Question

Figure shows equipotential concentric surfaces centred at $\text{O}$. The magnitude of electric field at a distance $r$ measured from $\text{O}$ is:

• A. $\frac{9}{r^2}{\text{Vm}}^{-1}$
• B. $\frac{6}{r^2}{\text{Vm}}^{-1}$
• C. $\frac{2}{r^2}{\text{Vm}}^{-1}$
• D. $\frac{16}{r^2}{\text{Vm}}^{-1}$

Answer 1

The correct option is B $\frac{6}{r^2}{\text{Vm}}^{-1}$

Potential at distance $r$ from $\text{O}$, $V=\frac{KQ}{r}$
For first equipotential surface, $60=\frac{KQ}{{10}^{-1}}$
i.e $Q=\frac{6}{K}$

Now, electric field at distance $r$ from $\text{O}$
$E=\frac{KQ}{r^2}$
$E=\frac{K\left(\frac{6}{K}\right)}{r^2}$
$E=\frac{6}{r^2}{\text{Vm}}^{-1}$