Conceptual Capacitor Question (s)


by BrianConlee
Tags: capacitor, charge, conceptual, direct current, volatge
BrianConlee
BrianConlee is offline
#1
Nov14-10, 02:11 AM
P: 65
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.)
Attached Thumbnails
Capacitors.jpg  
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
tiny-tim
tiny-tim is offline
#2
Nov14-10, 05:17 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,167
Hi Brian!
Quote Quote by BrianConlee View Post
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?
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.
Question 2: What if I started with the plates apart and turned on the voltage?
A tiny amount of charge.
goodnight!
DaleSpam
DaleSpam is offline
#3
Nov14-10, 08:52 AM
Mentor
P: 16,476
I agree with tiny tim
Quote Quote by BrianConlee View Post
Question 1: Do the plates maintain their positive and negative charges?
...
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.
The answer is no. You have
[tex]C=\frac{Q}{V}[/tex]
and
[tex]C=\epsilon\frac{A}{d}[/tex]

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.

Quote Quote by BrianConlee View Post
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.
They would charge, but since d is large Q would be small.

BrianConlee
BrianConlee is offline
#4
Nov14-10, 10:30 AM
P: 65

Conceptual Capacitor Question (s)


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?
sophiecentaur
sophiecentaur is offline
#5
Nov14-10, 10:49 AM
Sci Advisor
PF Gold
sophiecentaur's Avatar
P: 11,352
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.
tiny-tim
tiny-tim is offline
#6
Nov14-10, 11:14 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,167
Yes.
BrianConlee
BrianConlee is offline
#7
Nov14-10, 08:29 PM
P: 65
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.


Register to reply

Related Discussions
Conceptual Question Introductory Physics Homework 5
Conceptual Parallel Plate Capacitor Introductory Physics Homework 0
Simple conceptual question - rolling object question: Introductory Physics Homework 12
Conceptual Question Introductory Physics Homework 3
capacitor conceptual problem Classical Physics 5