Energy needed to pull parallel plate capacitorby kishor7km Tags: capacitor, energy stored, parallel plate 

#1
Nov1711, 03:04 PM

P: 4

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 



#2
Nov1711, 03:28 PM

P: 99

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.




#3
Nov1811, 03:54 AM

P: 4

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 



#4
Nov1811, 04:29 AM

P: 939

Energy needed to pull parallel plate capacitor 



#5
Nov1811, 04:46 AM

P: 4

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. 



#6
Nov1811, 04:56 AM

P: 939

I think it is correct.
One can look at it from another angle. 



#7
Nov1811, 12:49 PM

P: 99

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. 



#8
Nov1811, 01:39 PM

P: 4

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



#9
Nov1811, 03:03 PM

P: 939




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