Respuesta :
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, Â Â Â Â Â Â Â Â Â Â Â Â k = 8.99*10^9 Â ..... 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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