# Conceptual Capacitor Question (s)

1. Nov 14, 2010

### BrianConlee

I have a capacitor. Two plates, air dielectric: nothing fancy... in a simple circuit.

A simple circuit containing a dc voltage source and the capacitor.

I turn on the voltage source to the capacitor. When it is charged fully, I pull the two plates apart. (see image) I leave the voltage source on.

Question 1: Do the plates maintain their positive and negative charges?

If the answer is yes, I think the explanation is obvious. (Nothing changed the charge on the plates.

If the answer is no, why? What changed the charge on the plates. Not the capacitance of the capacitor, I know that is inversely proportional to the plate distance. Just the actual charges on the plates.

Question 2: What if I started with the plates apart and turned on the voltage?

If the answer is the plates would charge, I think the explanation is obvious. The same thing that happened when the plates were close and parallel is happening with the plates in a different position.

If the answer is "nothing would happen," why not? The only real difference is the plate distance.

Thanks. (This is why I'm awake at 3 A.M.)

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2. Nov 14, 2010

### tiny-tim

Hi Brian!
No, because the plates of the capacitor are still connected.

If the plates aren't connected, no charge can flow, but since they are connected, the only thing keeping the charges there is the potential difference (the voltage) created by the battery (like the pressure from a pump or a spring).

But if they are connected through a battery, the potential difference between them stays the same (because it comes from the battery) …

and potential difference = potential energy per charge = work done per charge = electric field times distance …

and since the electric field between two charged plates is proportional to charge and independent of distance, that means …

if p.d stays the same and distance increases, charge decreases.
A tiny amount of charge.

goodnight! :zzz:​

3. Nov 14, 2010

### Staff: Mentor

I agree with tiny tim
The answer is no. You have
$$C=\frac{Q}{V}$$
and
$$C=\epsilon\frac{A}{d}$$

Combining and solving for Q in terms of d you get that Q is inversely proportional to d (with everything else held constant). So if you double d you will halve Q.

They would charge, but since d is large Q would be small.

Last edited: Nov 14, 2010
4. Nov 14, 2010

### BrianConlee

Ok, now I'm starting to understand this a lot better. ( I did fall asleep, but thank you!)

I kept seeing that equation also, but I didn't connect the dots like that. So, as I increase distance, I have to increase voltage in the equation to maintain any given Q.

This does lead to a new question however.

I charge the capacitor in it's original setup... then I disconnect it from the voltage source. I then pull the plates apart.

Do they maintain the Q in this case?

Step 1. Charge capacitor
Step 2. Disconnect capacitor plates
Step 3. Pull plates apart.

Leakage aside. I should have a positively charged plate and a negatively charged one with the same magnitude as when they were together right?

That's the key here... disconnect the plates before seperation?

5. Nov 14, 2010

### sophiecentaur

Yes: If you disconnect the plates then, because the total charge must remain the same and the Capacity decreases, the Voltage must increase because Q (constant) = CV.

This suggests a way of generating high voltages; charge two plates and then separate them. But I don't think I've come across this as a practical source of HT, except I think the Cockroft Walton voltage multiplier may effectively do this by charging capacitors in parallel and discharging in series. - Nope, I don't think that's right even.

Last edited: Nov 14, 2010
6. Nov 14, 2010

### tiny-tim

Yes.

7. Nov 14, 2010

### BrianConlee

Woo hoo...

Finally I'm getting somewhere. :)

Today Capacitors, tomorrow the world!

Thank you all for responding and keep a young scientist/inventor/lover going strong.