sheet charge is *xA *= 3 − 2 = 1. The potential
at *A *is then

*VA *= 10
−6

4*π*0
√

45 − 30 × 10

−9

2*π*0
ln
√

8 − 8 × 10 −9

20
*(*1*)*+ *C
*

At point *B*, *rB *=
√
*(*4 − *(*−1*))*2 + *(*0 − *(*−1*))*2 + *(*1 − 2*)*2 = √27,

*ρB *=
√
*(*4 − 1*)*2 + *(*0 − 2*)*2 = √13, and the distance from the sheet charge is *xB *= 4 − 2 = 2.

The potential at *A *is then

*VB *= 10
−6

4*π*0
√

27 − 30 × 10

−9

2*π*0
ln
√

13 − 8 × 10 −9

20
*(*2*)*+ *C
*

Then

*VA *− *VB *= 10
−6

4*π*0

[ 1√ 45

− 1√ 27

] − 30 × 10

−9

2*π*0
ln

(√ 8

13

) − 8 × 10

−9

20
*(*1 − 2*) *= 193 V

4.11. Let a uniform surface charge density of 5 nC*/*m2 be present at the *z *= 0 plane, a uniform line charge
density of 8 nC*/*m be located at *x *= 0, *z *= 4, and a point charge of 2*µ*C be present at *P(*2*, *0*, *0*)*.
If *V *= 0 at *M(*0*, *0*, *5*)*, find *V *at *N(*1*, *2*, *3*)*: We need to find a potential function for the combined
charges which is zero at *M *. That for the point charge we know to be

*Vp(r) *= *Q
*4*π*0*r
*

Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have

*Vl(ρ) *= −
∫

*ρl
*

2*π*0*ρ
dρ *+ *C*1 = − *ρl
*

2*π*0
ln*(ρ)*+ *C*1

For the sheet charge, we have

*Vs(z) *= −
∫

*ρs
*

20
*dz*+ *C*2 = − *ρs
*

20
*z*+ *C*2

The total potential function will be the sum of the three. Combining the integration constants, we obtain:

*V *= *Q
*4*π*0*r
*

− *ρl
*2*π*0

ln*(ρ)*− *ρs
*20

*z*+ *C
*

46

4.11. (continued) The terms in this expression are not referenced to a common origin, since the charges are
at different positions. The parameters *r *, *ρ*, and *z *are *scalar distances *from the charges, and will be
treated as such here. To evaluate the constant, *C*, we first look at point *M *, where *VT *= 0. At *M *,
*r *= √22 + 52 = √29, *ρ *= 1, and *z *= 5. We thus have

0 = 2 × 10 −6

4*π*0
√

29 − 8 × 10

−9

2*π*0
ln*(*1*)*− 5 × 10

−9

20
5 + *C *⇒ *C *= −1*.*93 × 103 V

At point *N *, *r *= √1 + 4 + 9 = √14, *ρ *= √2, and *z *= 3. The potential at *N *is thus

*VN *= 2 × 10
−6

4*π*0
√

14 − 8 × 10

−9

2*π*0
ln*(
*

√
2*)*− 5 × 10

−9

20
*(*3*)*− 1*.*93 × 103 = 1*.*98 × 103 V = 1*.*98 kV

4.12. Three point charges, 0*.*4*µ*C each, are located at *(*0*, *0*,*−1*)*, *(*0*, *0*, *0*)*, and *(*0*, *0*, *1*)*, in free space.
a) Find an expression for the absolute potential as a function of *z *along the line *x *= 0, *y *= 1:

From a point located at position *z *along the given line, the distances to the three charges are
*R*1 =

√
*(z*− 1*)*2 + 1, *R*2 =

√
*z*2 + 1, and *R*3 =

√
*(z*+ 1*)*2 + 1. The total potential will be

*V (z) *= *q
*4*π*0

[ 1

*R*1
+ 1

*R*2
+ 1

*R*3

]

Using *q *= 4 × 10−7 C, this becomes

*V (z) *= *(*3*.*6 × 103*)
*[

1√
*(z*− 1*)*2 + 1

+ 1√
*z*2 + 1 +

1√
*(z*+ 1*)*2 + 1

] V

b) Sketch *V (z)*. The sketch will show that *V *maximizes to a value of 8*.*68 × 103 at *z *= 0, and then
monotonically decreases with increasing |*z*| symmetrically on either side of *z *= 0.

