Stored Energy In a Capacitor

[yankeekd25]

Homework Statement

A parallel plate is connected to one battery and has a material with a dielectric constant of 1.57 filling the space between the plates. The stored energy for the parallel plate capacitor is 21.6 Joules. While connected to the same battery the distance between the plates is decreased by a multiple of 2.3 and the dielectric is removed so that there is nothing between the plates. What is the stored energy in the capacitor now in Joules?

Homework Equations

C= K Enot A / d
Co= C/ K
U = 1/2 QV, or Q^2 / 2c

The Attempt at a Solution

I'm not really sure where to go on this problem. Do I have to somehow find the voltage of the battery and do something with that? Or maybe the C or Q? thanks.

 

[kuruman]

Yes, you need to find the voltage of the battery first. Then you need to find the new capacitance.

 

[yankeekd25]

Yes, you need to find the voltage of the battery first. Then you need to find the new capacitance.

Using the U= 1/2 Q V formula? Or something else? If that formula, how do I find Q?

 

[kuruman]

When the dielectric is removed and the plates are moved closer together does Q change or does it remain the same?

 

[yankeekd25]

When the dielectric is removed and the plates are moved closer together does Q change or does it remain the same?

Would Q change?. is there something you can do with the equations then to find an equation for V?

 

[Adjuster]

You do not have to find the actual voltage or capacitance - in any case you do not have all the information required for this.

What IS required is the proportion by which capacitance is changed.

Te remainder of this post was in error due to misreading of the original question, and so has been deleted.

 

[kuruman]

Please think again. The battery remains connected. If C changes and if Q remains constant (according to you) what happens to V?

 

[yankeekd25]

Please think again. The battery remains connected. If C changes and if Q remains constant (according to you) what happens to V?

V will increase?

 

[Adjuster]

I can only apologise for not having read the initial post carefully enough - I had assumed that the capacitor was initially charged, then isolated from the battery.

If the battery remains connected then the problem is simpler, as the voltage must remain constant and only the capacitance changes .

The charge (CV) and the stored energy (CV²/2) will each change in the same proportion as the capacitance, so you need to find that proportion, and multiply it by the initial energy.