Question

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

Answer 1

Step 1: Given data

The radius of the sphere is $r$.

The potential of the sphere is $V$.

Step 2: The outward pull on a charged sphere

The outward pull is the force exerted on a charge due to the electric field.

So, the outward pull of a sphere is defined by the form, $F=QE$, where, E is the electric field and Q is the total charge in the sphere.

So, pull per unit area is $F_u=\frac{QE}{4{\mathrm{\pi r}}^2}=E.\sigma$, where, $4{\mathrm{\pi r}}^2$ is the area of the surface and $\sigma$ is the charge per unit area.

Step 3: Electric field on a charged sphere

The electric field on a sphere of radius r is given by $E=\frac{\sigma }{{\epsilon }_o}$, where, $\sigma$ is the charge per unit area in the sphere and ${\epsilon }_o$is the permittivity of free space.

The relation between potential V and electric field E is $E=\frac{V}{r}$, where, r is the radius of the sphere.

Step 4: Finding the pull per unit area

As we know, pull per unit area $F_u=E.\sigma$.

So,

$$\begin{gathered} F_u=\sigma .E={\epsilon }_{o}E.E={\epsilon }_o{E}^2 \\ or{F}_u={\epsilon }_o{\left(\frac{V}{r}\right)}^2 \\ or{F}_u={\epsilon }_o\frac{V^2}{r^2} \end{gathered}$$

Therefore, the outward pull per unit area of its surface is ${\epsilon }_o\frac{V^2}{r^2}$.