Question

The electric potential at a point in free space due to a charge $Q$ coulomb is $Q\times {10}^{11}V$. The electric field at that point is

• A. $12\pi {\epsilon }_{0}Q\times {10}^{22}V{m}^{-1}$
• B. $4\pi {\epsilon }_{0}Q\times {10}^{22}V{m}^{-1}$
• C. $12\pi {\epsilon }_{0}Q\times {10}^{20}V{m}^{-1}$
• D. $4\pi {\epsilon }_{0}Q\times {10}^{20}V{m}^{-1}$

Answer 1

Answer: (B)

$4\pi {\epsilon }_{0}Q\times {10}^{22}V{m}^{-1}$

Here $V=\frac{Q}{4\pi {\epsilon }_{0}r}=Q\times {10}^{11}$
$\therefore 4\pi {\epsilon }_{0}r={10}^{-11}...\left(i\right)$
Now, $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=\frac{Q\times 4\pi {\epsilon }_0}{(4\pi {\epsilon }_0{r)}^2}=\frac{Q\times 4\pi {\epsilon }_0}{({10}^{-11}{)}^2}=4\pi {\epsilon }_{0}Q\times {10}^{22}V{m}^{-1}$ (Using (i))