The energy stored in a capacitor of capacitance 5\muF is 40J. Calculate the voltage applied across its terminals?
A
4000V
B
200V
C
16V
D
6V
E
4V
Verified Correct Answer: A.
Physical Principles & Numerical Solution
\(E = \frac{CV^{2}}{2}\)
\(V^{2} = \frac{2E}{C}\)
\(V^{2} = \frac{2\times 40}{5 \times 10^{-6}}\)
\(V^{2} = 16 \times 10^{6}\)
\(V = 4 \times 10^{3} = 4000v\)
Source: Myschool Past Questions Archive (Record #25095)