An isolated conducting sphere has a 20 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a current of 1.0000000 A out of it. How long would it take for the sphere to increase in potential by 1020 V?

11.33 ms

Explanation:

Using the formula q = It where I is current in Ampere, and t is in seconds

q = (1.000002 - 1.0000000) t

q = 0.000002t

Voltage on the surface of the sphere V = Kq / r

where K is 9.0 * 10⁹ N². m² / c² and R = 20 cm = 20 / 100 = 0.2m

V = Kq / r

substitute the value into q

V = K (0.000002t) / r

cross multiply

rV = K * 0.000002t

make t subject of the formula

t = 0.2 * 1020 / (9*10⁹ * 0.000002)

t = 204 / (18 * 10³)

t = 0.0113 s = 11.33 ms