Capacitance and electric field

[Bradracer18]

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
D. quadrouples

---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
D. quadrouples

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

 

[Andrew Mason]

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
D. quadrouples

---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.

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

AM

 

[Bradracer18]

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.

 

[Andrew Mason]

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.

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

 

[Bradracer18]

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?

 

[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

[tex]C = \frac{\epsilon_{0}A}{d}[/tex]

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

[tex]Q = CV[/tex]

you are actually saying that

[tex]Q \alpha V[/tex]

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, [itex]V[/itex] changes to [itex]2V[/itex] so obviously the charge must double. However, you aren't changing the geometry so C does not change at all. So you are right.

 

[Bradracer18]

Thanks maverick...and you too Andrew. I really appreciate it. That does make a lot more sense now though...I kinda forgot they were perportional...thanks again!