Q. A sphere of radius $r$ is charged to a potential V. The outward pull per unit area of its surface is given by
A) $\frac{4\pi\varepsilon_{0}V^{2}}{r^{2}}$
B) $\frac{\varepsilon_{0}V^{2}}{2r^{2}}$
C) $\frac{2\varepsilon_{0}V^{2}}{r^{2}}$
D) $\frac{\varepsilon_{0}V^{2}}{{4r}^{2}}$
Explanation
(b) Electrical pressure (force/area)
$\Rightarrow \ p = \frac{1}{2}\varepsilon_{0}E^{2}\ and\ E = \frac{V}{r}\ \therefore p = \frac{1}{2}\varepsilon_{0}\frac{V^{2}}{r^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.