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Chapter 2
Electric Fields
2.1 The Important Stuff
2.1.1 The Electric Field
Suppose we have a point charge q 0 located at r and a set of external charges conspire so as to exert a force F on this charge. We can define the electric field at the point r by:
E =
F
q 0
The (vector) value of the E field depends only on the values and locations of the external charges, because from Coulomb’s law the force on any “test charge” q 0 is proportional to the value of the charge. However to make this definition really kosher we have to stipulate that the test charge q 0 is “small”; otherwise its presence will significantly influence the locations of the external charges. Turning Eq. 2.1 around, we can say that if the electric field at some point r has the value E then a small charge placed at r will experience a force
F = q 0 E (2.2)
The electric field is a vector. From Eq. 2.1 we can see that its SI units must be N C. It follows from Coulomb’s law that the electric field at point r due to a charge q located at the origin is given by
E = k
q r^2
ˆr (2.3)
where ˆr is the unit vector which points in the same direction as r.
2.1.2 Electric Fields from Particular Charge Distributions
An electric dipole is a pair of charges of opposite sign (±q) separated by a distance d which is usually meant to be small compared to the distance from the charges at which we
18 CHAPTER 2. ELECTRIC FIELDS
E
q
r
(a)
E
q
r
(b)
Figure 2.1: The E field due to a point charge q. (a) If the charge q is positive, the E field at some point a distance r away has magnitude k|q|/r^2 and points away from the charge. (b) If the charge q is negative, the E field has magnitude k|q|/r^2 and points toward the charge.
want to find the electric field. The product qd turns out to be important; the vector which points from the −q charge to the +q charge and has magnitude qd is known as the electric dipole moment for the pair, and is denoted p. Suppose we form an electric dipole by placing a charge +q at (0, 0 , d/2) and a charge −q at (0, 0 , −d/2). (So the dipole moment p has magnitude p = qd and points in the +k direction.) One can show that when z is much larger than d, the electric field for points on the z axis is
Ez =
2 π 0
p z^3
= k
2 qd z^3
A linear charge distribution is characterized by its charger per unit length. Linear charge density is usually given the symbol λ; for an arclength ds of the distribution, the electric charge is dq = λds
For a ring of charge with radius R and total charge q, for a point on the axis of the ring a distance z from the center, the magnitude of the electric field (which points along the z axis) is
E =
qz 4 π 0 (z^2 + r^2 )^3 /^2
- Charged Disk & Infinite Sheet
A two-dimensional (surface) distribution of charge is characterized by its charge per unit area. Surface charge density is usually given the symbol σ; for an area element dA of the distribution, the electric charge is dq = σdA
For a disk or radius R and uniform charge density σ on its surface, for a point on the axis of the disk at a distance z away from the center, the magnitude of the electric field (which points along the z axis) is
E =
σ 2 0
( 1 −
z √ z^2 + r^2
) (2.6)
20 CHAPTER 2. ELECTRIC FIELDS
mg
F elec = qE
E
q = 24 mC
Figure 2.2: Forces acting on the charged mass in Example 1.
- An object having a net charge of 24 μC is placed in a uniform electric field of 610 N C directed vertically. What is the mass of this object if it “floats” in the field? [Ser4 23-16]
The forces acting on the mass are shown in Fig. 2.2. The force of gravity points downward and has magnitude mg (m is the mass of the object) and the electrical force acting on the mass has magnitude F = |q|E, where q is the charge of the object and E is the magnitude of the electric field. The object “floats”, so the net force is zero. This gives us:
|q|E = mg
Solve for m:
m =
|q|E g
(24 × 10 −^6 C)(610 N C )
(9. (^80) sm 2 )
= 1. 5 × 10 −^3 kg
The mass of the object is 1. 5 × 10 −^3 kg = 1.5 g.
