A 1.0-μF capacitor and a 2.0-μF capacitor are connected in series across a 1200-V source. The charge on each capacitor is

Question

Quizplus · Accepted Answer

The voltage across each capacitor will be different as they have different capacitances. Let the voltage across the 1.0 μF capacitor be V1 and the voltage across the 2.0 μF capacitor be V2. We know that the total voltage across the two capacitors is 1200 V, so we can write the equation:
V1 + V2 = 1200
We also know that the charge on each capacitor is given by Q = CV, where C is the capacitance and V is the voltage across the capacitor. So the charge on the 1.0 μF capacitor is Q1 = (1.0 μF)(V1) and the charge on the 2.0 μF capacitor is Q2 = (2.0 μF)(V2). 
Since the capacitors are in series, the charge on each capacitor must be the same. So we can write:
Q1 = Q2
Substituting Q = CV into this equation and rearranging, we get:
C1V1 = C2V2
We can solve these two equations for V1 and V2 to get: 
V1 = 800 V and V2 = 400 V
The charges on each capacitor are therefore:
Q1 = (1.0 μF)(800 V) = 0.80 mC
Q2 = (2.0 μF)(400 V) = 0.80 mC
So the answer is B.

A 10-μF Capacitor and a 2