Example 18-1 Charging a Sphere
A large, hollow metal sphere is electrically isolated from its surroundings, except for a wire that can carry a current
to charge the sphere. Initially, the sphere is uncharged and is at electrical potential zero. If the sphere has a radius of 0.150
m and the current is a steady 5.00 mA = 5.00 * 1026 A, how long does it take for the sphere to attain a potential of 4.00
* 105 V?
Set Up
The electric potential at the surface of a charged
sphere of radius R depends on the amount of
excess charge Q on the sphere. In Section 16-7
we learned that the electric field outside a
uniformly charged sphere is identical to that of
a particle with the same charge, ­located at the
center of the sphere. So the electric potential at
a point just outside the sphere must be given by
Equation 17-8, the equation for the potential of
a charged particle with total charge Q located
at the center of the sphere. We’ll rearrange this
equation to solve for the amount of charge
required to generate the given electric potential.
Then we’ll find the time required for this charge
to reach the sphere.
Electric potential of a charged sphere:
kQ
R
V =
(17-8)
0.150 m
Definition of current:
i =
q
t
i = 5.00 µA
(18-1)
hollow sphere
Solve
Use Equation 17-8 to solve for the final charge
on the sphere.
From Equation 17-8,
14.00 * 105 V2 10.150 m2
VR
=
k
18.99 * 109 N # m2 >C 2 2
V # C2
= 6.67 * 10-6 #
N m
Q =
Since 1 V = 1 J>C = 1 N # m>C, this becomes
V # C 2 1 N # m>C
ba
b
N#m
1V
= 6.67 * 1026 C = 6.67 mC
Q = a6.67 * 10-6
Our result Q = 6.67 mC is the amount of
charge q that must flow onto the sphere in
a time t. Use Equation 18-1 to solve for t.
Recall that 1 A = 1 C>s.
Reflect
From Equation 18-1,
q
6.67 * 10-6 C
6.67 * 10-6 C
=
=
i
5.00 * 10-6 A
5.00 * 10-6 C>s
= 1.33 s
t =
A charged sphere like this is used in a Van de Graaff generator, which you may have seen demonstrated in your physics
class. If you have, you know that when the generator is turned on to charge the sphere, the sphere can begin to throw off
sparks within a second or so. So a result for t on the order of 1 s is reasonable.