# Homework Help: Charge distribution on the surfaces of parallel conducting s

Tags:
1. Nov 15, 2015

### amind

Problem:
Consider two parallel and large sheets with a surface area . One has a charge and the other is uncharged.
Code (Text):

q
|       |
|       |
|       |
|       |
|       |

What would be the electric fields on the three regions as divided by the sheets ?

General solution to problems like as told by my teacher:
Using this principle, it is trivial to find a solution to this problem, distribution of charge:
Code (Text):

+q/2 +q/2    -q/2  +q/2
|             |
|             |
|             |
|             |
|             |

The surface charge densities and thus the electric field dictated by this distribution is indeed the correct answer.
However, I really don't understand how/why this method works and my naive attempt at solving this problem comes out to be very wrong.

Here's my attempt:
The charge on one sheet would induce some charge of opposite polarity on the opposite end. I recall a similar situation where two conducting sheets have opposite charges, the charges are concentrated only on the inner surface resulting in 0 electric fields outside the sheets. Thus, I reason that the charge on the outer surfaces of both the sheets in this case would be 0, and on the inner surfaces it would be and respectively.

So, someone please help, where did I go wrong ? How (or should I say 'Why') does my teacher's rule follow ?

2. Nov 15, 2015

### J Hann

One sheet was specified to be uncharged.
Then you can't have a charge q (or -q) on one surface and zero charge on the
other surface without creating charge from nothing.

3. Nov 15, 2015

### amind

Oh yeah, I messed that up while posting.
"....Thus, I reason that the charge on the outer surface of the charged sheet in this case would be 0, and on the inner surfaces of both would be and respectively and charge on the outer surface of the uncharged sheet would be +q".
i.e.
Code (Text):

0  +q        -q  +q
|             |
|             |
|             |
|             |
|             |

How about now ?

4. Nov 16, 2015

### ehild

No, it is not correct. You can think a couple of reasons why the teacher's approach is correct.
The charge redistributes in a way that makes the potential energy lowest.

From very far away, the two conducting plates look as one with a net charge q. You know that the electric field of a single plate is symmetric, with magnitude σ/(2ε0) on both sides. That means, that q/2 charge is on both outer surfaces.
You can apply the superposition principle to solve such problems. A plane distribution of charge with q/A=σ charge per unit area has electrix field σ/(2ε0) pointing away of the plane, and it goes through the plane of the other plate. The uncharged plate has no contribution to the electric field. So the field of the arrangement will look like as shown in the figure,

5. Nov 16, 2015

### amind

Thanks! You made it really clear!