Calculating Energy Stored in a Capacitor with Varying Gap

[sabak22]

Homework Statement

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

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 don't know what steps should be done before i reach the last step for calculating the energy stored. Please help!

 

[cepheid]

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.

 

[sabak22]

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

 

[cepheid]

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

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

 

[sabak22]

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

 

[gneill]

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

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?

 

[sabak22]

Yes gneill there is more stuff to it but i thought it wouldn't 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 :(

 

[gneill]

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

 

[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

 

[gneill]

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

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.

 

[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.

 

[gneill]

I think you meant (1/2)Q²/C.

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