Q123 / 199 · NEET 2008 (Mains) · Physics
The electric potential at a point in free space due to a charge $\mathrm{Q}$ coulomb is $\mathrm{Q} \times 10^{11}$ volts. The electric field at that point is
- A$12 \pi \varepsilon_0 Q \times 10^{22}$ volt/m
- Correct answerB$4 \pi \varepsilon_0 Q \times 10^{22}$ volt/m
- C$12 \pi \varepsilon_0 \mathrm{Q} \times 10^{20} \mathrm{volt} / \mathrm{m}$
- D$4 \pi \varepsilon_0 \mathrm{Q} \times 10^{20}$ volt/m
Solution
$$
\begin{aligned}
V & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}=Q \times 10^{11} \text { volt } \\
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R^2}=\frac{V}{R}=Q \times 10^{11} \times 4 \pi \varepsilon_0 \times 10^{11}[\text { from ...(i) }] \\
& =4 \pi \varepsilon_0 Q \times 10^{22} \text { volt } / \mathrm{m}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.