Q. A variable condenser is permanently connected to a 100 V battery. If capacity is changed from 2 $\mu$F and 10 $\mu$F, then energy change is equal to
A) $2 \times 10^{- 2\ }$J
B) $2.5 \times 10^{- 2\ }$J
C) $6.5 \times 10^{- 2\ }$J
D) $4 \times 10^{- 2\ }$J
Explanation
(d) $E_{1} = \frac{1}{2}C_{1}V^{\begin{array}{r}
2 \\
\
\end{array}}$
$= \frac{1}{2} \times 2 \times 10^{- 6} \times 100^{2} = 0.01J$
$E_{1} = \frac{1}{2}C_{2}V^{\begin{array}{r}
2 \\
\
\end{array}}$
$\frac{1}{2} \times 10 \times 10^{- 6} \times (100)^{2} = 0.05$ J
Energy change $= E_{2} - E_{2}$
$= 0.05 - 0.01 = 0.04J = 4 \times 10^{- 2}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.