Question

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

• A. $250/27$
• B. $27/250$
• C. $250$
• D. $27$

Answer 1

Answer: (A)

$250/27$

given V=500v
$\Rightarrow \frac{kq}{r}=500volts$
$\Rightarrow r=\frac{9\times {10}^9\times 3\times {10}^{-6}}{500}=\frac{27\times 1000}{500}$ $=54$

$weknowE=\frac{kq}{r^2}$

$\Rightarrow E=\frac{9\times {10}^9\times 3\times {10}^{-6}}{54\times 54}$

$\Rightarrow E=\frac{27\times 250}{27}$

$\Rightarrow E=\frac{250}{27}$