- An electron is released from rest in a uniform electric of magnitude 2. 00 × 10 4 N C. Calculate the acceleration of the electron. (Ignore gravitation.) [HRW6 23-29]
The magnitude of the force on a charge q in an electric field is given by F = |qE|, where E is the magnitude of the field. The magnitude of the electron’s charge is e = 1. 602 × 10 −^19 C, so the magnitude of the force on the electron is
F = |qE| = (1. 602 × 10 −^19 C)(2. 00 × 10 4 N C ) = 3. 20 × 10 −^15 N
Newton’s 2nd^ law relates the magnitudes of the force and acceleration: F = ma, so the acceleration of the electron has magnitude
a =
F
m
(3. 20 × 10 −^15 N)
(9. 11 × 10 −^31 kg)
= 3. 51 × 10 15 m s 2
That’s the magnitude of the electron’s acceleration. Since the electron has a negative charge the direction of the force on the electron (and also the acceleration) is opposite the direction of the electric field.
2.2. WORKED EXAMPLES 21
+2.0q (^) +q
+q
a
a
P
Figure 2.3: Charge configuration for Example 4.
- What is the magnitude of a point charge that would create an electric field of
- 00 N C at points 1 .00 m away? [HRW6 23-4]
From Eq. 2.3, the magnitude of the E field due to a point charge q at a distance r is given by
E = k
|q| r^2 Here we are given E and r, so we can solve for |q|:
|q| =
Er^2 k
(1. 00 N C )(1.00 m)^2 (
- 99 × 10 9 N·m 2 C^2
)
= 1. 11 × 10 −^10 C
The magnitude of the charge is 1. 11 × 10 −^10 C.
- Calculate the direction and magnitude of the electric field at point P in Fig. 2.3, due to the three point charges. [HRW6 23-12]
Since each of the three charges is positive they give electric fields at P pointing away from the charges. This is shown in Fig. 2.4, where the charges are individually numbered along with their (vector!) E–field contributions. We note that charges 1 and 2 have the same magnitude and are both at the same distance from P. So the E–field vectors for these charges shown in Fig. 2.4(being in opposite directions) must cancel. So we are left with only the contribution from charge 3. We know the direction for this vector; it is 45◦^ above the x axis. To find its magnitude we note that the distance of this charge from P is half the length of the square’s diagonal, or: r = 12 (
2 a) =
a √ 2
and so the magnitude is
E 3 = k 2 q r^2
2 kq (a/
4 kq a^2
2.2. WORKED EXAMPLES 23
+
- -
-q +
+q -2.0q
+2.0q
a
a
a a
r
+
- -
-q +
+q -2.0q
+2.0q
a
a
a a
+
- -
-q +
+q -2.0q
+2.0q
a
a
a a
(a) (^) (b) (c)
+
- -
-q +
+q -2.0q
+2.0q
a
a
a a
(d)
Figure 2.6: Directions of E field at the center of the square due to three of the corner charges. (a) Upper left charge is at distance r = a/
√ 2 from the center (as are the other charges). E field due to this charge points away from charge, in − 45 ◦^ direction. (b) E field due to upper right charge points toward charge, in +45◦^ direction. (c) E field due to lower left charge points toward charge, in +225◦^ direction. (d) E field due to lower left charge points away from charge, in +135◦^ direction.
The directions of the contributions to the total E field are shown in Fig. 2.6(a)–(d). The E field due to the upper left charge points away from charge, which is in − 45 ◦^ direction (as measured from the +x axis, as usual. The E field due to upper right charge points toward the charge, in +45◦^ direction. The E field due to lower left charge points toward that charge, in 180◦^ + 45◦^ = +225◦^ direction. Finally, E field due to lower right charge points away from charge, in 180◦^ − 45 ◦^ = +135◦^ direction. So we now have the magnitudes and directions of four vectors. Can we add them together? Sure we can!
ETotal = (7. 13 × 10 4 N C )(cos(− 45 ◦)i + sin(− 45 ◦)j)
- (1. 43 × 10 5 N C )(cos(+45◦)i + sin(45◦)j)
- (7. 13 × 10 4 N C )(cos(225◦)i + sin(225◦)j)
- (1. 43 × 10 5 N C )(cos(+135◦)i + sin(135◦)j)
(I know, this is the clumsy way of doing it, but I’ll get to that.) The sum gives:
ETotal = 0. 0 i + (1. 02 × 10 5 N C )j
So the magnitude of ETotal is 1. 02 × 10 5 N C and it points in the +y direction. This particular problem can be made easier by noting the cancellation of the E’s con- tributed by the charges on opposite corners of the square. For example, a +q charge in the upper left and a +2. 0 q charge in the lower right is equivalent to a single charge +q in the lower right (as far as this problem is concerned).
