Calculate the energy stored in a 20\(\mu F\) capacitor. If the p.d between the plates is 40V
- 3.2 x 10-2J
-
1.6 x 10-2J
- 8.0 x 10-4J
- 4.0 x 10-4J
The correct answer is: B
Explanation
W = \(\frac{cv^2}{2} = \frac{20 \times 10^{-6} \times 40^{2}}{2}\)= 1.6 x 10-2J