Question

The electric potential on the surface of a sphere of radius $R$ due to a charge $3\times 10^{-6}C$ is $500V$. The intensity of electric field on the surface of the sphere in $\left(N{C}^{-1}\right)$ is $\left[\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2{C}^{-2}\right]$

• A. < 10
• B. > 20
• C. between 10 and 20
• D. < 5

Answer 1

Answer: (A)

Here, $V=\frac{q}{4\pi {\epsilon }_{0}R}=500V$;
$R=\frac{q}{4\pi {\epsilon }_0(500)}$
$E=\frac{q}{4\pi {\epsilon }_0{R}^2}$
$=\frac{q{\left[4\pi {\epsilon }_0(500)\right]}^2}{4\pi {\epsilon }_0\times q^2}$
$=\frac{4\pi {\epsilon }_0\times 25\times 10^4}{q}$
$E=\frac{25\times 10^4}{9\times 10^9\times 3\times 10^{-6}}$
$=9.26N{C}^{-1}$