Two capacitors with capacitances of 1.0 μF and 0.35 μF,respectively,are connected in parallel.The system is connected to a 40-V battery.What electrical potential energy is stored in the 1.0-μF capacitor? A) 0.1 × 10−³ J B) 0.3 × 10−³ J C) 0.8 × 10−³ J D) 1.1 × 10−³ J E) 2.8 × 10−³ J

Question

Quizplus · Accepted Answer

The total capacitance of the system in parallel is C1 + C2 = 1.0 μF + 0.35 μF = 1.35 μF. 
The charge stored in each capacitor can be found using Q = CV, where V is the voltage across the system (40V in this case): 
Q1 = (1.0 μF)(40V) = 40 μC 
Q2 = (0.35 μF)(40V) = 14 μC 
The total charge stored in the system is the sum of the charges stored in each capacitor: 
Qtot = Q1 + Q2 = 54 μC 
The electrical potential energy stored in a capacitor is given by U = 0.5CV^2, so we can find the energy stored in the 1.0 μF capacitor: 
U1 = 0.5(1.0 μF)(40V)^2 = 800 μJ 
Therefore, the answer is C, 0.8 × 10^(-6) J.

Two Capacitors with Capacitances of 1