If a capacitor stores an energy of 0.2 J, what should be its capacitance when it is powered by a 110 V battery?
A
57 μF
B
33 μF
C
70 μF
D
45 μF
Verified Correct Answer: B.
Physical Principles & Numerical Solution
The energy stored in a capacitor when V and E are given: to find C, we use
E = \(\frac{1}{2}CV^2\)
0.2 = \(\frac{1}{2}C(110)^2\)
0.2 = 6050C
C = \(\frac{0.2}{6050}\) = 3.306 \times 10\(^{−5}\) ≈ 33 \muF
Source: Myschool Past Questions Archive (Record #78271)