# A question regarding isolated charged plates vs. capacitors

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1. Apr 6, 2015

### FallenLeibniz

In an example in my book, the author poses the following question: Given two large plates
at a distance d from each other with arbitrary charges Q' and Q" (with Q' on the bottom plate
and Q" on the top plate), how are the charges distributed on the plates? The author lists
the following equations:

$Q_{1u}+Q_{1l}=Q'$
$Q_{2u}+Q_{2l}=Q"$
$\frac{Q_{1u}}{A}+\frac{Q_{2l}}{A}=0$
$\frac{Q_{1l}}{A}-\frac{Q_{2u}}{A}=0$

Where the subscripts designate the side of the plate and A is the area of both plates.

Now the reasoning for the first two equations is obvious. In his explanation of the reasoning
behind the last two equations, he suggests that the charge distributes such as to create an
electric field in the conductor such that it cancels the field from the other plates.

My question is this. The solutions to the questions above suggest that when the plates just
have arbitrary charges, the charge that is on the sides of the plates that do not compose
the gap between them may not necessarily be zero. However, when the plates are connected via
a battery to form a capacitor, those charges on those surfaces must be zero. Why is this so?

P.S.
I would like to note that this scenario is in the context of Electrostatics and that the dimensions
composing A are way larger than the separation distance d.

2. Apr 7, 2015

### Simon Bridge

... what makes you think that?
 oh I think I see what you mean... its because the usual capacitor has equal and opposite charges.

3. Apr 7, 2015

### FallenLeibniz

@Simon Bridge: Yes. My wonder then becomes why in the case of the capacitor hooked to a battery must the charge from the outer surfaces become entirely distributed on the inner surfaces when this is not neccessarily the case with the isolated plates with arbitrary charges scenario.

4. Apr 8, 2015

### Simon Bridge

Its because the usual capacitor has equal and opposite charges.
If the charges are equal and the same, the charge ends up on the outer surfaces.
partly this is an artifact of the idealization though... for an arbitrary shape, clearly some charge must be closer than others.

5. Apr 8, 2015

### FallenLeibniz

What is the physical reason though why the charges migrate to the inner surfaces? Is it just specifically due the need for the fields to counterbalance?

6. Apr 8, 2015

### Simon Bridge

In terms of charges: Charges migrate to the inner surface only if they are oppositely charged - the physical reason is that opposite charges attract.
In terms of fields - the static configuration of charges is the one that produces a net zero field inside the conductor.
Basically you picks your model. Physics tells you what happens, but does not tell you why.

You can work it out for yourself...

Two opposite charges will try to be as close as possible to each other, like charges try to get as far away from each other as possible.
It is the balance of attraction and repulsion that produces the particular distribution.

Work it out imagining that there is a a single + unit charge on a conducting sphere and two -1 unit charges on another sphere: the two -1 charges try to be as far apart as possible while also trying to be as close as possible to the +1 charge. What is the stable configuration? You should be able to work it out from a sketch - use a pencil.

7. Apr 8, 2015

### FallenLeibniz

Even just thinking about that scenario, the first possible solution that comes to mind is that the negative charges take positions on the far points of one of the axes through the second sphere.

8. Apr 8, 2015

### Simon Bridge

Well done - so the first thing you thought of has the unpaired charge on the opposite side to the paired charge.
This is what is happening with the unequally but opposite charged plates.

Play around and see if you can find any other stable configurations.

9. Apr 8, 2015

### FallenLeibniz

Ok. I actually ended up using the solutions to the equations above combined with the idea of trying to visualize to get an idea of what is happening in the case of isolated plates with arbitrary charges. I see that when the two plates are introduced, they both work to end up at that "sweet spot" where the charges are comfortable with the Coloumbic forces and allow for the balancing of the fields. Thank you Simon for the guidance. It helped me out a lot.

10. Apr 9, 2015

### Simon Bridge

No worries. Often breaking something down to a smaller scale helps piece things together.