Solve 10.0 µF Capacitor Problem: Find Value of C

[SamTsui86]

Homework Statement

A 10.0 µF capacitor is fully charged across a 10.0 V battery. The capacitor is then disconnected from the battery and connected across an initially uncharged capacitor with capacitance C. The resulting voltage across each capacitor is 4.00 V. What is the value of C?

Homework Equations

V= Q/C

The Attempt at a Solution

My logic is that since the voltage changes, it is connected in series, so the charge is same. So I solved for Q for the first one and used it to plug solve for C for the second one. It is wrong, please help.

 

[berkeman]

Yes, the charge is conserved, it just gets distributed across the two capacitors insead of the one. What is the value of charge you got for the first single cap?

But also don't think of the caps as being in series in this problem. They are in parallel for the purposes of figuring out what the final voltage is...

 

[SamTsui86]

The charge for the first single cap is .0001 C

 

[SamTsui86]

what's next?

 

[berkeman]

You tell me. When two capacitors are connected in parallel, and they have that charge distributed across them, and the resulting voltage is 4V, then what is the total capacitance of the parallel combination? And what does that tell you for the value of the 2nd capacitor?

 

[SamTsui86]

since the voltage is 4 V and the charge is .0001 then the total capacitance would be .000025 F. Since it's parallel C eq = C1 +C2. so C2 s 15 uF. I got it now, thank you.