Question

A spherical conductor of radius $10\text{cm}$ has a charge of $3.2\times {10}^{-7}\text{C}$ distributed uniformly. What is the magnitude of electric field at point $15\text{cm}$ from centre of sphere$?$
$(\frac{1}{4\pi {ϵ}_0}=\text{9}\times {\text{10}}^{\text{9}}\text{N}{\text{m}}^2/{\text{C}}^2)$

• A. $1.28\times {10}^6\text{N/C}$
• B. $1.28\times {10}^7\text{N/C}$
• C. $1.28\times {10}^4\text{N/C}$
• D. $1.28\times {10}^5\text{N/C}$

Answer 1

The correct option is D $1.28\times {10}^5\text{N/C}$

For any spherical conductor, if it has any charge, the charge will reside on the surface of the conductor.

Given, radius of conductor is $10\text{cm}$ and we have to find electric field at distance of $15\text{cm}$.

So, we can assume the sphere as point charge to calculate the electric field at distance of $15\text{cm}$.

$Q=3.2\times {10}^{-7}\text{C}$
$r=15\text{cm}=0.15\text{m}$

$\text{|E|}=\frac{1}{4\pi {ϵ}_0}\times \text{Q}\times \frac{1}{r^2}$
$\text{|E|}=\text{9}\times {\text{10}}^{\text{9}}$$\times$$3.2\times {10}^{-7}$ $\times$$\frac{1}{{\text{0.15}}^2}$
$\text{|E|}=1.28\times {10}^5\text{N/C}$