Question

A sphere of radius $r$ is charged to a potential $V$. The outward pull per unit area of its surface is given by:

• A. $\frac{4\pi {ϵ}_0{V}^2}{r^2}$
• B. $\frac{{ϵ}_0{V}^2}{r^2}$
• C. $\frac{2\pi {ϵ}_0{V}^2}{r^2}$
• D. $\frac{{ϵ}_0{V}^2}{4r^2}$

Answer 1

The correct option is B $\frac{{ϵ}_0{V}^2}{r^2}$

The pull per unit area is given as Electric field * Charge per unit area.

Let the charge on the sphere be Q.

Surface charge density = $\sigma =\frac{Q}{4\pi r^2}$

Electric field at the surface can be found from Gauss law

$E=\frac{\sigma }{{\epsilon }_0}$

where ${\epsilon }_0$ is the permitivity of free space.

The above equation can be written as $\sigma ={\epsilon }_{0}E$

We know that $E=\frac{V}{r}$

Pull per unit area $P=E\times \sigma$

$P=\frac{V}{r}\times \left(E{\epsilon }_0\right)$

$P=\frac{V}{r}\times \left(\frac{V}{r}{\epsilon }_0\right)$

$P={\epsilon }_0\frac{V^2}{r^2}$