Question

A solid sphere of radius $R$ is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the center?
A. $R$
B. $\dfrac{R}{2}$
C. $\dfrac{R}{3}$
D. $2R$

Answer 1

electrostatic potential is defined as the amount of work done to move a point charge from the reference point to the given point. Here, we will use the expression of potential at the center of the sphere and any point in the sphere. Now, to calculate the distance from the surface we will subtract the electric potential at a distance $R$ from the center from the electric potential at a distance $x$ from the center.

Formula used:
The expression of the electrostatic potential at the center of the sphere is given by
$V = \dfrac{3}{2}\dfrac{{KQ}}{R}$
Here, $V$ is the electrostatic potential, $K$ is the constant of proportionality, $Q$ is the charge, and $R$ is the radius of the sphere.
Also, the expression of the electrostatic potential at a distance $x$ is given by
$V = \dfrac{{KQ}}{x}$

Complete step by step answer:
Consider a solid sphere which is of radius $R$ and this sphere is uniformly charged.
Now, electrostatic potential is defined as the amount of work done to move a point charge from the reference point to the given point.
Now, the potential at the center of the sphere is given by
${V_c} = \dfrac{3}{2}\dfrac{{KQ}}{R}$
Also, we will calculate the potential at some point which is at a distance $x$ from the center.
Therefore, the potential of a charged particle at a distance $x$ from its center is given by
${V_x} = \dfrac{{KQ}}{x}$
Now, as given in the question, we have to calculate the electrostatic potential at a point that is half of the potential at the center. Therefore, we will consider that the electrostatic potential at a distance $x$ is equal to half of the electrostatic potential at the center and is given by
$\dfrac{{KQ}}{x} = \dfrac{1}{2}\left( {\dfrac{3}{2}\dfrac{{KQ}}{R}} \right)$
$\Rightarrow \,\dfrac{1}{x} = \dfrac{3}{{4R}}$
$\Rightarrow \,x = \dfrac{4}{3}R$
Now, to calculate the distance of electrostatic potential from the surface $X$ , we will use the condition which is shown below
$X = x - R$
$\Rightarrow \,X = \dfrac{4}{3}R - R$
$\Rightarrow \,X = \dfrac{1}{3}R$
Therefore, the distance of the electrostatic potential from the surface is $\dfrac{1}{3}R$ .

Note:
Here the student should not get confused about the distance $x$ from the center and the distance $X$ from the surface. Just remember that $x$ is the distance of the point, where the electric potential is to be calculated, from the center and $X$ is the distance of that point from the surface. Also, the values of both the distances will be in terms of radius $R$ .