4.13. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0.5 mm on a side in free space. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them? This will be the magnitude of the charge times the potential difference between the finishing and starting positions, or

*W *= *(*4 × 10
−12*)*2

2*π*0

[ 1

2*.*5
− 1

5

]
× 104 = 5*.*76 × 10−10 J = 576 pJ

4.14. two 6-nC point charges are located at *(*1*, *0*, *0*) *and *(*−1*, *0*, *0*) *in free space.
a) Find *V *at *P(*0*, *0*, z)*: Since the charges are positioned symmetrically about the *z *axis, the potential

at *z *will be double that from one charge. This becomes:

*V (z) *= *(*2*) q
*4*π*0

√
*z*2 + 1 =

*q
*

2*π*0
√
*z*2 + 1

b) Find *Vmax *: It is clear from the part *a *result that *V *will maximize at *z *= 0, or *vmax *= *q/(*2*π*0*) *=
108 V.

47

4.14. (continued)
c) Calculate |*dV/dz*| on the *z *axis: Differentiating the part *a *result, we find

∣∣∣*dV
dz
*

∣∣∣ = *qz
π*0*(z*2 + 1*)*3*/*2 V*/*m

d) Find |*dV/dz*|*max *: To find this we need to differentiate the part *c *result and find its zero:

*d
*

*dz
*

∣∣∣*dV
dz
*

∣∣∣ = *q(*1 − 2*z*2*)
π*0*(z*2 + 1*)*5*/*2 = 0 ⇒ *z *= ±

1√ 2

Substituting *z *= 1*/*√2 into the part *c *result, we find
∣∣∣*dV
dz
*

∣∣∣
*max
*

= *q*√
2*π*0*(*3*/*2*)*3*/*2

= 83*.*1 V*/*m

4.15. Two uniform line charges, 8 nC/m each, are located at *x *= 1, *z *= 2, and at *x *= −1, *y *= 2 in free
space. If the potential at the origin is 100 V, find *V *at *P(*4*, *1*, *3*)*: The net potential function for the two
charges would in general be:

*V *= − *ρl
*2*π*0

ln*(R*1*)*− *ρl
*2*π*0

ln*(R*2*)*+ *C
*

At the origin, *R*1 = *R*2 =
√

5, and *V *= 100 V. Thus, with *ρl *= 8 × 10−9,

100 = −2 *(*8 × 10
−9*)
*

2*π*0
ln*(
*

√
5*)*+ *C *⇒ *C *= 331*.*6 V

At *P(*4*, *1*, *3*)*, *R*1 = |*(*4*, *1*, *3*)*− *(*1*, *1*, *2*)*| =
√

10 and *R*2 = |*(*4*, *1*, *3*)*− *(*−1*, *2*, *3*)*| =
√

26. Therefore

*VP *= − *(*8 × 10
−9*)
*

2*π*0

[
ln*(
*

√
10*)*+ ln*(
*

√
26*)
*]
+ 331*.*6 = −68*.*4 V

48

4.16. Uniform surface charge densities of 6, 4, and 2 nC*/*m2 are present at *r *= 2, 4, and 6 cm, respectively,
in free space.

a) Assume *V *= 0 at infinity, and find *V (r)*. We keep in mind the definition of absolute potential
as the work done in moving a unit positive charge from infinity to location *r *. At radii outside all
three spheres, the potential will be the same as that of a point charge at the origin, whose charge
is the sum of the three sphere charges:

*V (r) (r > *6 cm*) *= *q*1 + *q*2 + *q*3
4*π*0*r
*

= [4*π(.*02*)
*2*(*6*)*+ 4*π(.*04*)*2*(*4*)*+ 4*π(.*06*)*2*(*2*)*] × 10−9

4*π*0*r
*

= *(*96 + 256 + 288*)π *× 10
−13

4*π(*8*.*85 × 10−12*)r *=
1*.*81

*r
*V where r is in meters

As the unit charge is moved inside the outer sphere to positions 4 *< r < *6 cm, the outer sphere
contribution to the energy is fixed at its value at *r *= 6. Therefore,

