Q. A piece of cloud having area $25 \times 10^{6}m^{2}$ and electric potential of $10^{5}$ volts. If the height of cloud is $0.75\ km,$ then energy of electric field between earth and cloud will be
A) $250\ J$
B) $750\ J$
C) $1225\ J$
D) $1475\ J$
Explanation
(d) Energy $= \frac{1}{2}\varepsilon_{0}E^{2} \times (A \times d) = \frac{1}{2}\varepsilon_{0}\left( \frac{V^{2}}{d^{2}} \right)Ad$
$= \frac{1}{2} \times \frac{8.85 \times 10^{- 12} \times \left( 10^{5} \right)^{2} \times 25 \times 10^{6}}{0.75 \times 10^{3}} = 1475J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.