Question

The potential inside a point in a solid sphere will be:
A) Same as that seen at the surface
B) Will be less than what was seen at the surface
C) Will be more than what was seen at the surface
D) Will be equal to the potential at the centre

Answer 1

Hint: This problem can be solved by using the direct formula for the electric potential at a point inside the sphere in terms of the distance of the point from the centre of the sphere, the total charge of the sphere and the radius of the sphere.

Formula used:
$V=\dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{x}^{2}} \right)$ $\left( x\le R \right)$

Complete step by step answer:
We will use the direct formula for the electric potential inside a point of a solid sphere to compare the value for the electric potential for an inside point with that on the surface and at the centre.
The electric potential $V$ for a point inside a solid sphere of radius $R$ and total charge $Q$ is given by
$V=\dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{x}^{2}} \right)$ $\left( x\le R \right)$ --(1)
Where $x$ is the distance of the point from the centre of the sphere and $K=9\times {{10}^{9}}kg.{{m}^{3}}{{s}^{-2}}{{C}^{-2}}$ is the universal electric constant.
For the electric potential at the centre ${{V}_{centre}}$, $x=0$.
Putting this in (1), we get
${{V}_{centre}}=\dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{0}^{2}} \right)=\dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-0 \right)=\dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}} \right)=\dfrac{3KQ}{2R}>\dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{x}^{2}} \right)$
Hence, the potential at the centre is greater than the potential at any other inside point.
Now, for the electric potential at the surface ${{V}_{surface}}$, $x=R$.
Putting this in (1), we get
${{V}_{surface}}=\dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{R}^{2}} \right)=\dfrac{KQ}{2{{R}^{3}}}\left( 2{{R}^{2}} \right)=\dfrac{KQ}{R} < \dfrac{KQ}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{x}^{2}} \right)$
Hence, the electric potential at the surface is lesser than that at any inside point.
Therefore, the correct option is C) Will be more than what was seen at the surface.

Note: The same result holds for the magnitude of gravitational potential also. This is because Coulomb’s law of electrostatic attraction’ mathematical form and Newton’s law of gravitation’s mathematical form are similar in nature and both are inverse square laws that vary inversely with the distance from the centre of a body. In fact the mathematical formula of (1) can be written for the magnitude of the gravitational potential by replacing the charge $Q$ with the mass $M$ of the body and the universal electric constant $K$ with the universal gravitational constant $G=6.67\times {{10}^{-11}}N.{{m}^{2}}k{{g}^{-2}}$.