2.2.2 Electric Fields from Particular Charge Distributions
- Electric Quadrupole Fig. 2.7 shows an electric quadrupole. It consists of two dipole moments that are equal in magnitude but opposite in direction. Show that the value of E on the axis of the quadrupole for points a distance z from its center (assume z d) is given by
E =
3 Q
4 π 0 z^4
24 CHAPTER 2. ELECTRIC FIELDS
+q -q -q +q^ P
z
d d
- p +p
Figure 2.7: Charges forming electric quadrupole in Example 6.
in which Q (defined by Q ≡ 2 qd^2 ) is known as the quadrupole moment of the charge distribution. [HRW6 23-17]
We note that as the problem is given we really have three separate charges in this con- figuration: A charge − 2 q at the origin, a charge +q at z = −d and a charge +q at z = +d. (Again, see Fig. 2.7.) We are assuming that q is positive; for now let us also assume that the point P (for which we want the electric field) is located on the z axis at some positive value of z, as indicated in Fig. 2.7. We will now find the contribution to the electric field at P for each of the three charges. The center charge (− 2 q) lies at a distance z from the point P. So then the magnitude of the E field due to this charge is k (^2) z 2 q , but since the charge is negative the field points toward the charge, which in this case in the −z direction. So then the contribution Ez (at point P ) by the center charge is
E(center) z = −k
2 q z^2 The charge on the left (+q) lies at a distance z + d from the point P. So then the magnitude of the E field due to this charge is k (^) (z+qd) 2. Since this charge is positive, this field points away from the charge, namely in the +z direction. So the contribution to Ez by the left charge is
E z(left) = +k
q (z + d)^2 The charge on the right (+q) lies at a distance z − d from the point P. The magnitude of the field due to this charge is k (^) (z−qd) 2 and this charge is also positive so we get a contribution
E z(right) = +k
q (z − d)^2
The field at point P is the sum of the three contributions. We factor out kq from each term to get:
Ez = kq
[ −
z^2
(z + d)^2
(z − d)^2
] .
At this point we are really done with the physics of the problem; the rest of the work is doing the mathematical steps to get a simpler (approximate) expression for Ez.
26 CHAPTER 2. ELECTRIC FIELDS
0
-e m
z
q
R
Figure 2.8: Electron oscillates on z axis through center of charged ring or radius R and total charge q, as in Example 7.
- An electron is constrained to the central axis of the ring of charge with radius R and total charge q. Show that the electrostatic force exerted on the electron can cause it to oscillate through the center of the ring with an angular frequency
ω =
√ eq 4 π 0 mR^3
where m is the electron’s mass. [HRW6 23-19]
A picture of this problem is given in Fig. 2.8; included is the electron which oscillates through the center. For this to happen, the charge q must be positive so that the electron is always attracted back to the ring (i.e. the force is restoring.) From Eq. 2.5 we have the magnitude of the E field for points on the axis of the ring (which we will call the z axis, with its origin at the center of the ring):
E =
q|z| 4 π 0 (z^2 + R^2 )^3 /^2
Note, we need an absolute value sign on the coordinate z to get a positive magnitude. Now, the E field points along the ±z axis; when z is positive it goes in the +z direction and when z is negative it goes in the −z direction. So the z component of the electric field which the electron “sees” at coordinate z is in fact
Ez =
qz 4 π 0 (z^2 + R^2 )^3 /^2 From Eq. 2.2 we get force on the electron as it moves on the z axis:
Fz = (−e)Ez =
−eqz 4 π 0 (z^2 + R^2 )^3 /^2
2.2. WORKED EXAMPLES 27
This expression for the force on the electron is rather messy; it is a restoring force since its direction is opposite the displacement of the electron from the center, but it is not linear , that is, it is not simply proportional to z. (Our work on harmonic motion assumed a linear restoring force.) We will assume that the oscillations are small in the sense that the maximum value of |z| is much smaller than R. If that is true, then in the denominator of Eq. 2.11 we can approximate:
