Question

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

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

Answer 1

The correct option is B

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

Step 1: Given Data

The given charge is $Q$

The electric potential at a point is $Q\times {10}^{11}V$

Step 2: Formula Used

The formula of electric potential $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$, where $r$ is the distance of the point from the charge.

The value of $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9$ in SI units

Step 3: Finding $r$

upon substituting $Q\times {10}^{11}=9\times {10}^9\frac{Q}{r}$

⇒ $r=0.09m$

Step 4: Formula of electric field and calculation

Electric field $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

⇒ $E=9\times {10}^9\frac{Q}{{(9\times {10}^{-2})}^2}$

⇒ $E=Q\times \frac{{10}^{13}}{9}$

⇒ $E=\frac{Q\times {10}^{22}}{9\times {10}^9}$

⇒ $E=\frac{Q\times {10}^{22}}{{\displaystyle \frac{1}{4\pi {\epsilon }_0}}}$

⇒ $E=4\pi {\epsilon }_{0}Q\times {10}^{22}$

Hence, the correct answer is option B.