Q. A sphere of radius $1\ cm$ has potential of $8000\ V,$ then energy density near its surface will be
A) $64 \times 10^{5}\ J/m^{3}$
B) $8 \times 10^{3}\ J/m^{3}$
C) $32\ J/m^{3}$
D) $2.83\ J/m^{3}$
Explanation
(d) Energy density $u_{e} = \frac{1}{2}\varepsilon_{0}E^{2} = \frac{1}{2} \times 8.86 \times 10^{- 12} \times \left( \frac{V}{r} \right)^{2}$
$= 2.83\ J/m^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.