(z^2 + R^2 )^3 /^2 ≈
( R^2
) 3 / 2 = R^3
Making this replacement in Eq. 2.11, we find:
Fz ≈
−eqz 4 π 0 R^3
( (^) eq
4 π 0 R^3
) z
This is a force “law” which just like the Hooke’s Law, Fz = −kz with the role of the spring constant k being played by
k ⇔
eq 4 π 0 R^3
By analogy with the harmonic motion of a mass m on the end of a spring of force
constant k, where the result for the angular frequency was ω =
√ k m , using Eq. 2.12 we find that angular frequency for the electron’s motion is
ω =
√ eq 4 π 0 R^3 m
- A thin nonconducting rod of finite length L has a charge q spread uniformly along it. Show that the magnitude E of the electric field at point P on the perpendicular bisector of the rod (see Fig. 2.9) is given by
E =
q 2 π 0 y
(L^2 + 4y^2 )^1 /^2
[HRW6 23-24]
We first set up a coordinate system with which to do our calculation. Let the origin be at the center of the rod and let the x axis extend along the rod. In this system, the point P is located at (0, y) and the ends of the rod are at (−L/ 2 , 0) and (+L/ 2 , 0). If the charge q is spread uniformly over the rod, then it has a linear charge density of
λ =
q L
Then if we take a section of the rod of length dx, it will contain a charge λdx. Next, we consider how a tiny bit of the rod will contribute to the electric field at P. This is shown in Fig. 2.10. We consider a piece of the rod of length dx, centered at the coordinate
2.2. WORKED EXAMPLES 29
x. This element is small enough that we can treat it as a point charge... and we know how to find the electric field due to a point charge. The distance of this small piece from the point P is
r =
√ x^2 + y^2
and the amount of electric charge contained in the piece is
dq = λdx
Therefore the electric charge in this little bit of the rod gives an electric field of magnitude
dE = k
dq r^2
kλdx (x^2 + y^2 )
This little bit of electric field dE points at an angle θ away from the y axis, as shown in Fig. 2.10. Eventually we will have to add up all the little bits of electric field dE due to each little bit of the rod. In doing so we will be adding vectors and so we will need the components of the dE’s. As can be seen from Fig. 2.10, the components are:
dEx = dE sin θ = −
kλdx (x^2 + y^2 )
sin θ and dEy = cos θ dE =
kλdx (x^2 + y^2 )
cos θ (2.13)
Also, some basic trigonometry gives us:
sin θ =
x r
x (x^2 + y^2 )^1 /^2
cos θ =
y r
y (x^2 + y^2 )^1 /^2
Using these in Eqs. 2.13 gives:
dEx = −
kxλdx (x^2 + y^2 )^3 /^2
dEy =
kyλdx (x^2 + y^2 )^3 /^2
The next step is to add up all the individual dEx’s and dEy ’s. The result for the sum of the dEx’s is easy: It must be zero! By considering all of the little bits of the rod, we can see that dEx will be positive just as often as it is negative, and when the sum is taken the result is zero. (We sometimes say that this result follows “from symmetry”.) The same is not true for the dEy ’s; they are always positive and we will have to do some work to add them up. Eq. 2.14 gives the contribution to Ey arising from an element of length dx centered on x. The bits of the rod extend from x = −L/2 to x = +L/2, so to get the sum of all the little bits we do the integral:
Ey =
∫
rod
dEy =
∫ (^) +L/ 2
−L/ 2
kyλdx (x^2 + y^2 )^3 /^2
Eq. 2.15 gives the result for the E field at P (which is what the problem asks for) so we are now done with the physics of the problem. All that remains is some mathematics to work out the integral in Eq. 2.15.
30 CHAPTER 2. ELECTRIC FIELDS
First, since k, λ and y are constants as far as the integral is concerned, they can be taken outside the integral sign:
Ey = kyλ
∫ (^) +L/ 2
−L/ 2
(x^2 + y^2 )^3 /^2
dx
The integral here is not difficult; it can be looked up in a table or evaluated by computer. We find:
Ey = kλy
x y^2
x^2 + y^2
∣∣ ∣∣ ∣
− L 2
Evaluate!
