Two capacitors with capacitances of 1.0 μF and 0.50 μF,respectively,are connected in series.The system is connected to a 50-V battery.What electrical potential energy is stored in the 1.0-μF capacitor? A) 5.6 × 10−⁴ J B) 4.3 × 10−³ J C) 0.065 × 10−³ J D) 1.4 × 10−⁴ J

Question

Quizplus · Accepted Answer

The potential difference across a capacitor in a series connection is given by V=Q/C, where V is the potential difference, Q is the charge stored on the capacitor, and C is the capacitance. Since both capacitors have the same charge, we can find the potential difference across each capacitor by using V=Q/C.

For the 1.0 μF capacitor, V = Q/1.0μF = Q*10^6 C/V

For the 0.50 μF capacitor, V = Q/0.50μF = 2Q*10^6 C/V

The total potential difference across the series connection is V_total = V_1 + V_2.

Using V_total = 50V, we can solve for Q, which is the same for both capacitors:

Q = CV = (1.0 μF + 0.50 μF) × (50 V) = 75 μC

The potential energy stored in a capacitor is given by U=1/2CV^2.

For the 1.0 μF capacitor, U = 1/2(1.0 μF)(25 V)^2 = 0.31 × 10^-6 J

Therefore, the answer is D: 1.4 × 10^-14 J (rounded to two significant figures).

Two Capacitors with Capacitances of 1