91Ó°ÊÓ

The charge, $q=30nC$

The coulomb’s constant, $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}=k=9×{10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$C

A solid conducting sphere of radius,$R=\text{3cm}$ ,

The distance of point A ,$R=\text{3cm}$

The distance of point B , $R_B=\text{5.0cm}$

After reading the question ,the expression for the electric potential is,$\mathit{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{R}}$.

Here,${\mathbf{Î\mu }}_0$is the permittivity of the free space, q is the charge, and R is the distance.

The solid conducting is as shown in the following figure.

The electric field throughout the conducting volume is zero. Thus, the potential is constant and is equal to the potential on the surface of the charged sphere.

Formulae are as follow:

$V=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R}$

Where, V is electric potential and R is radius.

Potential at the surface of the sphere:

$\begin{array}{c}{V}_s=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{o}R}\\ =\frac{\left(\text{30nC}\right)\left(\frac{\text{1C}}{{\text{10}}^{\text{-9}}\text{nC}}\right)(\text{9}×{10}^{\text{9}}\text{N}/\text{C})}{\left(\text{3cm}\right)\left(\frac{\text{1m}}{{\text{10}}^{\text{2}}\text{cm}}\right)}\\ =\text{9,000V}\end{array}$

Potential at point A ,

$V_A=V_s=\text{9000V}$(because point A is inside the sphere.)

Potential at point B ,

$\begin{array}{c}{V}_B=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}_B}\\ =\frac{\left(\text{30nC}\right)\left(\frac{\text{1C}}{{10}^{−9}\text{nC}}\right)(9×{10}^9\text{N}/\text{C})}{\left(\text{5.0cm}\right)\left(\frac{\text{1m}}{{\text{10}}^{\text{2}}\text{cm}}\right)}\\ =5\text{,400V}\end{array}$

The electric potential difference between the points S and B is,

$\begin{array}{c}{V}_S−V_B=\text{9,000V}−\text{5400V}\\ =\text{3600V}\end{array}$

Hence, the electric potential difference between the points S and B is $V_S−V_B=3\text{600V}$ .

The electric potential difference between the points A and B is,

$\begin{array}{c}{V}_A−V_B=9,\text{000V}−\text{5400V}\\ =\text{3600V}\end{array}$

Hence, the electric potential difference between the points A and B is $V_A−V_B=\text{3600V}$ .