*V (r) (*4 *< r < *6 cm*) *= *q*1 + *q*2
4*π*0*r
*

+ *q*3
4*π*0*(.*06*)
*

= 0*.*994
*r
*

+ 13*.*6 V

In moving inside the sphere at *r *= 4 cm, the contribution from that sphere becomes fixed at its
potential function at *r *= 4:

*V (r) (*2 *< r < *4 cm*) *= *q*1
4*π*0*r
*

+ *q*2
4*π*0*(.*04*)
*

+ *q*3
4*π*0*(.*06*)
*

= 0*.*271
*r
*

+ 31*.*7 V

Finally, using the same reasoning, the potential inside the inner sphere becomes

*V (r) (r < *2 cm*) *= 0*.*271
*.*02

+ 31*.*7 = 45*.*3 V

b) Calculate *V *at *r *= 1*, *3*, *5*, *and 7 cm: Using the results of part *a*, we substitute these distances (in
meters) into the appropriate formulas to obtain: *V (*1*) *= 45*.*3 V,
*V (*3*) *= 40*.*7 V, *V (*5*) *= 33*.*5 V, and *V (*7*) *= 25*.*9 V.

c) Sketch *V *versus *r *for 0 *< r < *10 cm.

49

4.17. Uniform surface charge densities of 6 and 2 nC*/*m2 are present at *ρ *= 2 and 6 cm respectively, in free
space. Assume *V *= 0 at *ρ *= 4 cm, and calculate *V *at:

a) *ρ *= 5 cm: Since *V *= 0 at 4 cm, the potential at 5 cm will be the potential difference between
points 5 and 4:

*V*5 = −
∫ 5

4
**E **· *d***L **= −

∫ 5 4

*aρsa
*

0*ρ
dρ *= − *(.*02*)(*6 × 10

−9*)
*0

ln

( 5

4

)
= −3*.*026 V

b) *ρ *= 7 cm: Here we integrate piecewise from *ρ *= 4 to *ρ *= 7:

*V*7 = −
∫ 6

4

*aρsa
*

0*ρ
dρ *−

∫ 7 6

*(aρsa *+ *bρsb)
*0*ρ
*

*dρ
*

With the given values, this becomes

*V*7 = −
[
*(.*02*)(*6 × 10−9*)
*

0

] ln

( 6

4

)
−
[
*(.*02*)(*6 × 10−9*)*+ *(.*06*)(*2 × 10−9*)
*

0

] ln

( 7

6

)
= −9*.*678 V

4.18. A nonuniform linear charge density, *ρL *= 8*/(z*2 + 1*) *nC/m lies along the *z *axis. Find the potential at
*P(ρ *= 1*, *0*, *0*) *in free space if *V *= 0 at infinity: This last condition enables us to write the potential
at *P *as a superposition of point charge potentials. The result is the integral:

*VP *=
∫ ∞
−∞

*ρLdz
*

4*π*0*R
*

where *R *= √*z*2 + 1 is the distance from a point *z *on the *z *axis to *P *. Substituting the given charge
distribution and *R *into the integral gives us

*VP *=
∫ ∞
−∞

8 × 10−9*dz
*4*π*0*(z*2 + 1*)*3*/*2 =

2 × 10−9
*π*0

*z*√
*z*2 + 1

∣∣∣∞−∞ = 144 V

4.19. The annular surface, 1 cm *< ρ < *3 cm, *z *= 0, carries the nonuniform surface charge density *ρs *=
5*ρ *nC*/*m2. Find *V *at *P(*0*, *0*, *2 cm*) *if *V *= 0 at infinity: We use the superposition integral form:

*VP *=
∫ ∫

*ρs da
*

4*π*0|**r **− **r**′|
where **r **= *z***a***z *and **r**′ = *ρ***a***ρ *. We integrate over the surface of the annular region, with *da *= *ρ dρ dφ*.
Substituting the given values, we find

