Question

The electric potential on the surface of a sphere of radius $R$ due to a charge $3 \times 10^{-6} C$ is $500 \,V$. The intensity of electric field on the surface of the sphere in $\left( N C ^{-1}\right)$ is $\left[\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \,N \,m ^{2} C ^{-2}\right]$

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

Answer 1

Answer: (A)

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