Question

At distance $r$ from a point charge, the ratio $\frac{u}{v^2}$ (where $u$ is energy density and $v$ is potential) is best represented by

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We know that $u=\frac{1}{2}{ϵ}_0{E}^2$
(where E is the electric field due to point charge)
$⟹u=\frac{1}{2}{ϵ}_0(\frac{k{q}_1}{r^2}{)}^2$
$⟹u=\frac{1}{2}{ϵ}_0\frac{k^2{q}^2}{r^4}$
Potential $v$ due to point charge at a distance $r$ is given by
$v=\frac{kq}{r}$----(i)
$⟹v^2=\frac{k^2{q}^2}{r^2}$----(ii)
Dividing (i) from (ii) we get,
$\frac{u}{v^2}=\frac{1}{2}{ϵ}_0\times \frac{1}{r^2}$
$⟹\frac{u}{v^2}\propto \frac{1}{r^2}$
The graph of this will be same as that of $y\propto \frac{1}{x^2}$ (i.e hyberbola).