Question

The electrical 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 is $ \left[ \frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right] $ $ (in\text{ }N{{C}^{-1}}) $ :

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

Answer 1

Answer: (A)

given $V=500 v$
$
\begin{array}{l}
\Rightarrow \frac{ kq }{ r }=500 volts \\
\Rightarrow r =\frac{9 \times 10^{9} \times 3 \times 10^{-6}}{500}=\frac{27 \times 1000}{500}=54
\end{array}
$
we know $E=\frac{k q}{r^{2}}$
$
\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}
$