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. $4\pi {\epsilon }_{0}Q\times {10}^{22}V/m$
• B. $127\pi {\epsilon }_{0}Q\times {10}^{20}V/m$
• C. $4\pi {\epsilon }_{0}Q\times {10}^{20}V/m$
• D. $12\pi {\epsilon }_{0}Q\times {10}^{22}V/m$

Answer 1

Answer: (A)

$V=Q\times {10}^{11}$

$E=?$

$E=\frac{V}{r}=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

$V=\frac{Q}{4\pi {\epsilon }_{0}r}=Q\times {10}^{11}$

$r=\frac{Q}{4\pi {\epsilon }_0\times Q\times {10}^{11}}$

$r=\frac{1}{4\pi {\epsilon }_0\times {10}^{11}}$

$E=\frac{V}{r}=\frac{Q\times {10}^{11}}{\frac{1}{4\pi {\epsilon }_0\times {10}^{11}}}=4\pi {\epsilon }_{0}Q\times {10}^{22}V/m$