Q. A $6\mu F$ capacitor is charged from $10\ volts$ to $20\ volts.$ Increase in energy will be
A) $18 \times 10^{- 4}\ J$
B) $9 \times 10^{- 4}\ J$
C) $4.5 \times 10^{- 4}\ J$
D) $9 \times 10^{- 6}\ J$
Explanation
(b) $\mathrm{\Delta}E = E_{Final} - E_{Initial} = \frac{1}{2}C\left( V_{Final}^{2} - V_{Initial}^{2} \right)$
$= \frac{1}{2} \times 6 \times (20^{2} - 10^{2}) \times 10^{- 6}$
$= 3 \times (400 - 100) \times 10^{- 6} = 3 \times 300 \times 10^{- 6} = 9 \times 10^{- 4}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.