# Homework Help: Capacitors and varying gap

1. Oct 19, 2011

### sabak22

1. The problem statement, all variables and given/known data

A capacitor is completely charged with 670 nC by a voltage source that had 200 V.
What is its capacitance?==== 3.35×10-9 F 9(I got this y doing Q/V)

The initial air gap of the capacitor above was 7 mm. What is the stored energy if the air gap is now 10 mm?

Now this is the part with which I am stuck

3. The attempt at a solution

I know that Energy = (1/2)CV^2 or QV/2, as well as Q^2/2C
But the problem is, I dont know what steps should be done before i reach the last step for calculating the energy stored. Please help!

2. Oct 19, 2011

### cepheid

Staff Emeritus
It sounds to me like you have to figure out how capacitance depends on the air gap, use this information to compute the new capacitance, and use the result of that to compute the stored energy.

3. Oct 19, 2011

### sabak22

yup I know, as distance increases Capacitance decreases. so in this case my capacitance will decrease because gap is now 10mm.

4. Oct 19, 2011

### cepheid

Staff Emeritus
Yeah, but you need to actually compute by how much it decreases, so that you know the new capacitance value.

5. Oct 19, 2011

### sabak22

umm, would it make sense if my new capacitance was now 2.345*10^-9 F?

6. Oct 20, 2011

### Staff: Mentor

Sure. But it would make even more sense if you showed how you arrived there

Moving on, the question is vague about whether or not the battery remains connected while the plates of the capacitor are adjusted. Are there more details in the question statement that we haven't seen?

7. Oct 20, 2011

### sabak22

Yes gneill there is more stuff to it but i thought it wouldnt be important, but i guess it is:

b)Now the plates of the charged capacitor are pulled apart with the voltage source already disconnected. - The energy stored in the capacitor: Increases.

an then in c theyre asking for teh new capacitance, but whne i entered 2.345*10^-9 F? it says incorrect :(

8. Oct 20, 2011

### Staff: Mentor

Your new capacitance looks okay. Perhaps they're complaining about the significant figures?

9. Oct 20, 2011

### sabak22

wait gneill, its not the capacitance theyre asking me to enter theyre asking me to enter the energy stored after the gap has changed! I tried applying U=(1/2)CV^2 Which gives me 4.96*10^-5 J. But its still incorrect. what am i doing wrong? seems right to me

10. Oct 20, 2011

### Staff: Mentor

You said that part c was asking for the new capacitance.

If they're asking for the energy then you need to realize that if the voltage source is no longer connected when the plates are moved, then the voltage across the capacitor will change. What remains the same is the charge (it can't go anywhere if the capacitor is isolated!).

Use you new capacitance value and the known charge to find the energy stored.

11. Oct 20, 2011

### sabak22

Yay I got it, so i did U=(1/2)Q/C^2 which gave me = 9.57*10^-5

Thank you very much gneill. just one quick question though, how will I know what the new voltage across the capacitor is after the battery has been disconnected? Just for my own knowledge.

12. Oct 20, 2011

### Staff: Mentor

I think you meant (1/2)Q2/C.

With the charge and the new capacitance value you can find the voltage. V = Q/C.