Stupid Capacitor question

[jinyong]

Q=CV for a capacitor.

The Q is for the amount of charge on one plate right(for parallel capacitor case)? So is the total charge 2Q on the capacitor?

Does it always have +Q on one plate and -Q on the other? Will it ever happen when it's +Q on one plate and not -Q on the other or vice versa?

Thanks.

 

[FrogPad]

I think a more strict notation on the expression you showed above sheds more light on the capacitance.

$$C = \frac{Q}{V_{12}}$$
$$V_{12} = - \int_2^1 \vec E \cdot d\vec \ell$$

$$V_{12}$$ is the voltage from the capacitor (1) with positive charge to the capacitor (2) with negative charge.

Also when you calculate the capacitance $$C$$ you assume charges of +Q on one conductor and -Q on the other conductor.

So let me sum this um, and write it even stricter.
$$C=\frac{Q_1}{V_{12}}$$

C: the capacitance of the conducting body
Q_1: the amount of charge (this is positive since we are referring to conductor one, which we defined to have positive charge).
V_12: the voltage between conductor 1 and 2

To be honest I don't know about your +Q_1 on one plate and -Q_2, where |Q_1| =\= |Q_2| question. My intuition says that it has to be the case. If you think about a capacitor as follows:

1) Place a conducting body in free space, it can be any shape or size.
2) Now place another conducting body of any shape or size (maybe not the same as the original even) in freespace
3) Apply a voltage (potential difference) between the conducting bodies.
4) Charges will begin to arrange themselves on each of the conducting bodies
5) Given some time the first conductor will have +Q, and the second conductor will have -Q... this will happen VERY quickly.

But wait a few, and someone will come on here and give you more details. Hopefully I cleared up how you use Q=CV a bit though :)

 

[berkeman]

When the capacitor is connected to a circuit, then yes, the + charge on one plate equals the - charge on the other plate. The only way you could get a mismatch in charge would be if the capacitor leads were floating (not connected to anything), and you deposited some static charge on one of the plates.

Oh, and don't call capacitors stupid. They hate it when you do that.