What would happen to the charge and capacitance?

[YMMMA]

Homework Statement

In the attached file..
The question asks about the charge and capacitance after being separated by twice the original separation. I also need to know if it is kept connected to the battery( spearted to twice the distance,too) , what would happen to both the charge and capacitance? What would happen if the distance remained the same?

Homework Equations

Capacitance is proportional to the area of the plates and inversely proportional to the distance between them.
C= charge/ potential difference

The Attempt at a Solution

Well, if it disconnected from the battery, the charge will stay the same. However, the capacitance will be halved since the distance is doubled . Now, if it is kept connected, I think that because charge increases with increasing voltage, so it will keep attaining all the potential and then be constant; additionally, the capacitance is not changed if the distance remained the same, but it will be halved if the distance is doubled. So, capacitance is independent on the charge. Am I right?

 

[TSny]

The Attempt at a Solution

Well, if it disconnected from the battery, the charge will stay the same. However, the capacitance will be halved since the distance is doubled .

Yes.

Now, if it is kept connected, I think that because charge increases with increasing voltage, so it will keep attaining all the potential and then be constant;

I'm not sure what you are saying here. Can you clarify?

additionally, the capacitance is not changed if the distance remained the same, but it will be halved if the distance is doubled. So, capacitance is independent on the charge. Am I right?

Yes.

 

[YMMMA]

I mean that charge increases at first then remain constant, is this right?

 

[TSny]

I mean that charge increases at first then remain constant, is this right?

Are you saying that as the separation between the plates increases, the charge on the plates increases (while the capacitor remains connected to the voltage source)? If so, why would the charge increase?

 

[CWatters]

Now, if it is kept connected, I think that because charge increases with increasing voltage...

If it remains connected to the battery all the time can/will the voltage change?

..the capacitance is not changed if the distance remained the same, but it will be halved if the distance is doubled.

Correct.

So now look at your relevant equation. It has three variables Q,C and V.

When the plates move which do you know stays constant? Which is changing? What is the implication for the remaining one?

You don't need to try and guess what's happening in the capacitor just treat it as a maths problem for the moment. Then afterwards you can work out what's going on physically.

 

[YMMMA]

Are you saying that as the separation between the plates increases, the charge on the plates increases (while the capacitor remains connected to the voltage source)? If so, why would the charge increase?

Alright, the voltage in the battery is actually constant, so the charge remains the same.

 

[TSny]

Alright, the voltage in the battery is actually constant, so the charge remains the same.

Yes, V remains constant. But keep in mind that C is changing while the plates are being separated. Q will not stay constant.

 

[YMMMA]

If it remains connected to the battery all the time can/will the voltage change?
Correct.

So now look at your relevant equation. It has three variables Q,C and V.

When the plates move which do you know stays constant? Which is changing? What is the implication for the remaining one?

You don't need to try and guess what's happening in the capacitor just treat it as a maths problem for the moment. Then afterwards you can work out what's going on physically.

Ahaa, if Q=CV, V constant, C changes with distance, so Q also changes with C

 

[TSny]

Ahaa, if Q=CV, V constant, C changes with distance, so Q also changes with C

Yes. Good.

 

[YMMMA]

I appreciate all your help, Thanks!