Question

A spherical conductor of radius $3\text{m}$ carries a charge of $2\mu \text{C}$ on its surface. What is the value of electric field just outside the surface of the sphere?

• A. $2\times {10}^3\text{N/C}$
• B. $4\times {10}^3\text{N/C}$
• C. $2\times {10}^6\text{N/C}$
• D. $4\times {10}^6\text{N/C}$

Answer 1

The correct option is A $2\times {10}^3\text{N/C}$

Given :
Radius of sphere $\left(R\right)=3\text{m}$
Charge on the surface $\left(q\right)=2\mu \text{C}$

Surface charge density of the sphere,
$\sigma =\frac{q}{4\pi R^2}$

Therefore, electric field just outside the surface of the sphere is given by :
$E=\frac{\sigma }{{ϵ}_o}$

$E=\frac{q}{4\pi {ϵ}_o{R}^2}$

$E=\frac{1}{4\pi {ϵ}_o}\times \frac{2\times {10}^{-6}}{3^2}$

$E=9\times {10}^9\times \frac{2\times {10}^{-6}}{9}$

$\Rightarrow E=2\times {10}^3\text{N/C}$