Phys 203A
Homework 1 Solutions
1. (a) An Ampere is a Coulomb per second, so
84 A  h  84 C  h3600 s   3.0 10 5 C.

s 
h
(b) The change in potential energy is U = qV = (3.0  105 C)(12 V) = 3.6  106 J.
13. (a) The potential difference is
VA V B  k
q
q
k
rA
rB
 8.99 10 9 N  m2 C2 1.0  106 C 1  1   4500 V.
2.0 m 1.0 m
(b) The electric potential at each point depends only on the distance from the point charge, not
the direction, so the result is unchanged.
15. First, we observe that V (x) cannot be equal to zero for x > d. In fact V (x) is always negative
for x > d. Now we consider the two remaining regions on the x axis: x < 0 and
0 < x < d. For x < 0 the separation between q1 and a point on the x axis whose coordinate is x is
given by d1 = –x; while the corresponding separation for q2 is d2 = d – x. We set
q
q 
V(x)  k  1  2   kq  1  3   0 V
d
x d  x 
 1 d2 
to obtain x = –d/2. Similarly, for 0 < x < d we have d1 = x and d2 = d – x. Let
q
q 
V(x)  k  1  2   kq 1  3   0 V
x d  x 
d1 d2 
and solve: x = d/4.
23. We choose the zero of electric potential to be at infinity. The initial electric potential energy
U 1 of the system before the particles are brought together is therefore zero. After the system is set
up the final potential energy is found from Eq. 25-26, where we write the distance between
charges qi and q j as ri  j :
U2 

k
all pairs

kq 2
a

 1 1
1
1 1
1 
 kq 2    
  

a
a
a
a
2a
2a 

2.58q 2
2 4  
.
a
qi q j
ri  j

Thus the amount of work required to set up the system is given by

W ext  U  U 2 U 1  


kq 2
4 2 .
a
33. We take the initial potential energy of the moving electron to be zero (the usual convention
that U   0 J ). Each of the fixed electrons contributes equally to the electric potential at the
moving electron’s final position, so the final potential energy is U2  2ke2 d, where d  1.0 cm is
half the distance between the fixed electrons. The initial kinetic energy is K1  12 mev12 , where me is
the mass of an electron and v1 is the initial speed of the moving electron. The final kinetic
energy is zero. Conservation of energy yields K1  U 2 or 12 mev12  2ke2 d . Hence
48.99 10 9 N  m2 C2  1.60 10 19 C2
4ke 2
v  dm 
 3.2 10 2 m s .
0.010 m9.11 10 31 kg
e
41. We use Eq. 25-38:
V

Ex (x, y)   x   x 2.0 V / m2 x 2  3.0 V / m2  y 2   (4.0 V / m2 )x;
V

E y (x, y)  
  2.0 V/ m2 x 2  3.0 V/ m2  y 2  (6.0 V/ m2 )y .
y
y
Evaluated at x = 3.0 m and y = 2.0 m, the components of the electric field are
Ex3.0 m,2.0 m  12 V / m;
E y3.0 m,2.0 m  12 V / m.

From the components, we obtain the magnitude of E :
E  Ex2  E 2y  17 V / m.


E makes an


angle  with thepositive x axis, where
E y 

  tan1 E   135.

x

43. (a) The charge on every part of the ring is the same distance from any point P on the axis.
This distance is r  z 2  R 2 , where R is the radius of the ring and z is the distance from the center
of the ring to P. It must be stressed that in the integral that follows, this distance is a constant for
all charge elements dq. The electric potential at P is
dq
dq
V  k r  k  2

z  R2
kq
k
 dq  z 2  R 2 .
z 2  R2
(b) The z-component of the electric field is given by
Ez   V  kq  (z 2  R 2 ) 1/2
z
z
 12 kq(z 2  R 2)  3/2 (2z ) 
kqz
.
(z 2  R 2 ) 3/2
This agrees with the calculation from Section 23-7. The other components of the electric field
can not be directly calculated in this way, because we have only considered the potential for
points on the axis, but they are seen to be zero via the symmetry arguments of Section 23-7.
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