# Homework Help: Charge on metallic plates

1. May 28, 2013

### Saitama

The problem statement, all variables and given/known data

Following are the questions based on the above paragraph

Q1)The charge appearing on the outer surface of plate 1, when switches K1 and K2 are open
A)zero
B)Q
C)-Q
D)-3Q

Q2)If K1 is closed and K2 is open, the charge appearing on the right surface of plate 2 is
A) $Q/2+(\epsilon_0A/d)V/4$
B)$(\epsilon_0A/d)V/4+3Q/2$
C)$(\epsilon_0A/d)V/4-Q/2$
D)$3Q/2$

Q3)If both switches are closed, the charge appearing on plate 4 is
A)$(\epsilon_0A/d)V$
B)$(\epsilon_0A/d)V/2+Q$
C)$Q-(\epsilon_0A/d)V/2$
D)$(\epsilon_0A/d)V/3$

2. Relevant equations

3. The attempt at a solution
For Q1), it can be easily done by equating the electric field at any point inside the plate equal to zero but the solution solves it in a single line. It is written that "Charge on outermost surface=(net charge on system/2)=0". How did it arrive at this result?

About Q2) and Q3), I have no idea. I am clueless on how to even begin with them.

Any help is appreciated. Thanks!

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2. May 28, 2013

### haruspex

For (2), when the switch is closed you will get a movement of charge between plates 1 and 4, right? And the quantity moved will be so as to produce a potential difference of V between them?

3. May 28, 2013

### rude man

I am just posting to be kept in touch on this. Hope our Smart Ones chime in!

On Question 1 I would have said that the surface charge on the left side of plate 1 = Q/2. If there is charge on plate 1 the charges have to be on the surface. If plate 1 were isolated then certainly half the excess charge woud reside on the left side. Why would the proximity of plates 2 et alia change that? How small does d have to be to change all that? Certainly, a gaussian "box" running from just outside the left side of plate 1 to just outside the right side of plate 4 would yield zero net flux as required if the right side of plate 4 held surface charge -Q/2.

4. May 28, 2013

### haruspex

For part 1, the close proximity allows you to treat the fields as orthogonal to the plates everywhere.

5. May 28, 2013

### Saitama

I am still not sure what to do. When the switch is closed, let an extra charge $q_1$ flow to plate 1, so the new charges on plate 1 and plate 4 are $Q+q_1$ and $-Q-q_1$. Should I do the usual practice of equating electric field to zero inside the plate? And what about the solution to Q1?

6. May 29, 2013

### rude man

That was implicit in my statement about the gaussian box. No leakage out the sides, and the flux pointing to the left on both the left and right ends.

7. May 29, 2013

### haruspex

You need to get an expression for the potential inside each of the end plates, since you need to use the fact that the difference between them is V.
For Q1, I don't think there's really an easier way than the way you did it. But it's easy to see that you can generalise it to an arbitrary stack of plates with some given total charge, and it may be that the author considers this a known standard result.

8. May 29, 2013

### Saitama

I am still unsure what to do here. Can you please explain a bit more?

9. May 29, 2013

### haruspex

In terms of the initial charges and the unknown transfer of charge q between the end plates, you can determine the charge on each face of each sheet, right? Taking the plates to be thin compared with d, and setting the potential of the left hand plate as V0, say, you can compute in these terms the potential inside the other end plate. The difference in potential between the two end plates is V, giving you an equation for q. You can now deduce the charge on each face of each sheet.

10. May 30, 2013

### Saitama

The charge on the left face of plate 1 and right face of plate 4 is zero from the result of previous solution. The new charge distribution is shown in the following figure.

Looks good?

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11. May 30, 2013

### haruspex

Right, that's the new charge distribution. So what's the potential difference (in terms of the Qs and ds) between plates 1 and 4?

12. May 30, 2013

### Saitama

$$\frac{q_1}{C}+\frac{2Q+q_1}{C}+\frac{q_1}{C/2}=V$$
where $C=A\epsilon_o/d$

Looks correct?

13. May 30, 2013

### utkarshakash

There is another simpler way to solve these problems. Ok I'm giving you a hint. Transform this arrangement to a circuit consisting of 3 capacitors, 2 switches and 2 batteries. Now by looking at the circuit you can easily find the charge distribution on the capacitors.

14. May 30, 2013

### haruspex

Sorry for the delay - was busy.
Yes, that's what I get. Unfortunately, it doesn't correspond to any of the offered answers. Seems to me there has to be a factor 1/3 in the answer.

15. May 30, 2013

### Saitama

I do get option B) as my answer. Solving the above equation,
$$q_1=\frac{CV}{4}-\frac{Q}{2}$$
We require the charge on the right surface of plate 2 which is $2Q+q_1$. Substituting $q_1$, I get B.

16. May 30, 2013

### haruspex

Just realised I didn't notice the 2d width - I thought they were all just d.
So, on to the last part

17. May 30, 2013

### Saitama

Here's how I think the charges arrange,

Correct?

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18. May 31, 2013

### haruspex

With both switches closed, Q cannot matter any more. The charges will rebalance between 1 and 4 and between 2 and 3 in a way that's entirely driven by the applied potentials. Your diagram shows the total charge on plate 2 being 2Q, so I don't think it can be right.
You do know that plates 1 and 4 will have opposite total charges, and plates 2 and 3 will have opposite total charges. Try assigning unknowns to those and solving.

19. May 31, 2013

### Saitama

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20. May 31, 2013

### haruspex

Yes, that looks right. Now deduce the relationship between q and q2 from the potential differences.

21. May 31, 2013

### Saitama

I get
$$V-\frac{q}{C}-2V-\frac{q}{C/2}=0$$
Solving this,
$$q=\frac{-CV}{3}$$
Hence, the charge on plate 4 is -q=CV/4.

Thank you, haruspex!