Ey = kλy
L/ 2
y^2
( L^2 4 +^ y
2
) 1 / 2 −^
(−L/2)
y^2
( L^2 4 +^ y
2
) 1 / 2
= kλy
L
y^2
(L 2 4 +^ y 2
) 1 / 2
We can cancel a factor of y; also, make the replacements k = (^4) π^10 and λ = q/L. This gives:
Ey =
4 π 0
q L
L
y
( L^2 4 +^ y
2
) 1 / 2
q 4 π 0
y
( L^2 4 +^ y
2
) 1 / 2
We’re getting close! We can make the expression look a little neater by pulling a factor of 14 out of the parentheses in the denominator. Use:
( L^2 4
) 1 / 2
( L^2 + 4y^2
) 1 / 2
in our last expression and finally get:
Ey =
q 4 π 0
y (L^2 + 4y^2 )^1 /^2
=
q 2 π 0 y
(L^2 + 4y^2 )^1 /^2
Since at point P , E has no x component, the magnitude of the E field is
E =
q 2 π 0 y
(L^2 + 4y^2 )^1 /^2
(and the field points in the +y direction for positive q.)
32 CHAPTER 2. ELECTRIC FIELDS
Since the electron starts from rest (v 0 = 0), we have v = at and so the magnitude of its velocity 48 ns after being released is
v = at = (9. 13 × 10 13 m s 2 )(48 × 10 −^9 s) = 4. 4 × 10 6 m s.
So the final speed of the electron is 4. 4 × 10 6 m s. We do a similar calculation for the proton; the only difference is its larger mass. Since the magnitude of the proton’s charge is the same as that of the electron, the magnitude of the force will be the same: F = 8. 32 × 10 −^17 N
But as the proton mass is 1. 67 × 10 −^27 kg, its acceleration has magnitude
a =
F
m
(8. 32 × 10 −^17 N)
(1. 67 × 10 −^27 kg)
= 4. 98 × 10 10 m s 2
And then the magnitude of the velocity 48 ns after being released is
v = at = (4. 98 × 10 10 m s 2 )(48 × 10 −^9 s) = 2. 4 × 10 3 m s.
So the proton’s final speed is 2. 4 × 10 3 m s.
- Beams of high–speed protons can be produced in “guns” using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun’s electric field were 2. 00 × 10 4 N C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1 .00 cm? [HRW6 23-35]
(a) The proton has charge +e = 1. 60 × 10 −^19 C so in the given (uniform) electric field, the force on the protons has magnitude
F = |q|E = eE = (1. 60 × 10 −^19 C)(2. 00 × 10 4 N C ) = 3. 20 × 10 −^15 N
Then we use Newton’s second law to get the magnitude of the protons’ acceleration. Using the mass of the proton, mp = 1. 67 × 10 −^27 kg,
a =
F
mp
(3. 20 × 10 −^15 N)
(1. 67 × 10 −^27 kg)
= 1. 92 × 10 12 m s 2
(b) The protons start from rest (this is assumed) and move in one dimension, accelerating with ax = 1. 92 × 10 12 m s 2. If they move through a displacement x − x 0 = 1.00 cm, then one of our favorite equations of one–dimensional kinematics gives us:
v^2 x = v^2 x 0 + 2ax(x − x 0 ) = 0^2 + 2(1. 92 × 10 12 m s 2 )(1. 00 × 10 −^2 m) = 3. 83 × 10 10 m
2 s^2
so that the final velocity (and speed) is
vx = 1. 96 × 10 5 m s
2.2. WORKED EXAMPLES 33
Mg
Felec
E
E = 462 N/C
Figure 2.12: Forces acting on the water drop in Example 12.
