Find Capacitance From Charged Capacitors

[McCoy13]

Homework Statement

A 13-µF capacitor and a capacitor of unknown capacitance are both charged to 4.00 kV. After charging, the two capacitors are disconnected from the voltage source. The capacitors are then connected to each other--positive plate to negative plate and negative plate to positive plate. The final voltage across the terminals of the 13-µF capacitor is 1.00 kV.

Homework Equations

There are a couple things I assumed about the scenario. First, the initial arrangement is immaterial (since it's not described I think this is a safe assumption). Second, once the capacitors are connected to each other, their voltages are the same. So not only is the 13-µF capacitor at 1kV but so is the unknown capacitor. Third, charge is conserved in the circuit. So my relevant equation would be:

C=Q/V where V is the same for both capacitors. Therefore we only need to find the charge.

The Attempt at a Solution

We can easily compute the charge on the first capacitor before the two are connected. It is 52mC. Now for it to drop from 4kV to 1kV this corresponds with a drop to 13mC in charge on the known capacitor. So I set up the following system of equations:

52mC + a = Q
13mC + b = Q
where a and b are the charge held by the unknown capacitor before (a) and after (b) the connection.

We also know:
a/4000kV = C
b/1000kV = C
where C is the capacitance of the unknown capacitor.

Now the first time I tried solving it I assumed the second capacitor had a charge of 39mC and just stuck it in the equation to get 9.75µF, but this is incorrect. When I solved it using substitution with the equations I've given I got 13µF which is obviously also wrong. I feel completely lost.

 

[LowlyPion]

Homework Statement

A 13-µF capacitor and a capacitor of unknown capacitance are both charged to 4.00 kV. After charging, the two capacitors are disconnected from the voltage source. The capacitors are then connected to each other--positive plate to negative plate and negative plate to positive plate. The final voltage across the terminals of the 13-µF capacitor is 1.00 kV.

Homework Equations

There are a couple things I assumed about the scenario. First, the initial arrangement is immaterial (since it's not described I think this is a safe assumption). Second, once the capacitors are connected to each other, their voltages are the same. So not only is the 13-µF capacitor at 1kV but so is the unknown capacitor. Third, charge is conserved in the circuit. So my relevant equation would be:

C=Q/V where V is the same for both capacitors. Therefore we only need to find the charge.

The Attempt at a Solution

We can easily compute the charge on the first capacitor before the two are connected. It is 52mC. Now for it to drop from 4kV to 1kV this corresponds with a drop to 13mC in charge on the known capacitor. So I set up the following system of equations:

52mC + a = Q
13mC + b = Q
where a and b are the charge held by the unknown capacitor before (a) and after (b) the connection.

We also know:
a/4000kV = C
b/1000kV = C
where C is the capacitance of the unknown capacitor.

Now the first time I tried solving it I assumed the second capacitor had a charge of 39mC and just stuck it in the equation to get 9.75µF, but this is incorrect. When I solved it using substitution with the equations I've given I got 13µF which is obviously also wrong. I feel completely lost.

If you write the equation correctly you can easily solve.
Q1 = 4kV*13uf
Q2 = 4kV*C2

When you reverse the Caps you get the further equation that
Q1 - Q2 = 1kV*(C1+C2)

Since you know Q1 and C1 and you have Q2 in terms of C2 then all you have to do is solve.

 

[McCoy13]

Thanks.