At t = 0 the switch S is closed with the capacitor uncharged. If C = 40 μF, ε = 50 V, and R = 5.0 kΩ, how much energy is stored by the capacitor when I = 2.0 mA?
A) 20 mJ
B) 28 mJ
C) 32 mJ
D) 36 mJ
E) 40 mJ

Question

At t = 0 the switch S is closed with the capacitor uncharged. If C = 40 μF, ε = 50 V, and R = 5.0 kΩ, how much energy is stored by the capacitor when I = 2.0 mA? 
A) 20 mJ 
B) 28 mJ 
C) 32 mJ 
D) 36 mJ 
E) 40 mJ

Quizplus · Accepted Answer

The energy stored by a capacitor can be calculated using the formula:
$$ E = \frac{1}{2} CV^{2} $$
where C is the capacitance and V is the voltage across the capacitor. Since the current I is given, we can find the voltage across the capacitor using Ohm's Law:
$$ V = IR $$
Substituting the values given in the problem:
$$ V = (2.0	imes10^{-3}~\mathrm{A})(5.0	imes10^{3}~\Omega) = 10~\mathrm{V} $$
Now we can calculate the energy stored:
$$ E = \frac{1}{2}(40	imes10^{-6}~\mathrm{F})(10~\mathrm{V})^{2} \approx \boxed{\mathrm{(C)~32~mJ}} $$

At T = 0 the Switch S Is Closed with the Capacitor