*VP *=
∫ 2*π
*

0

∫ *.*03
*.*01

*(*5 × 10−9*)ρ*2 *dρ dφ
*4*π*0

√
*ρ*2 + *z*2

Substituting *z *= *.*02, and using tables, the integral evaluates as

*VP *=
[
*(*5 × 10−9*)
*

20

] [
*ρ
*

2

√
*ρ*2 + *(.*02*)*2 − *(.*02*)
*

2

2
ln*(ρ *+

√
*ρ*2 + *(.*02*)*2*)
*

]*.*03
*.*01

= *.*081 V

50

4.20. Fig. 4.11 shows three separate charge distributions in the *z *= 0 plane in free space.
a) find the total charge for each distribution: Line charge along the *y *axis:

*Q*1 =
∫ 5

3
*π *× 10−9*dy *= 2*π *× 10−9 C = 6*.*28 nC

Line charge in an arc at radius *ρ *= 3:

*Q*2 =
∫ 70◦

10◦
*(*10−9*) *3 *dφ *= 4*.*5 × 10−9 *(*70 − 10*) *2*π
*

360
= 4*.*71 × 10−9 C = 4*.*71 nC

Sheet charge:

*Q*3 =
∫ 70◦

10◦

∫ 3*.*5
1*.*6

*(*10−9*) ρ dρ dφ *= 5*.*07 × 10−9 C = 5*.*07 nC

b) Find the potential at *P(*0*, *0*, *6*) *caused by each of the three charge distributions acting alone: Line
charge along *y *axis:

*VP*1 =
∫ 5

3

*ρLdL
*

4*π*0*R
*=
∫ 5

3

*π *× 10−9*dy
*4*π*0

√
*y*2 + 62

= 10 3

4 × 8*.*854 ln*(y *+
√
*y*2 + 62*)
*

∣∣∣5
3
= 7*.*83 V

Line charge in an arc a radius *ρ *= 3:

*VP*2 =
∫ 70◦

10◦

*(*1*.*5 × 10−9*) *3 *dφ
*4*π*0

√ 32 + 62 =

*Q*2

4*π*0
√

45
= 6*.*31 V

Sheet charge:

*VP*3 =
∫ 70◦

10◦

∫ 3*.*5
1*.*6

*(*10−9*) ρ dρ dφ
*4*π*0

√
*ρ*2 + 62

= 60 × 10 −9

4*π(*8*.*854 × 10−12
(

2*π
*

360

)∫ 3*.*5
1*.*6

*ρ dρ*√
*ρ*2 + 36

= 9*.*42
√
*ρ*2 + 36

∣∣∣3*.*5
1*.*6

= 6*.*93 V

c) Find *VP *: This will be the sum of the three results of part *b*, or

*VP *= *VP*1 + *VP *2 + *VP *3 = 7*.*83 + 6*.*31 + 6*.*93 = 21*.*1 V

4.21. Let *V *= 2*xy*2*z*3 + 3 ln*(x*2 + 2*y*2 + 3*z*2*)*V in free space. Evaluate each of the following quantities at
*P(*3*, *2*,*−1*)*:

a) *V *: Substitute *P *directly to obtain: *V *= −15*.*0 V
b) |*V *|. This will be just 15*.*0 V.
c) **E**: We have

**E
**∣∣∣
*P
*= −∇*V
*

∣∣∣
*P
*= −

[(
2*y*2*z*3 + 6*x
*

*x*2 + 2*y*2 + 3*z*2
)

**a***x *+
(

4*xyz*3 + 12*y
x*2 + 2*y*2 + 3*z*2

)
**a***y
*

+ (

6*xy*2*z*2 + 18*z
x*2 + 2*y*2 + 3*z*2

)
**a***z
*

]
*P
*

= 7*.*1**a***x *+ 22*.*8**a***y *− 71*.*1**a***z *V*/*m

51

4.21d. |**E**|*P *: taking the magnitude of the part *c *result, we find |**E**|*P *= 75*.*0 V*/*m.
e) **a***N *: By definition, this will be

**a***N
*∣∣∣
*P
*= − **E**|