# How capacitance, charge, and energy stored changes on a capacitor?

## Homework Statement

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same?

(a) C
(b) Q
(c) E between the plates
(d) V

## Homework Equations

$$C = \frac{Q}{V}$$

$$C = \frac{\epsilon_0A}{d}$$

$$E = \frac{\sigma}{\epsilon_0}$$

## The Attempt at a Solution

[/B]
I'm doing some self study here and I'm trying to understand this problem. I'd love some thoughts on areas I'm confused on.

To start off, we are charging up this capacitor so that it holds charge Q and has a potential difference of V.

After disconnecting the battery, I know that the charge, Q, must stay the same because there's physically nowhere for the charge to go. So (b) remains the same.

Now, I wanted to tackle the potential difference, V, across the plates as we pull them apart. The potential difference between the two plates is equal to the work needed per unit charge in moving from one plate to the other. That is, if I physically move a charge from one plate to another, the work I do (per unit charge) is equal to the potential difference. This work will increase (because we're moving over a longer distance) if we increase the spacing on the plates. And that's why the potential difference will increase as we increase the space between the plates. So (d) increases.

The electric field, E, is given by the equation above, and is constant so it will remain the same. (c) remains the same. CONFUSION HERE: I don't have a good intuition on why it is the electric field remains the same when the potential difference increases. Any thoughts here?

And finally, the capacitance. I think I have a hard time reconciling the geometry relationship for capacitance with the formula that relates it to charge/voltage. From the formulas above, we see that if the potential difference increases, that the capacity decreases. CONFUSION HERE: But I always see this as the capacity to store charge and if the capacitance decreased, then its capacity to hold charge has decreased and so it should have less charge on it. But the Q can't go anywhere so I'm unclear as to how the capacitance decreases.

Any insight into the last two paragraphs would be great.

gleem
First what is the definition of electric field.(how is it related to the potential difference. Note that the electric field for a parallel plate capacitor is considered uniform.

Second If the charge is fixed as you already have noted then as the capacitance (a geometrically determined characteristic) is decreased what is the only thing that can happen as you increase the distance.?

CWatters
Homework Helper
Gold Member
From the formulas above, we see that if the potential difference increases, that the capacity decreases.

No, in the equation...

Q = CV................. (1)

..C (Capacitance) is a constant, in that it's not dependent on Q or V (The size of a bucket doesn't depend on the volume of water in it). As gleem said it's a geometrically determined characteristic as evident from:

C = εA/d................(2)

If you substitute for C in (1) you get..

Q = (εA/d) * V

You correctly stated that Q is constant so what's the relationship between d and V ?

rude man
Homework Helper
Gold Member
The electric field, E, is given by the equation above, and is constant so it will remain the same. (c) remains the same. CONFUSION HERE: I don't have a good intuition on why it is the electric field remains the same when the potential difference increases. Any thoughts here?
You know voltage is the displacement integral of the E field, so V = Ed if d is the distance between plates. You have already correctly deduced that V goes up as d goes up, so E = V/d = constant should not surprise.

If you have covered Gauss's law that is a nice alternative way to find E.
And finally, the capacitance. I think I have a hard time reconciling the geometry relationship for capacitance with the formula that relates it to charge/voltage. From the formulas above, we see that if the potential difference increases, that the capacity decreases. CONFUSION HERE: But I always see this as the capacity to store charge and if the capacitance decreased, then its capacity to hold charge has decreased and so it should have less charge on it. But the Q can't go anywhere so I'm unclear as to how the capacitance decreases.
It's capacity to store charge at a given voltage. So since V went up it shouldn't surprise that C had to go down.

No, in the equation...

Q = CV................. (1)

..C (Capacitance) is a constant, in that it's not dependent on Q or V (The size of a bucket doesn't depend on the volume of water in it). As gleem said it's a geometrically determined characteristic as evident from:

C = εA/d................(2)
In the problem, the distance between the plates increases, so capacitance is not constant here.

After disconnecting the battery, I know that the charge, Q, must stay the same because there's physically nowhere for the charge to go. So (b) remains the same.
Yes, exactly.
Now, I wanted to tackle the potential difference, V, across the plates as we pull them apart. The potential difference between the two plates is equal to the work needed per unit charge in moving from one plate to the other. That is, if I physically move a charge from one plate to another, the work I do (per unit charge) is equal to the potential difference. This work will increase (because we're moving over a longer distance) if we increase the spacing on the plates. And that's why the potential difference will increase as we increase the space between the plates. So (d) increases.
That's one way to look at it. If you consider the plates to be very large, then the electric field is constant between the plates; since ##E=-\Delta V/d##, if ##d## increases then so must ##V## in order to keep ##E## constant.
The electric field, E, is given by the equation above, and is constant so it will remain the same. (c) remains the same. CONFUSION HERE: I don't have a good intuition on why it is the electric field remains the same when the potential difference increases. Any thoughts here?
See above. ##E## must be constant due to Gauss's Law, again assuming the plates are large.
And finally, the capacitance. I think I have a hard time reconciling the geometry relationship for capacitance with the formula that relates it to charge/voltage. From the formulas above, we see that if the potential difference increases, that the capacity decreases. CONFUSION HERE: But I always see this as the capacity to store charge and if the capacitance decreased, then its capacity to hold charge has decreased and so it should have less charge on it. But the Q can't go anywhere so I'm unclear as to how the capacitance decreases.
##C=Q/V##. If ##Q## is constant and ##V## is increasing, then ##C## must be decreasing as you stated. Capacitance is the ability to hold charge at a given voltage. A high value capacitor can store a lot of charge using only a small potential difference while the opposite is true of a small value capacitor. Here the charge has stayed the same but the voltage of that charge has increased.

CWatters
Homework Helper
Gold Member
In the problem, the distance between the plates increases, so capacitance is not constant here.

I know that but the OP wrote..

From the formulas above, we see that if the potential difference increases, that the capacity decreases.

The capacitance is not dependant on Q or V.

The capacitance is not dependant on Q or V.
Why not? It is related to those vales by definition. At any point in time ##C=Q/V##. The voltage is changing and the charge is constant. So ##C## must change to satisfy this definition at all times. The capacitance of a device does not have to be constant. Older radios used variable capacitors for the tuning knob.

rude man
Homework Helper
Gold Member
Why not? It is related to those vales by definition. At any point in time ##C=Q/V##. The voltage is changing and the charge is constant. So ##C## must change to satisfy this definition at all times. The capacitance of a device does not have to be constant. Older radios used variable capacitors for the tuning knob.
He never said capacitance has to be constant. Vary d and you change C just as your "old radios" did. He said capacitance is not a function of Q or V.
I take a capacitor and put V on it. Q = CV. I then change V to V'. Q' = CV' but C does not change.

CWatters
Homework Helper
Gold Member
Why not? It is related to those vales by definition. At any point in time ##C=Q/V##. The voltage is changing and the charge is constant. So ##C## must change to satisfy this definition at all times. The capacitance of a device does not have to be constant. Older radios used variable capacitors for the tuning knob.
I know, I built my first about 47 years ago when I was 11.

I'm also reasonably familiar with variable capacitance diodes (which do indeed vary their capacitance when the voltage is changed), but none of this helps the OP understand the problem.

I take a capacitor and put V on it. Q = CV. I then change V to V'. Q' = CV' but C does not change.
Except that ##Q=Q^{\prime}## for this problem, so which variable must change?

CWatters