An isolated conducting sphere has a 20 cm radius. one wire carries a current of 1.20 a into it. another wire carries a current of 1. 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

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