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 1.0

Question

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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Gia Bảo 4 years 2021-07-26T16:03:43+00:00 1 Answers 120 views 0

Answers ( )

  1. thienan
    0
    2021-07-26T16:04:59+00:00
    Reply

    Answer:

    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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