Two capacitors with capacitances of 2 μF and 0.4 μF,respectively,are connected in series.The system is connected to a 100-V battery.What electrical potential energy is stored in the 2-μF capacitor? A) 10.0 × 10−³ J B) 2.0 × 10−³ J C) 3.3 × 10−⁴ J D) 2.8 × 10−⁴ J E) 0.56 × 10−⁴ J

Question

Quizplus · Accepted Answer

First, we need to find the equivalent capacitance of the system:
$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2\mu F} + \frac{1}{0.4\mu F} = \frac{5}{2\mu F}$$
$$C_{eq} = \frac{2\mu F}{5} = 0.4\mu F$$

Since the capacitors are connected in series, the charge on each capacitor is the same. Therefore, we can use the formula for the electrical potential energy stored in a capacitor:
$$U = \frac{1}{2}CV^2$$
Substituting the known values:
$$U_{2\mu F} = \frac{1}{2}(2\mu F)(100V)^2 = 10	imes10^{-13}J$$

Therefore, the answer is (D) 2.8 × 10−14 J.

Two Capacitors with Capacitances of 2 μF and 0