1. The problem statement, all variables and given/known data 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. 2. Relevant equations c=εA/d ; c=Q/V ; 3. The attempt at a solution
You will get more help if you show some attempt at the problem, or at least give us an idea of what part you are struggling with. People don't want to just give you the answers to your homework.
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 stored before modification) Is this correct??? What will happen to capacitor,C if the distance(d) increases?? Whether both capacitance and voltage will change?? I know C=Q/V ; So how to calculate energy in this case?? I am little confused... Please help
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 = 0.0005 joules Second case: after plates are pulled away 10 times Energy,W2 =1/2 * (0.1*10pF) * (10*10kV)^2 = 0.005 joules So the energy required to pull the plates is = W2 - W1 = 0.0045 joules. Is it correct??? or am I missing something?? because this answer is not there in available choices.
This is exactly right. Well you've already stated that the voltage will change. You just found how much it changes when the distance is increased in the previous part. Then according to the definition of capacitance which you have written there, how does the capacitance change if the voltage changes? To find the work, try writing your work equation in terms of just voltage and charge, or just capacitance and charge rather than having both of them in your equation at once.
@ grzz and CanIExplore, thanks for your help guys. I think now I got the clear idea..thanks again for your time.
Exactly. That was the other point of view that I had in mind. It is very conventient to have only one variable to deal with. Hence it is profitable to involve the charge because since the capacitor is isolated this charge remains constant.