Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $Q$ be charge,
Here energy density at a point ${}^{\prime }{r}^{\prime }\text{m}$ away from charge is
$u=\frac{1}{2}{ϵ}_0{E}^2=\frac{1}{2}\frac{{ϵ}_0{K}^2{Q}^2}{2r^2}$
Potential at a point ${}^{\prime }{r}^{\prime }\text{m}$ away from charge is
$v=\frac{KQ}{r}$
$\frac{u}{v^2}=\frac{\frac{1}{2}{ϵ}_0{K}^2\frac{Q^2}{r^4}}{\frac{K^2{Q}^2}{r^2}}=\frac{1}{2}\frac{{ϵ}_0}{r^2}$
$\therefore \frac{u}{v^2}\propto \frac{1}{r^2}$
so the correct option is B.

Why this Question? Energy density is the computation of the amount of energy that can be stored in a given system. So, the more the energy density of a system, the greater will be the amount of energy stored in it.