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Problem
- A 2-g ping-pong ball rubbed against a wool jacket acquires a net positive charge of 1 μC. Estimate the
fraction of the ball’s electrons that have been removed.
Solution
If half the ball’s mass is protons, their number (equal to the original number of electrons) is 1 g= m p
. The
number of electrons removed is 1 μC= e ,so the fraction removed is
( 1 μC= e )
( 1 g = m p
! 6 C " 1. 67 " 10
!2 4 g
!1 9 C " 1 g
!1 1
(a hundred billionth).
Problem
- Find the ratio of the electrical force between a proton and an electron to the gravitational force between
the two. Why doesn’t it matter that you aren’t told the distance between them?
Solution
At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is
stronger than the gravitational force by a factor of:
F
elec
F
grav
ke
2
r
2
r
2
Gm p
m e
9 N " m
2 /C
2 )(1.6! 10
#1 9 C)
2
#1 1
N " m
2
/kg
2
)(1. 67! 10
#2 7
kg)(9. 11! 10
#3 1
kg)
3 9
.
The spacial dependence of both forces is the same, and cancels out.
Problem
- Two charges, one twice as large as the other, are located 15 cm apart and experience a repulsive force
of 95 N. What is the magnitude of the larger charge?
Solution
The product of the charges is
q 1
q 2
= r
2
F Coulomb
= k = (0. 15 m)
2
(95 N)=(9! 10
9
N " m
2
/C
2
) = 2.38! 10
#1 0
C
2
. If one charge is twice the
other, q 1 =^2 q 2
, then
1
2
q 1
2
= 2.38! 10
"1 0
Cand q 1 =^ ±21.^8 μC.
Problem
- A proton is on the x- axis at x = 1. 6 nm.An electron is on the y - axis at y = 0. 85 nm.Find the net
force the two exert on a helium nucleus (charge + 2 e ) at the origin.
Solution
A unit vector from the proton’s position to the origin is! î ,so the Coulomb force of the proton on the
helium nucleus is F P,He
= k ( e )( 2 e )(! î )=(1.6 nm)
2
= !0. 180 î nN.(Use Equation 23-1, with q 1
for the
proton, q 2
for the helium nucleus, and the approximate values of k and e given.) A unit vector from the
electron’s position to the origin is!
j , so its force on the helium nucleus is
F
e,He
= k (! e )( 2 e )(!
j )=( 0. 85 nm)
2
= 0. 638
j nN. The net Coulomb force on the helium nucleus is the sum
of these. (The vector form of Coulomb’s law and superposition, as explained in the solution to Problems 15
and 19, provides a more general approach.)
Problem
- A proton is at the origin and an electron is at the point x = 0.41 nm, y = 0.36 nm.Find the electric
force on the proton.
Solution
The magnitude of the force is
F
p
ke
2
r
2
9 N " m
2 /C
2 )(1.6! 10
#1 9 C)
2
2
2
)! 10
#1 8
m
2
#1 0
N,
and its direction is from the proton (at r p
= 0)to the electron (at r e
= ( 0. 41 î + 0. 36
j ) nm),for an
attractive force, at an angle! = tan
" 1
(0.36=0. 41 ) = 41. 3 °to the x - axis. The vector form of Coulomb’s law,
F
p
=! ke
2
( r p
! r e
)= r p
! r e
3
(see solution of next problem) gives the same result:
F
p
9
N # m
2
/C
2
)( 1. 6 " 10
!1 9
C)
2
(! 0. 41 î! 0. 36
j )=( 0. 41
2
2
)
3 = 2
( 10
! 9
m)
2
= ( 5. 82 î + 5. 11
j ) " 10
!1 0 N.
Problem
- A charge 3 q is at the origin, and a charge! 2 q is on the positive x - axis at x = a .Where would you
place a third charge so it would experience no net electric force?
Solution
The reasoning of Example 23-3 implies that for the force on a third charge Q to be zero, it must be placed
on the
x - axis to the right of the (smaller) negative charge, i.e., at x > a .The net Coulomb force on a third charge
so placed is F x
= kQ [3 qx
! 2
! 2 q ( x! a )
! 2
],so F x
= 0 implies that 3 ( x! a )
2
= 2 x
2
,or
x
2
! 6 xa + 3 a
2
= 0. Thus, x = 3 a ± 9 a
2
! 3 a
2
= (3 ± 6 ) a .Only the solution (3 + 6 ) a = 5. 45 a is
to the right of x = a.
