Question

A 11.8 uF capacitor is fully charged across a 12.0 V battery. The capacitor is then disconnected from the battery and connected across an initially uncharged capacitor, C. The resulting voltage across each capacitor is 2.66 V. What is the capacitance C?

Answer 1

Use the formula Q = CV.

We can get the charge on the first capacitor: CV = (11.8 * 10^-6 F)(12.0 V) = 1.42 * 10^-4 C.

Now that charge is shared among two capacitors, and since we know the new voltage is 2.66 V, we can figure the new capacitance: C = Q/V = (1.42 * 10^-4 C)/(2.66 V) = 5.32 * 10^-5 F.

The formula for capacitors in parallel is to just add their individual capacitances. So we know:

C + (11.8 * 10^-6 F) = 5.32 * 10^-5 F

C = 4.14 * 10^-5 F or 41.4 μF