Question

A spherical conductor of radius $R$ carries charge $Q$ on its surface. The value of electric field just outside the surface of the conductor is $E$. If the radius of the counductor is doubled while keeping the charge same, what will be the value of new electric field just outside the surface of the conductor?

• A. $E$
• B. $2E$
• C. $\frac{E}{4}$
• D. $\frac{E}{2}$

Answer 1

The correct option is C $\frac{E}{4}$

We know that electric field just outside the surface of a conductor is given by :
$E=\frac{\sigma }{{ϵ}_o}$

Surface charge density$\left(\sigma \right)$ for spherical surface is given by :
$\sigma =\frac{Q}{4\pi R^2}$

Therefore,
$E=\frac{Q}{4\pi {ϵ}_o{R}^2}$

When the radius of the sphere is doubled, the electric field just outside the surface of the sphere is given by :
$E^{\prime }=\frac{Q}{4\pi {ϵ}_o(2R{)}^2}$

$E^{\prime }=\frac{Q}{16\pi {ϵ}_o{R}^2}$

$\Rightarrow E^{\prime }=\frac{E}{4}$

Hence, option $\left(c\right)$ is the correct answer.