Problem
- In Fig. 23-39 take q 1
= 68 μC, q 2
=! 34 μC,and q 3
= 15 μC.Find the electric force on q 3
Solution
Denote the positions of the charges by r 1
j , r 2
= 2 î ,and r 3
= 2 î + 2
j (distances in meters). The vector
form of Coulomb’s law (in the solution to Problem 15) and the superposition principle give the net electric
force on q 3
as:
F
3
= F
1 3
+ F
2 3
kq 1
q 3
( r 3
! r 1
r 3
! r 1
3
kq 2
q 3
( r 3
! r 2
r 3
! r 2
3
9 N)( 15 " 10
! 6 )[( 68 " 10
! 6 )( 2 î +
j )= 5 5 + (! 34 " 10
! 6 ) 2
j = 8 ]
= ( 1. 64 î! 0. 326
j ) N,
or F 3
= F
3 x
2
2
= 1. 67 Nat an angle of! = tan
" 1
( F 3 y
= F
3 x
) = "11. 2 °to the x - axis.
Problem
- Two identical small metal spheres initially carry charges q 1
and q 2
, respectively. When they’re 1.0 m
apart they experience a 2.5-N attractive force. Then they’re brought together so charge moves from one
to the other until they have the same net charge. They’re again placed 1.0 m apart, and now they repel
with a 2.5-N force. What were the original values of q 1
and q 2
Solution
The charges initially attract, so q 1
and q 2
have opposite signs, and2.5 N =! kq 1
q 2
= 1 m
2
.When the spheres
are brought together, they share the total charge equally, each acquiring
1
2
( q 1
).The magnitude of their
repulsion is2.5 N = k
1
4
( q 1
2
= 1 m
2
.Equating these two forces, we find a quadratic equation
1
4
( q 1
2
=! q 1
q 2
, or q 1
2
q 2
2
= 0,with solutions q 1
= (! 3 ± 8 ) q 2
.Both solutions are
possible, but since 3 + 8 = (3! 8 )
! 1
,they merely represent a relabeling of the charges. Since
! q 1
q 2
= 2. 5 N " m
2
=( 9 # 10
9
N " m
2
/C
2
) = ( 16. 7 μ C)
2
, the solutions are
q 1
= ± 3 + 8 (16. 7 μC) = ±40. 2 μ C and q 2
= m40. 2 μC=( 3 + 8 ) = m6.90 μC,or the same values with
q 1
and q 2
interchanged.
Problem
- A 65- μCpoint charge is at the origin. Find the electric field at the points (a) x = 50 cm, y = 0;(b)
x = 50 cm, y = 50 cm; (c) x =! 25 cm, y = 75 cm.
Solution
The electric field from a point charge at the origin is E ( r ) = kq r ˆ= r
2
= kq r = r
3
,since r ˆ = r = r. (a) For
r = 0.5 î m
and q = 65 μ C, E = (9! 10
9
N " m
2
/C
2
)( 65 μ C) î =(0.5 m)
2
= 2.34 î MN/C.(b) At r^ =^0.^5 m^ ( î^ +^
j ),
E = ( 9! 65! 10
3
N " m
2
/C)( 0 .5 m)( î +
j )=( 0. 5 2 m)
3
= ( 827 kN/C)( î +
j ). (The field strength is
- 17 MN/C at 45°
to the x axis.) (c) When
r = (! 0. 25 î + 0. 75
j ) m, E = ( 5. 85 " 10
5
N # m
2
/C)(! 0. 25 î + 0. 75
j ) m=[(! 0. 25 )
2
2
]
3 = 2
m
3
=
(! 296 î + 888
j ) kN/C ( E = 936 kN/C, " x
Problem
- A 1.0- μCcharge and a 2.0- μCcharge are 10 cm apart, as shown in Fig. 23-41. Find a point where
the electric field is zero.
FIGURE 23 - 41 Problem 32 Solution.
Solution
The field can be zero only along the line joining the charges (the x - axis). To the left or right of both
charges, the fields due to each are in the same direction, and cannot add to zero. Between the two, a
distance x > 0 from the 1 μCcharge, the electric field is E = k [ q 1
î = x
2
(! î )=(10 cm! x )
2
],which
vanishes when 1 μC= x
2
= 2 μC=(10 cm! x )
2
, or x = 10 cm=( 2 + 1 ) = 4. 14 cm.
Problem
- A dipole lies on the y axis, and consists of an electron at (^) y = 0. 60 nmand a proton at y = !0. 60 nm.
Find the electric field (a) midway between the two charges, (b) at the point x = 2.0 nm, y = 0 ,and
(c) at the point x =! 20 nm, y = 0.
