Solve "Charging a Capacitor" Homework Problem

[clickcaptain]

Homework Statement

A 4.6 mu or micro FF capacitor is charged to a potential difference of 15.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 4.6 mu or micro FF capacitor then drops to 7 V. What is the capacitance of the second capacitor?

Homework Equations

Q = CV

Capacitor in series
1/C_eq= 1/C₁+1/C₂

The Attempt at a Solution

I found the charge of the capacitor in the first scenario when the potential difference is 15 V

Q = CV
= 4.6E-6 F *15 V
= 7.35E-5 C

Since the potential drops to 7 V, you know the capacitors are in a series so that means the charge remains the same on both capacitors but potential changes

Capacitor in series
1/C_eq= 1/C₁+1/C₂

Q = C_eq * V

7.35E-5 C = (1/(4.6E-6 F) +1/C₂)*7 V

I solved for C₂...

C₂ = -4.6E-6 C

- which is negative so its obvious that its wrong, but could someone tell me where I went wrong that would be greatly appreciated. Thank you.

 

[w3390]

I think there is an error in your calculation of the charge on the first capacitor. It seems that your work is correct, but when I multiply (4.6 microF)*(15V) I get the charge to be
6.9e-5 C. Since you know that the voltage drops to 7V across the first capacitor, you can treat the circuit as if there is initially a 7V battery connected to only the second capacitor, which is unknown. In this case, you know that V=7V and Q= 6.9e-5C. From here you can manipulate the equation so that C=(Q/V). Therefore, C should equal 9.9 microF.

 

[clickcaptain]

Your way makes sense, and its a lot simpler than what I tried. The answer didn't work though so I'm going to try to keep double checking figures, i can't figure out what else would be wrong.

 

[mplayer]

Aren't the capacitors connected in parallel? And isn't the amount of charge calculated in the first part the total amount of charge in the 2-capacitor system in part two?

 

[mplayer]

clickcaptain,
I'm pretty sure you were doing it correctly in your first post (except for the first charge calculation, I got 6.9*10^-5 C like w3390). Draw out the 2-capacitor circuit in the way it's described and I think you'll see that they are in parallel. When calculating C₂, your C_eq should be C_eq = (C₁ + C₂) and not 1/C_eq = (1/C₁) + (1/C₂).

Are you arriving at the correct solution now?