Understanding Capacitance: Answers to Common Questions Explained

[Noesis]

I have thought about these questions, and although they seem easy I have different answers to them:

1) Consider a parallel plate capacitor with some capacitance C. The plates of the capacitor are connected to a battery that maintains a constant potential. If the plates are moved farther apart, does the charge on each plate increase, decrease, or remain the same?

I don't know whether to say that the charge remains the same since the electric potential will remain constant...or that the distance will alter the potential and thus the charge will increase.

Mathematically, I also get another answer:
If C = Q/V, where V is the potential between the capacitors and it doesn't change, and we know the capacitance must go down since the distance is increased, then logically the charge must decrease.

So I don't know whether to go with decrease, same, or increase.

2) Consider the same parallel plate capacitor connected to a battery. After moving the plates apart, the battery is disconnected so that the plates are electrically isolated. The plates are then moved back to their original separation. Is the voltage between the plates larger, smaller, or the same as the potential of the battery?

Here, the charges will remain constant since it was electrically isolated. So I imagine that this means the potential will remain the same...and thus once brought back together the potential is the same.

Am I thinking about this wrong? I believe I don't have a very good understanding regarding potentials and capacitance.

If anybody could shed some light on this I would greatly appreciate it.

 

[Dmitri]

The formula for parallel plate capacitor is C = E*A/d where A - area of the plate, d - distance between the plates, E - constant.
Use this equation and Q = C*V for both problems.

 

[Noesis]

Thank you.

So from the equations, and from some logic..I get that the charge decreases for the first one, and in the second one the voltage is the same.

Charge decreases in the first one, because the capacitance goes down when you move them farther apart, and thus the charge must go down.

But what I am worried about here is, does the voltage go up when the capacitance goes down?

From V = Q/C ... it would. So does the charge really decrease or does it stay the same?

For the second one..since the charge is the same, I figure the electric field, will be same, and obviously so will the area and distance.

So from V = Q/C .. it will still be the same.

Can anyone confirm or dispel these ideas? Thank you.

 

[Dmitri]

For the second one. The charge does not change but the capacitance does change since you decrease the distance between the plates. From Q=V*C you can see that in order for charge to stay constant, voltage has to increase in order to accommodate decrease in capacitance.