Question

A solid conducting sphere of radius $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{cm}$has a charge of$\mathbf{30}\mathbf{nc}$ distributed uniformly over its surface. Let $\mathbf{A}$ be a point$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{cm}$ from the center of the sphere,$\mathbf{S}$ be a point on the surface of the sphere, and $\mathbf{B}$be a point$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{cm}$ from the center of the sphere.What are the electricpotential differences

(a) ${\mathit{V}}_{\mathbf{S}}\mathbf{-}{\mathit{V}}_{\mathbf{B}}$and

(b) ${\mathit{V}}_{\mathbf{A}}\mathbf{-}{\mathit{V}}_{\mathbf{B}}$?

Short answer

The electric potential difference between the points S and B is $V_S-V_B=3600\text{V}$.

The electric potential difference between the points A and B is $V_A-V_B=3\text{600V}$ .

Step-by-step solution

Step 1: Given data:

The charge, $q=30nC$

The coulomb’s constant, $\frac{1}{4\pi {\epsilon }_o}=k=9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$C

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

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

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

Determining the concept:

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

Here,${\mathbf{\epsilon }}_0$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:

$V=\frac{q}{4\pi {\epsilon }_{0}R}$

Where, V is electric potential and R is radius.

(a) Determining thepotential at the surface of the sphere:

Potential at the surface of the sphere:

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

Potential at point A ,

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

Potential at point B ,

$\begin{array}{c}{V}_B=\frac{q}{4\pi {\epsilon }_o{R}_B}\\ =\frac{\left(\text{30nC}\right)\left(\frac{\text{1C}}{{10}^{-9}\text{nC}}\right)(9\times {10}^9\text{N}/\text{C})}{\left(\text{5.0cm}\right)\left(\frac{\text{1m}}{{\text{10}}^{\text{2}}\text{cm}}\right)}\\ =5\text{,400V}\end{array}$

(b) Determining theelectric potential difference between the points  S and  B :

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 $V_S-V_B=3\text{600V}$ .

(c) Determining the electric potential difference between the points  A and B  :

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 $V_A-V_B=\text{3600V}$ .