Question

(a) If an isolated conducting sphere$10 \text{cm}$in radius has a net charge of$4.0\mu C$and if$V=0$at infinity, what is the potential on the surface of the sphere? (b) Can this situation actually occur, given that the air around the sphere undergoes electrical breakdown when the field exceeds$3.0MV/m$?

Short answer

a) The potential on the surface of the sphere is, $3.6\times {10}^5 \text{V}$.

b) The situation is not possible

Step-by-step solution

Step 1: Identification of the given data

For the isolated conducting sphere

Its radius $r=10 \text{cm}$and its charge $q=4.0\mu C$, $V=0$at infinity.

Understanding the concept

Electric Potential is given by,$V=\frac{1}{4\pi {\epsilon }_o}·\frac{q}{r}$

(a) Calculate the potential on the surface of the sphere

The electric potential is expressed as,

$V=\frac{1}{4\pi {\epsilon }_o}·\frac{q}{r}$

Substitute all the value in the above equation.

$$\begin{gathered} V=\left(9.0\times {10}^{9}N.m^2/C^2\right)\times \frac{4.0\times {10}^{-6} \text{C}}{0.10 \text{m}} \\ V=3.6\times {10}^5 \text{V} \end{gathered}$$

Hence the potential on the surface of the sphere is, $3.6\times {10}^5 \text{V}$.

(b) Find out if this situation canactually occur, given that the air around the sphere undergoes electrical breakdown when the field exceeds 3.0MV/m

The field just outside the sphere is expressed as,

$$\begin{gathered} E=\frac{1}{4\pi {\epsilon }_o}·\frac{q}{r^2} \\ E=\frac{V}{r} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} E=\frac{3.6\times {10}^5 \text{V}}{0.10 \text{m}} \\ =3.6\times {10}^{6}V/m \end{gathered}$$

Hence the value is exceeds$3.0MV/m$, so this situation cannot occur.