Solution
We can use the result of Example 23-6, with y replaced by x , and x by! y (or equivalently,
j^ by^ î , and
î by!
j ). Then E ( x ) = 2 kqa
j ( a
2
2
)
! 3 = 2
,where q = e = 1.6! 10
"1 9
Cand a = 0. 6 nm.(Look at
Fig. 23-18 rotated 90° CW.) The constant
2 kq = 2(9! 10
9
N " m
2
/C
2
)(1. 6! 10
#1 9
C) = (2.88 GN/C)(nm)
2
. (a) At x = 0 , E ( 0 ) = 2 kq
j = a
2
=
( 2. 88 GN/C)
j =( 0. 6 )
2
= ( 8. 00 GN/C)
j. (b) For x = 2 nm,
E = ( 2. 88 GN/C)
j ( 0. 6 )( 0. 6
2
2
)
! 3 = 2
= ( 190 MN/C)
j.
(c) At x^ =^20 nm,^ E^ =^ (^2.^88 GN/C)^
j ( 0. 6 )( 0. 6
2
2
)
! 3 = 2
= ( 216 kN/C)
j.
Problem
- The dipole moment of the water molecule is6.2! 10
"3 0
(^) C# m.What would be the separation distance
if the molecule consisted of charges ± e? (The effective charge is actually less because electrons are
shared by the oxygen and hydrogen atoms.)
Solution
The distance separating the charges of a dipole is d = p = q = 6. 2! 10
"3 0
C # m= 1. 6! 10
"1 9
C = 38. 8 pm.
Problem
- You’re 1.5 m from a charge distribution whose size is much less than 1 m. You measure an electric
field strength of 282 N/C.You move to a distance of 2.0 m and the field strength becomes 119N/C.
What is the net charge of the distribution? Hint: Don’t try to calculate the charge. Determine instead
how the field decreases with distance, and from that infer the charge.
Solution
Taking the hint, we suppose that the field strength varies with a power of the distance, E^ '^ r^
n
.Then
n
, or n = ln(282=119)=ln(0. 75 ) = !3.00.A dipole field falls off like r
! 3
,hence the net
charge is zero.
Problem
- A 30-cm-long rod carries a charge of 80 μCspread uniformly over its length. Find the electric field
strength on the rod axis, 45 cm from the end of the rod.
Solution
Applying the result of Example 23-7, at a distance a = 0. 45 mfrom the near end of the rod, we get
E = kQ = a ( a + l) = (9! 10
9
N " m
2
/C
2
)( 80 μ C)=(0. 45 m)(0.45 m + 0.30 m) = 2.13 MN/C.
Problem
- Two identical rods of length l lie on the x - axis and carry uniform charges ± Q ,as shown in Fig. 23-43.
(a) Find an expression for the electric field strength as a function of position x for points to the right of
dE x
= k ( dq ) x ( x
2
2
)
! 3 = 2
= 2 " k # xr ( x
2
2
)
! 3 = 2
dr ,which holds for x positive away from the ring’s
center. (d) Integrating from r = 0 to R , one finds E x
0
R
dE x
,or
E
x
= 2! k " x
r dr
( x
2
2 )
3 = 2 0
R
= 2! k " x
x
2
2
0
R
= 2! k "
x
x
x
( x
2
2 )
1 = 2
(Note: For x > 0, x = x and the field is E x
= 2! k "[1 # x ( x
2
2
)
1 = 2
].However, for
x < 0, x =! x and E x
= 2! k "[# 1 + x ( x
2
2
)
1 = 2
]. This is consistent with symmetry on the axis,
since E x
( x ) =! E x
(! x ).)
Problem
- A semicircular loop of radius a carries positive charge Q distributed uniformly over its length. Find the
electric field at the center of the loop (point P in Fig. 23-45). Hint: Divide the loop into charge
elements dq as shown in Fig. 23-45, and write dq in terms of the angle d !. Then integrate over !to
get the net field at P.
FIGURE 23 - 45 Problem 50 Solution.
Solution
This problem is the same as Problem 73, with! 0
= 0.Thus, E ( P ) = 2 kQ î =! a
2
.
Problem
- The electric field 22 cm from a long wire carrying a uniform line charge density is 1.9 kN/C. What will
be the field strength 38 cm from the wire?
Solution
For a very long wire (^) (l ¿ 38 cm), Example 23-9 shows that the magnitude of the radial electric field falls
off like 1 = r .Therefore, E (38 cm)= E ( 22 cm) = 22 cm= 38 cm;or E (38 cm) = (22=38)1.9 kN/C = 1.10 kN/C.