- A spherical water drop 1. 20 μm in diameter is suspended in calm air owing to a downward–directed atmospheric electric field E = 462 N C. (a) What is the weight of the drop? (b) How many excess electrons does it have? [HRW5 23-49]
(a) Not much physics for this part. The volume of the drop is
V = 43 πR^3 = 43 π(1. 20 × 10 −^6 ) m^3 = 9. 05 × 10 −^19 m^3
Assuming that the density of the drop is the usual density of water, ρ = 1. 00 × (^103) mkg 3 , we can get the mass of the drop from M = ρV. Then the weight of the drop is
W = Mg = ρV g = (1. 00 × (^103) mkg 3 )(9. 05 × 10 −^19 m^3 )(9. 80 m s 2 ) = 8. 87 × 10 −^15 N
(b) The forces acting on the water drop are as shown in Fig. 2.12, namely gravity with mag- nitude Mg directed downward and the electric force with magnitude Felec = |q|E, directed upward. Here, E is the magnitude of the electric field and |q| is the magnitude of the charge on the drop. The net force on the drop is zero, and so this allows us to solve for |q|:
Felec = |q|E = Mg =⇒ |q| =
Mg E
Plug in the weight W = Mg found in part (a) and the given value of the E:
|q| =
Mg E
(8. 87 × 10 −^15 N)
(462 N C )
= 1. 92 × 10 −^17 C
We know that the drop has a negative charge (electric field points down, but the electric force points up) so that the charge on the drop is
q = − 1. 92 × 10 −^17 C.
2.2. WORKED EXAMPLES 35
Newton’s second law the magnitude of the proton’s acceleration is (^) meEp , where mp is the mass of the proton. Then the x−acceleration of the proton is
ax,p = eE mp
and since the proton starts from rest, its position is given by
xp = (^12)
eE mp
t^2 (2.16)
The charge of the electron is −e. Then the force on the electron will have magnitude eE and point in the −x direction. The magnitude of the electron’s acceleration is eE me , where me is the mass of the electron, but the electron’s acceleration points in the −x direction. Then the x−acceleration of the electron is
ax,e = −
eE mp
and since the electron starts from rest and is initially at x = D, its position is given by
xe = D − (^12)
eE me
t^2 (2.17)
Clearly, there is some time at which xp and xe are equal. This happens when
1 2
eE mp
t^2 = D − (^12)
eE me
t^2
A little bit of algebra will allow us to solve for t. Regroup some terms:
1 2
eE mp
t^2 + (^12)
eE me
t^2 = D
1 2 eEt
2
( 1 mp
me
) = D
t^2 =
2 D
eE
( 1 mp
me
)− 1 (2.18)
Taking the square root to get t is not necessary because we want to plug t back into either one of the equations for the coordinates to find the value of x at which the meeting occurred, and both of those equations contain t^2. Putting 2.18 into 2.16 we find:
xmeet = (^12)
eE mp
t^2 meet
eE mp
2 D
eE
( 1 mp
me
)− 1
36 CHAPTER 2. ELECTRIC FIELDS
Then we keep doing algebra to get a beautiful, simple form!
xmeet =
D
mp
( 1 mp
me
)− 1
D
mp
( mpme (me + mp)
)
Dme (me + mp)
Now just plug numbers into Eq. 2.19. We are given D, and we also need:
mp = 1. 67 × 10 −^27 kg me = 9. 11 × 10 −^31 kg
These give:
D =
(5. 0 × 10 −^2 m)(9. 11 × 10 −^31 kg) (9. 11 × 10 −^31 kg + 1. 67 × 10 −^27 kg) = 2. 73 × 10 −^5 m = 27 μm
As for the concluding question... We note that the final answer did not depend on the value of the electric field strength E (a good thing, since it wasn’t given). We can think about why this happened: Since at the meeting place both particles had been traveling for the same amount of time, the distances traveled by each are proportional to their accelerations. If we change both accelerations by the same factor, the meeting point will occur at the same place because the ratio of distances traveled by each particle is the same. So applying the same scale factor to both ap and ae does not change the answer. But changing the value of the electric field does apply the same scale factor to both accelerations; since this will not change the answer, we don’t expect to see E show up in the expression for xmeet. (That was long–winded... do you have a simpler way to see this?)
- The electrons in a particle beam each have a kinetic energy of 1. 60 × 10 −^17 J. What are the magnitude and direction of the electric field that will stop these electrons in a distance of 10 .0 cm? [Ser4 23-47]
Let’s think about the direction first. The acceleration of the electrons must be directly opposite their initial (beam) velocities in order for them to come to a halt. So the force on them is also opposite the beam direction. From the vector equation F = qE we see that if the charge q is negative —as it is for an electron— then the force and electric field have opposite directions. So the electric field must point in the same direction as the initial velocities of the electrons (the beam direction). See Fig. 2.15. It is easiest to use the work–energy theorem to solve the problem. Recall:
Wnet = ∆K
