Question

The electric field due to a charge at a distance of $3m$ from it is $500N/C$. The magnitude of the charge is$[\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2/C^2]$ :

• A. $2.5$ micro-coulomb
• B. $2.0$ micro-coulomb
• C. $1.0$ micro-coulomb
• D. $0.5$ micro-coulomb

Answer 1

The correct option is D $0.5$ micro-coulomb

We know $E$ due to charge $q$ $=\frac{kq}{r^2}$
$E=\frac{kq}{r^2}$
$\Rightarrow 500=\frac{(9\times {10}^9)\times q}{(3{)}^2}$

$\Rightarrow q=5\times {10}^{-7}C$
$\Rightarrow q=0\cdot 5\mu C$