Q. Two capacitors each of $1\mu F$ capacitance are connected in parallel and are then charged by $200\ volts\ d.c.$ supply. The total energy of their charges (in joules) is
A) 0.01
B) 0.02
C) 0.04
D) 0.06
Explanation
(c) $U = \frac{1}{2}CV^{2} = \frac{1}{2} \times 2 \times (200)^{2} \times 10^{- 6} = 0.04J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.