Thu June 21, 2012 By: Anisha Anand

how much work is to be done to charge a 24 micro farad capacitor when potential difference across the plates = 500 v

Expert Reply
Thu June 21, 2012
Capacitance: C= 24µF
Potential Difference: V= 500V
Charge stored: Q
C= Q/V
so that Q= CV
Work done, W to move charge Q through a Potential differencr V =  QV
W= QV= CV2
   = 24 x 10-6 x 5002
   = 6 J
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