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## 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/C

_{1}+1/C

_{2}

## 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/C

_{1}+1/C

_{2}

Q = C

_{eq}* V

7.35E-5 C = (1/(4.6E-6 F) +1/C

_{2})*7 V

I solved for C

_{2}...

**C**

_{2}= -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.