At distance r from a point charge- the ratio uv2 where u is energy density and v is potential is best represented by

We know that u = 1/2ϵ_0 E^2 (where E is the electric field due to point charge) u = 1/2ϵ_0 (kq_1/r^2)^2 u = 1/2ϵ_0 k^2q^2/r^4 Potential v due to point charge at ...

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Byju's Answer

Standard XIII

Physics

Electric Potential at a Point in Space Due to a Point Charge Q

At distance r...

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

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Solution

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
$v = \frac{k q}{r}$ ----(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).

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At distance r from a point charge, the ratio uv2 (where u is energy density and v is potential) is best represented by

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