Work to Move parallel capacitor Plates

[Typhon4ever]

Hi, in my book there are two capacitor plates distance x apart (they are kept connected to a constant voltage source). These plates are moved apart to distance 3x. In order to find the work done to move the capacitors apart (using W=F dl=QE dl), my book takes the charge Q of one plate and electric field felt on that one plate and integrates it from x to 3x. Why can't we take the take the charge and electric field felt on both plates and integrate half the distance of x to 3x?

 

[Doc Al]

In order to find the work done to move the capacitors apart (using W=F dl=QE dl), my book takes the charge Q of one plate and electric field felt on that one plate and integrates it from x to 3x.

OK, but Q and E are not constant as the distance changes.

Why can't we take the take the charge and electric field felt on both plates and integrate half the distance of x to 3x?

Who says you can't?

 

[Typhon4ever]

OK, but Q and E are not constant as the distance changes.


Who says you can't?

Okay, how would I set up the integration of both plates? My book says the Q on one plate is \frac{\epsilon_0 A V}{L} and the electric field felt on that one plate is \frac{V}{2L} how would I change these to be both plates?

 

[Doc Al]

Okay, how would I set up the integration of both plates? My book says the Q on one plate is \frac{\epsilon_0 A V}{L} and the electric field felt on that one plate is \frac{V}{2L} how would I change these to be both plates?

The only thing that would change would be your variable of integration. Since you want to move each plate half the distance, for each plate the position y would relate to the distance between them by L = 2y. Then just integrate from y = x/2 to y = 3x/2. You'll need to multiply your answer by 2, of course.

Compare that to just moving one plate (like in your book). In that case y would be the total distance, so L = y. And you'd integrate from y = x to y = 3x. (And not multiply by 2.)

Each method should give the same answer for the work done.

 

[Typhon4ever]

The only thing that would change would be your variable of integration. Since you want to move each plate half the distance, for each plate the position y would relate to the distance between them by L = 2y. Then just integrate from y = x/2 to y = 3x/2. You'll need to multiply your answer by 2, of course.

Compare that to just moving one plate (like in your book). In that case y would be the total distance, so L = y. And you'd integrate from y = x to y = 3x. (And not multiply by 2.)

Each method should give the same answer for the work done.

Oh, I see. You have to focus on one plate. There is no possible way to calculate work done on both plates at once? Is that because the Q between them both is 0?

 

[Doc Al]

Oh, I see. You have to focus on one plate. There is no possible way to calculate work done on both plates at once?

I'm not sure what you mean. You are calculating the work done on both plates--it's the same on both so that's why you multiply by 2.

 

[Typhon4ever]

I'm not sure what you mean. You are calculating the work done on both plates--it's the same on both so that's why you multiply by 2.

What I meant is that is there a way to calculate the total work done without multiplying by two? Or can we only calculate the work on one plate and have to double it.

 

[Doc Al]

What I meant is that is there a way to calculate the total work done without multiplying by two? Or can we only calculate the work on one plate and have to double it.

Well, you can always do the integration twice! Once for each side.

 

[Typhon4ever]

Well, you can always do the integration twice! Once for each side.

Isn't that the same thing as multiplying by 2?

 

[Doc Al]

Isn't that the same thing as multiplying by 2?

Yep.

But since the two plates experience the same force and movement, why not take advantage of symmetry?

 

[Typhon4ever]

Thanks!