Question

An isolated conducting sphere has a 12 cm radius. One wire carries a current of 1.0000049 A into it while another wire carries a current of 1.0000000 A out of it. How long in seconds would it take for the sphere to increase in potential by 1500 V

Answer 1

Answer:

it will take for the sphere to increase in potential by 1500 V, 503.71 s.

Explanation:

The charge on the sphere after t seconds is:

q = (1.0000049 - 1.0000000) t = 0.0000049 t

The voltage on the surface is

V = k * $\frac{q}{R}$ = k 0.0000049 t / R

solve for t

t = (R*V) / (0.0000049 k) = (0.12 * 1500) / (0.0000049 * $9e^{9}$) = 503.71 s