Recent content by kishor7km

@ grzz and CanIExplore, thanks for your help guys. I think now I got the clear idea..thanks again for your time.

Ok. So in this case capacitance becomes 0.1C, right?? hence the energy stored in the capacitor(after pulling the plates) will be 10 times more than that of initial condition. My solution is, First case: Energy,W1 = 1/2 * C * V^2 = 1/2 * 10pF * (10kV)^2...

Hi CanIExplore, Thanks for your advice. 3. The Attempt at a Solution :: I assume, the electric field will not change if distance between plate increase, but voltage will do. Voltage, V =Ed So If d -> 10d, V=E*10d = 10V . So New Energy, W2=1/2 * C * V ^2 = (10^2)W1 . ( W1 is energy...

Homework Statement a parallel plate capacitor of 10pF is charged to 10kV ( air dielectric). It is then isolated from battery.The plates are pulled away from each other until distance is 10 times more than before. what is the energy needed to pull the plates. Homework Equations c=εA/d ...