Question

Fig. shows equipotential surfaces concentric at O, the magnitude of electric field at a distance r measured from O is.

Answer 1

Correct option is B. $$ \frac {6}{r^2}(Vm^{-1}) $$
Equipotential Lines are concentric circles, These pattern of equipotential lines should be due to point charge at the centre of circles.
Let us consider few equipotential lines,
$$ 20= \frac{1}{4 \pi \varepsilon_0} \frac{q}{30 \times 10^{-2}} \Rightarrow \frac{q}{4 \pi \varepsilon_0}=6 $$
$$ 60= \frac{1}{4 \pi \varepsilon_0} \frac{q}{10 \times 10^{-2}} \Rightarrow \frac{q}{4 \pi \varepsilon_0}=6 $$
For point charge, electric field should be in radial direction.
The value is $$ E= \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}= \frac{6}{r^2}(Vm^{-1}) $$