Chapter 23, Problem 89P

(a)

To determine

The electric potential of each of the three shells.

Answer to Problem 89P

The electric potential $V\left(a\right)=V\left(b\right)$ is $kQ\left(\frac{1}{b}-\frac{1}{c}\right)$ and $V\left(c\right)=0$ .

Explanation of Solution

Given:

The inner shell is uncharged.

The middle shell has a positive charge $\text{'}+Q\text{'}$

The outer shell has a negative charge $\text{'}-Q\text{'}$

Formula used:

The expression for Gauss’s Law is given by,

  $E_r\left(4\pi r^2\right)=\frac{Q_{enc}}{{\epsilon }_0}$

The expression for potential is given by,

  $V=\frac{Kq}{r}$

Calculation:

The required diagram is shown in Figure 1.

  

Figure 1

Applying Gauss’s law to a spherical Gaussian surface of radius $r\ge c$ , the expression is written as,

  $\begin{array}{c}{E}_r\left(4\pi r^2\right)=\frac{Q_{enc}}{{\epsilon }_0}\\ =0\end{array}$

And $E_r=0$ because the net charge enclosed by the Gaussian surface is zero, which is expressed as,

  $\begin{array}{c}V={\displaystyle \int \overrightarrow{E}.\overrightarrow{dr}}\\ V\left(c\right)=0\end{array}$

Applying Gauss’s law to a spherical Gaussian surface of radius $b<r<c$ , the electric field is calculated as,

  $\begin{array}{c}{E}_r\left(4\pi r^2\right)=\frac{Q_{enc}}{{\epsilon }_0}\\ \text{and, }{E}_r\left(b<r<c\right)=\frac{kQ}{r^2}\end{array}$

The potential between $b$ and $c$ can be calculatedby integrating the expression for potential,

  $\begin{array}{c}V\left(b\right)-V\left(c\right)=-kQ{\displaystyle \underset{c}{\overset{b}{\int }}\frac{dr}{r^2}}\\ =kQ\left(\frac{1}{b}-\frac{1}{c}\right)\end{array}$

And since $V\left(c\right)=0$

Therefore,

  $V(b)=kQ\left(\frac{1}{b}-\frac{1}{c}\right)$

The inner shell too doesn’t carry charge, so the field between the points $a$ and $b$ is zero.

Conclusion:

Therefore, The electric potential $V\left(a\right)=V\left(b\right)$ is $kQ\left(\frac{1}{b}-\frac{1}{c}\right)$ and $V\left(c\right)=0$ .

(b)

To determine

The electric potential of each shell and the final charge on each shell.

Answer to Problem 89P

The electric potential of the shell $V\left(a\right)=V\left(c\right)$ is $0$ and $V\left(b\right)=kQ\frac{\left(c-b\right)\left(b-a\right)}{b^2\left(c-a\right)}$ . And the final charge on each shell is $-Q\frac{a\left(c-b\right)}{b\left(c-a\right)}$

Explanation of Solution

Calculation:

Given:

The inner shell is uncharged.

The middle shell has a positive charge $\text{'}+Q\text{'}$

The outer shell has a negative charge $\text{'}-Q\text{'}$

Formula used:

The expression for Gauss’s Law is given by,

  $E_r\left(4\pi r^2\right)=\frac{Q_{enc}}{{\epsilon }_0}$

The expression for potential is given by,

  $V=\frac{Kq}{r}$

Calculation:

When the inner and outer shells are connected their potentials become equal as a consequence of the distribution of charge.

The charges on the surface $a$ and $c$ are related according to,

  $Q_a+Q_c=-Q$ …… (1)

The value of $Q_b$ remains unchanged which is expressed as,

  $Q_b=Q$

The potential of shells a and c is given by,

  $\begin{array}{c}V\left(a\right)=V\left(c\right)\\ =0\end{array}$

The electric field between the points $a\text{and}b$ is $\frac{k{Q}_a}{r^2}$ and the potential at b is given by,

  $V\left(b\right)=k{Q}_a\left(\frac{1}{b}-\frac{1}{a}\right)$ …… (2)

The enclosed charge for $b<r<c\text{is}{Q}_a+Q$ . So the field in thie region as per the Gauss Law is written as,

  $E_{b<r<c}=\frac{k\left(Q_a+Q\right)}{r^2}$

The potential difference between b and ciscalculated as,

  $\begin{array}{c}V\left(c\right)-V\left(b\right)=k\left(Q_a+Q\right)\left(\frac{1}{c}-\frac{1}{b}\right)\\ =-V\left(b\right)\end{array}$

Solve further,

  $V\left(b\right)=k\left(Q_a+Q\right)\left(\frac{1}{b}-\frac{1}{c}\right)$ …… (3)

Equate equation (2) and (3).

  $Q_a=-Q\frac{a\left(c-b\right)}{b\left(c-a\right)}$ …… (4)

Substitute $-Q\frac{a\left(c-b\right)}{b\left(c-a\right)}$ for $Q_a$ in equation (1).

  $\begin{array}{l}{Q}_a+Q_c=-Q\\ -Q\frac{a\left(c-b\right)}{b\left(c-a\right)}+Q_c=-Q\\ Q_c=-Q\frac{c\left(b-a\right)}{b\left(c-a\right)}\end{array}$

The potential across $b$ is calculated as,

  $\begin{array}{c}V\left(b\right)=k\left(Q_a+Q\right)\left(\frac{1}{b}-\frac{1}{c}\right)\\ =k\left(-Q\frac{a\left(c-b\right)}{b\left(c-a\right)}+Q\right)\left(\frac{1}{b}-\frac{1}{c}\right)\\ =kQ\frac{\left(c-b\right)\left(b-a\right)}{b^2\left(c-a\right)}\end{array}$

Conclusion:

Therefore, the electric potential of the shell $V\left(a\right)=V\left(c\right)$ is $0$ and $V\left(b\right)=kQ\frac{\left(c-b\right)\left(b-a\right)}{b^2\left(c-a\right)}$ . And the final charge on each shell is $-Q\frac{a\left(c-b\right)}{b\left(c-a\right)}$