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23 February, 14:28

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?

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  1. 23 February, 16:08
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    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
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