The figure shows equipotential surfaces concentric at $O$. The magnitude of electric field at a distance $r$…
The figure shows equipotential surfaces concentric at $O$. The magnitude of electric field at a distance $r$ metres from $O$ is
- $\frac{9}{r^2} \mathrm{Vm}^{-1}$
- $\frac{16}{r^2} \mathrm{Vm}^{-1}$
- $\frac{2}{r^2} \mathrm{Vm}^{-1}$
- $\frac{6}{r^2} \mathrm{Vm}^{-1}$
Solution
Potential $v=\frac{k q}{r}$
For $30 \mathrm{~V}$ equipotential surface,
$\begin{aligned}
& 30=\frac{k q}{t} \text {, here } r=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m} \\
& 30=\frac{k q}{20 \times 10^{-2}} \\
& \therefore \quad \mathrm{kq}=60 \times 10 \times 10^{-2}=6
\end{aligned}$
Therefore, electric field at distance $r$ from charge $q$.
$\mathrm{E}=\frac{k q}{r^2}=\frac{6}{r^2} \mathrm{~V}^{-1}$
Asked in: AP EAMCET 2016
Practice more Electrostatics questions on Aicharya