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Capacitor+work energy theorem problem

  • #1
Krushnaraj Pandya
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Homework Statement


Plate a of a parallel-plate, air filled capacitor is connected to a spring having force
constant k, and plate b is fixed. They rest on a table top.If a charge +Q is placed on plate a and a charge −Q is placed on plate b, by how
much does the spring expand?

Homework Equations


All applicable

The Attempt at a Solution


I have already solved this and the answer is q^2/2k(epsilon)A by equating forces at equilibrium. I now wish to solve this using work energy theorem also. The work done is F*s=Q*E*s=Q^2 x/2(epsilon) where x is extension. Change in potential energy of spring is 1/2 kx^2 and change in electrostatic potential energy is -kq^2/x^2; am I wrong somewhere?
 

Answers and Replies

  • #2
Doc Al
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Using forces is the way to go. One must assume that equilibrium is achieved.

Realize that if you try to use energy methods, the work done by the field (to get to the equilibrium point) is more than the work done by the spring, so there is excess energy.

This is similar to the problem of hanging a mass on a spring. When you hang the mass, how much does the spring stretch? The equilibrium point is one answer, found by force methods. But that assumes you gently lowered the weight onto the spring. But what if you just let the weight fall? It will overshoot the equilibrium point. (And oscillate about it.)
 
  • #3
Krushnaraj Pandya
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Using forces is the way to go. One must assume that equilibrium is achieved.

Realize that if you try to use energy methods, the work done by the field (to get to the equilibrium point) is more than the work done by the spring, so there is excess energy.

This is similar to the problem of hanging a mass on a spring. When you hang the mass, how much does the spring stretch? The equilibrium point is one answer, found by force methods. But that assumes you gently lowered the weight onto the spring. But what if you just let the weight fall? It will overshoot the equilibrium point. (And oscillate about it.)
the original question mentions that charges are given to both plates very slowly, can we use energy methods now?
 
  • #4
Doc Al
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the original question mentions that charges are given to both plates very slowly, can we use energy methods now?
That is equivalent to lowering the mass onto the spring gently to the equilibrium point.

So, force balance is the way to go. Realize that the plates will always be in equilibrium if the charges are added slowly. (The force from the field is not constant.) So if you want, you can try calculating the work done and using energy methods. But setting the final forces equal is the easy way.
 
  • #5
Krushnaraj Pandya
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Alright, I probably shouldn't think of more complicated things when the answer is easily available through force. Thank you very much.
 
  • #6
Delta2
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BUT, suppose you have already calculated the extension of the spring, and then they ask you to calculate the work of the electric field during the extension process. What would you answer (I know this is a kind of silly tricked question hehe).
 
  • #7
CWatters
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  • #8
Delta2
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  • #9
Krushnaraj Pandya
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Ok fine, but wouldn't the work of the E-field also equal the increase in the potential energy of the spring?
1)I suppose work done by spring+work done by E-field=increase in potential energy of spring + decrease in potential energy of system of two plates
2)work done by spring is -0.5kx^2,
3) we don't know W by E-field but I thought it should be F*s=Q*E*s=Q^2*x/2A(epsilon) where x is extension and A is area of plate. ( since E is constant)
4)increase in potential energy of spring is 0.5kx^2
5) and decrease in potential energy- since change in electrostatic potential=E*(extension)=change in potential energy/q. Therefore decrease is E*x*q. We can put E as Q/A2(epsilon) if we so wish

Am I wrong anywhere?
 
  • #10
Doc Al
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3) we don't know W by E-field but I thought it should be F*s=Q*E*s=Q^2*x/2A(epsilon) where x is extension and A is area of plate. ( since E is constant)
The field is not constant, since charge is added slowly.
 
  • #11
Krushnaraj Pandya
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The field is not constant, since charge is added slowly.
oh, right...I totally forgot about that. So except 3) and 5), are the other parts correct. And how would we go about it now? It seems quite complicated...
 
  • #12
Krushnaraj Pandya
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Ok fine, but wouldn't the work of the E-field also equal the increase in the potential energy of the spring?
How did you get this?
 
  • #14
Krushnaraj Pandya
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BUT, suppose you have already calculated the extension of the spring, and then they ask you to calculate the work of the electric field during the extension process. What would you answer (I know this is a kind of silly tricked question hehe).
I guess I'd write the equations I numbered, except 3) and 5) since they're apparently wrong. I don't know how else we can do this...
 
  • #15
Delta2
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The way I understand your 1) step above is -W+W=W-W=0, that is since the charge is added slowly, the force from the spring is at any instant of time equal and opposite to the force of the electric field. So the work of electric field is minus the work of the force of the spring that is ##-1/2kx^2##.
And also the increase in potential energy of the spring equals minus the work of the force of the spring, and the decrease of the potential energy of the plates equals the work of the e-field.
 
  • #16
Krushnaraj Pandya
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The way I understand your 1) step above is -W+W=W-W=0, that is since the charge is added slowly, the force from the spring is at any instant of time equal and opposite to the force of the electric field. So the work of electric field is minus the work of the force of the spring that is ##-1/2kx^2##.
And also the increase in potential energy of the spring equals minus the work of the force of the spring, and the decrease of the potential energy of the plates equals the work of the e-field.
Ohh, I see
Ok fine, but wouldn't the work of the E-field also equal the increase in the potential energy of the spring?
that'd imply this and it'd also imply ##-1/2kx^2##. is equal to change in electric potential energy...how can we write the RHS of the equation?
...
 
  • #17
Krushnaraj Pandya
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Ohh, I see

that'd imply this and it'd also imply ##-1/2kx^2##. is equal to change in electric potential energy...how can we write the RHS of the equation?
...
is it just the energy stored by capacitor = 1/2 QV?
 
  • #18
Delta2
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  • #19
Krushnaraj Pandya
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Yes!.
ohh, ok! I got it, thank you very much for expanding my knowledge by twisting the question...
 

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