Charge and Potential Difference on 2 Capacitors

[nw0rbrolyat]

Homework Statement

A C_1 = 2.40 \mu(micro) F capacitor is charged to 861 V and a C_2 = 6.40 \mu(micro) F capacitor is charged to 662 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other.

What will be the potential difference across each? [Hint: charge is conserved.]

What will be the charge on each?

Homework Equations

1/CT = 1/C1+1/C2+...+1/CN
E = CV2 / 2

C = 2E / V2

V = sqrt(2E / C)

The Attempt at a Solution

I am confused as to what the question is saying. There will be no battery and the 2 capacitors are connected? Please help!

 

[ehild]

What are E and V? And what do you mean on E=CV2/2? 2/2=1. So is this C*V? Or CV₂/2 Or CV²/2?

The charge is conserved. What is the total charge on the connected plates?

What is the resultant capacitance?

The plates connected are on the same potential. How is the capacitance, potential difference (voltage) and charge related?


ehild

 

[nw0rbrolyat]

Im not sure if I have the right equations here. I am totally clueless on this problem

 

[ehild]

But you can think and look after the definition of capacitance, don't you? You can get the charge on the capacitor from the voltage and capacitance. First calculate the charge on both capacitors.

ehild

 

[Xerxes1986]

The capacitors are going to be connected in parallell, which means that the voltage over 1 will equal the voltage over the other one.

Also, charge will be conserved. So the total charge you put on each capacitor will equal to the total charge that is over BOTH capacitors when they are connected. You will find a simple equation in your book that looks like this:

Q=CV

So calculate the charge on each capacitor using the capacitance and the voltage applied, and add them together. That will be your Q_{final}. Then calculate the new capacitance of the 2 capacitors in parallel and your voltage will then be given by the same equation above :)