An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a current of 1.000 000 0 A out of it. How long would it take for the sphere to increase in potential by 1000 V?

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

qnet = (Iin - Iout) * t

qnet = t * (1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

V = k*qnet / r

Where, k = 8.99*10^9 ... Coulomb's constant

qnet = V*r / k

t = 1000*0.1 / (8.99*10^9 * 0.000002)

t = 5.56 ms