A 8.1 mu F capacitor is charged by a 155V battery (see the figure a) and then is disconnected from the battery. When this capacitor C_1 is then connected (see the figure b) to a second (initially uncharged) capacitor, C_2, the final voltage on each capacitor is 20V. What is the value of C_2?
Q = CV
The Attempt at a Solution
(8.1*10^-6 F)(155 V) = 1.256 * 10^-3 C
1.256*10^-3 C / 20 V = 6.3*10^-5 F
I did it this way because I thought charge was conserved.