Capacitance and electric field

1. Jul 3, 2006

I've got a few more I need help with.....I think I might have these ones, but not quite sure....as the concepts are still new to me.

1. If the voltage between the plates of a parallel plate capacitor is doubled, the capacitance of the capacitor....
A. is halved
B. remains the same
C. is doubled

---I think is is A(halved)....because Ceq = Q/V.....so doubling something on the bottom....would be the same as halving C on the other side.

2. If the voltage applied to a parallel plate capacitor is doubled, the electric field between the plates.....
A. is halved
B. remains the same
C. is doubled

----I'm thinking the answer is C(doubled), because E=V/d....

2. Jul 4, 2006

Andrew Mason

What has to occur to Q in order for the voltage to double?

AM

3. Jul 4, 2006

well if you take Ceq/Q and inverse them. So....then V=Q/Ceq. So....if you double Q, you double V. Likewise.....if you half Ceq, then V will double. Am I right, or not? And, in this question, I'm looking for C....right.

4. Jul 4, 2006

Andrew Mason

Let's assume that plate separation does not change. So if to double V you necessarily must double Q what happens to C? I think that is what the question is asking. Otherwise the question is ambiguous.

AM

Last edited: Jul 4, 2006
5. Jul 4, 2006

If to double V.....and you HAVE to double Q......then I'd say C stays the same.....or am I still off......this is confusing to me for some reason.

Andrew.....am I correct on the other question then?

6. Jul 4, 2006

maverick280857

You are right about the second problem.

To clear up any residual confusion, I will suggest you refer to your problem and then directly to Andrew's post and only if necessary read this:

The capacitance is given by

$$C = \frac{\epsilon_{0}A}{d}$$

It is clear that $C$ is not a function of charge $Q$ or potential difference $V$. Hence, when you write

$$Q = CV$$

you are actually saying that

$$Q \alpha V$$

that is the charge is directly proportional to the applied potential difference. It says no more.

Now when you have a capacitor you have figure out whether it is connected to a source (constant V) or is isolated (constant Q). In your first problem, $V$ changes to $2V$ so obviously the charge must double. However, you aren't changing the geometry so C does not change at all. So you are right.

7. Jul 4, 2006