Inserting a charged plate into a capacitor

[Kaguro]

Homework Statement

A parallel-plate capacitor is connected to a cell. Its plates A and B have charges +Q and -Q respectively. A third place C, identical to A and B and having charge +Q is inserted midway in between A and B,parallel to them. Which of the following are correct?
(1)The charge on the inner face of B is now -3Q/2
(2)There is no change in the potential difference between A and B.
(3)The potential difference between A and C is one-third of the potential difference between B and C.
(4) The charge on the inner face of A is now Q/2.

This is a Multiple Select Question so more than one option maybe correct.

Relevant Equations

Q=CV maybe for the capacitor...

Potential depends on the charge contained by the conducting plate. So the plate C should change the electric field and hence potential on both plates A and B. This should change the absolute value of potential,but since A and B are still connected to that cell, I think the potential difference between them shouldn't change. So maybe (2) is correct. What do you think?

 

[TSny]

since A and B are still connected to that cell, I think the potential difference between them shouldn't change. So maybe (2) is correct. What do you think?

That's correct. The cell maintains a constant potential difference between plates A and B.

How about statements (1), (3), and (4)?

 

[Kaguro]

That's correct. The cell maintains a constant potential difference between plates A and B.

How about statements (1), (3), and (4)?

Then I think, since this a parallel plate capacitor, E=$\frac{\sigma}{2\epsilon_0}$ away from charge. Since plates A and C have same charge,and same area, they should emit exact same electric field in exactly opposite direction. So net electric field between A and C is $\frac{\sigma}{2\epsilon_0}$+$\frac{\sigma}{2\epsilon_0}$-$\frac{\sigma}{2\epsilon_0}$=$\frac{\sigma}{2\epsilon_0}$ from A to B direction. And in the region between C and B,there should be $\frac{3\sigma}{2\epsilon_0}$ Since potential is $E \Delta x$ So potential difference between A and C is 1/3 of that of between C and B. So (3) should be correct as well.

I have no idea how will the induced charges be...

 

[TSny]

Bear with me. I'm not sure I'm following your argument.

Then I think, since this a parallel plate capacitor, E=$\frac{\sigma}{2\epsilon_0}$ away from charge. Since plates A and C have same charge,and same area, they should emit exact same electric field in exactly opposite direction.

Why do you say that plate A has the same charge as plate C?
Plate C has charge Q. Before plate C is inserted, plate A also has charge Q. How do you know whether or not the battery supplies some charge to plate A while C is inserted? If the battery does supply charge to A while C is inserted, then plate A would no longer have the same charge as C.

 

[TSny]

Suppose you let $\rm q_A$ denote the amount of charge on the inside surface of plate A.

It's interesting to see how far one can go in deducing the charges on the other surfaces in terms of $\rm q_A$. For example, can you express the amount of charge on the upper surface of plate C in terms of $\rm q_A$?

 

[Kaguro]

Okay, I somewhat solved it.

The conductors can't have charges inside. So all must be on surface. Because they're also equipotential, C has Q/2 on top part of it and Q/2 on bottom part. A has Q on bottom part and B has -Q at top part. These are all free charges.

Now, Q/2 of C induces -Q/2 on A. So net charge on A is Q/2.

Similarly Q/2 of X induces -Q/2 on B. So net charge on B is -3Q/2.

So options 1 and 4 are correct.

Now, free charge on A will induce -Q on A side of C. So net charge on that side is -Q/2

Now calculating charges induced on C:

Upper part of C. A induces -Q. This sends +Q to bottom part of C.
Bottom part of C, B induces +Q. This sends -Q to upper part of C.

So finally C has Q/2-Q-Q=-3Q/2 on top part
And Q/2+Q+Q=5Q/2 on bottom part.

Only free charges may induce charge on other bodies.
I have all the induced charges, now tell me how do I calculate the potential difference?

 

[Kaguro]

Should I just find E field due to each portion of each plate and find resultant field in the two regions?

That says the potential difference between A and C is just half of that between C and B.

So option 3 should be wrong.

 

[TSny]

Now, Q/2 of C induces -Q/2 on A. So net charge on A is Q/2.

Similarly Q/2 of X induces -Q/2 on B. So net charge on B is -3Q/2.

So options 1 and 4 are correct.

------------

So finally C has Q/2-Q-Q=-3Q/2 on top part

I can't follow your reasoning here. Some of your conclusions are correct while others are incorrect.
You conclude, I think, that the lower surface of A ends up with charge Q/2 while the upper surface of C ends up with charge -3Q/2. See the part of your post that I quoted above in green.

Have you studied Gauss' law? Or, have you studied the relation between $\sigma$ and the magnitude of the E field at the surface of a conductor?

Either one of these ideas can be used to show that the final charge on the upper surface of C must be equal in magnitude but opposite in sign of the final charge on the lower surface of A.

In post #5 I was trying to move the discussion in a direction which I believe will get to the answer without too much work. However, I don't want to make you work the problem in a certain way if that's not the way you want to work it.

 

[rude man]

Good advice from post #5!
Maybe another hint or 2:
1. 6 sides, 6 surface charges.
BUT - remember, the E field is constant from one inside surface to the opposite inside surface.
2. Put a test charge inside any plate. Does it move? If not, why not?

The above will reduce the number of unknowns from 6 to 3, with 3 independent equations.

And potential is just the integral of an E field.

EDIT: you also need a parameter obtained from before the "C" plate was inserted.

This is not a trivial problem (at least not the way I see it).

 

[vela]

Now calculating charges induced on C:

Upper part of C. A induces -Q. This sends +Q to bottom part of C.
Bottom part of C, B induces +Q. This sends -Q to upper part of C.

You double-counted here. The +Q that moves from the upper surface of C to the bottom surface of C is the same +Q on the bottom surface that you're saying is induced by B. You might find it easier to see if you consider the case where C is initially uncharged.

In any case, I'm not sure your method is correct. The charge on C is +Q. Why doesn't all of it induce a charge on A? After all, the charges on the bottom of C still do contribute to the E field in the region above C.

You do seem to be still assuming the net charge on plate A and the net charge on plate B don't change when C is inserted. Can you justify that? As @TSny noted back in post 4, it's possible that the cell provided or siphoned off some charge.