Chapter 23, Problem 57P

(a)

To determine

The expression for the charge $dQ$ on a spherical shell of radius $r\text{'}$ and thickness $dr\text{'}$ .

Answer to Problem 57P

The expression for the charge $dQ$ on a spherical shell of radius $r\text{'}$ and thickness $dr\text{'}$ is $\frac{3Q}{R^3}r{\text{'}}^{2}dr\text{'}$ .

Explanation of Solution

Given Data:

The radius of a spherical shell is $r\text{'}$

The thickness of spherical shell is $dr\text{'}$

Formula used:

The expression for the charge $dQ$ is given as,

  $\begin{array}{c}dQ=pdV\text{'}\\ =pdA\text{'}dr\text{'}\end{array}$

Here, $dA$ is the area of a spherical shell which is given by,

  $dA=4\pi r{\text{'}}^2$

Calculation:

The expression for the charge $dQ$ on a spherical shell of radius $r\text{'}$ and thickness $dr\text{'}$ is calculated as,

  $\begin{array}{c}dQ=4\pi r{\text{'}}^{2}pdr\text{'}\\ =4\pi r{\text{'}}^2\left(\frac{Q}{\frac{4}{3}\pi R^3}\right)dr\text{'}\\ =\frac{3Q}{R^3}r{\text{'}}^{2}dr\text{'}\end{array}$

Conclusion:

Therefore, the expression for the charge $dQ$ on a spherical shell of radius $r\text{'}$ and thickness $dr\text{'}$ is $\frac{3Q}{R^3}r{\text{'}}^{2}dr\text{'}$ .

(b)

To determine

The expression for the potential $dV$ at radius $r$ due to charge on a shell of radius $r\text{'}$ and thickness $dr\text{'}$ .

Answer to Problem 57P

The expression for the potential $dV$ at radius $r$ due to charge on a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is $\frac{3kQ}{R^3}r\text{'}dr\text{'}$

Explanation of Solution

Formula used:

The expression for the potential in the interval of $r\le r\text{'}\le R$ due to charge $dQ$ is given as,

  $dV=\frac{kdQ}{r\text{'}}$

Calculation:

The expression for the potential is calculated as,

  $\begin{array}{c}dV=\frac{kdQ}{r\text{'}}\\ =\frac{k}{r\text{'}}\left(\frac{3Q}{R^3}r{\text{'}}^{2}dr\text{'}\right)\\ =\frac{3kQ}{R^3}r\text{'}dr\text{'}\end{array}$

Conclusion:

Therefore, the expression for the potential $dV$ at radius $r$ due to charge on a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is $\frac{3kQ}{R^3}r\text{'}dr\text{'}$

(c)

To determine

The potential at $r$ due to all the charge in the region farther than $r$ from the centre of the sphere.

Answer to Problem 57P

The potential at $r$ due to all the charge in the region farther than $r$ from the centre of the sphere is $\frac{3kQ}{2R^3}\left(R^2-r^2\right)$ .

Explanation of Solution

Formula used:

The expression for the potential $dV$ at radius $r$ due to charge on a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is given as,

  $dV=\frac{3kQ}{R^3}r\text{'}dr\text{'}$

Calculation:

Integrate the above expression in the interval of $r\text{'}=r$ to $r\text{'}=R$ ,

  $\begin{array}{c}V=\frac{3kQ}{R^3}{\displaystyle \underset{r}{\overset{R}{\int }}r\text{'}dr\text{'}}\\ =\frac{3kQ}{2R^3}\left(R^2-r^2\right)\end{array}$

Conclusion:

Therefore, the potential at $r$ due to all the charge in the region farther than $r$ from the centre of the sphere is $\frac{3kQ}{2R^3}\left(R^2-r^2\right)$ .

(d)

To determine

The expression for the potential $dV$ at $r$ due to charge in a shell of radius $r\text{'}$ and thickness $dr\text{'}$ .

Answer to Problem 57P

The expression for the potential $dV$ at $r$ due to charge in a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is $\frac{3kQ}{R^{3}r}\left(r{\text{'}}^{2}dr\text{'}\right)$ .

Explanation of Solution

Formula used:

The expression for the potential at radius $r$ due to the charge in a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is given as,

  $dV=\frac{kdQ}{r}$

Calculation:

The potential at radius $r$ due to the charge in a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is calculated as,

  $\begin{array}{c}dV=\frac{kdQ}{r}\\ =\frac{kpdV\text{'}}{r}\\ =\frac{kp\left(4\pi r^2\right)dr\text{'}}{r}\\ =\frac{4\pi kp}{r}r{\text{'}}^{2}dr\text{'}\end{array}$ ...... (1)

The value of $p$ is given by,

  $p=\frac{Q}{\frac{4}{3}\pi R^3}$

Substitute the value of $p$ in equation (1)

  $\begin{array}{c}dV=\frac{4\pi kp}{r}r{\text{'}}^{2}dr\text{'}\\ =\frac{4\pi k}{r}\left(\frac{Q}{\frac{4}{3}\pi R^3}\right)r{\text{'}}^{2}dr\text{'}\\ =\left(\frac{3kQ}{R^{3}r}\right)r{\text{'}}^{2}dr\text{'}\end{array}$

Conclusion:

Therefore, the expression for the potential $dV$ at $r$ due to charge in a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is $\frac{3kQ}{R^{3}r}\left(r{\text{'}}^{2}dr\text{'}\right)$ .

(e)

To determine

The potential at $r$ due to all the charge in the region closer than $r$ from the centre of the sphere.

Answer to Problem 57P

The potential at $r$ due to all the charge in the region closer than $r$ from the centre of the sphere is $\frac{kQ}{R^3}{r}^2$ .

Explanation of Solution

Formula used:

The expression for the potential $dV$ at $r$ due to charge in a shell of radius $r\text{'}$ and thickness $dr\text{'}$ is given as,

  $\frac{3kQ}{R^{3}r}\left(r{\text{'}}^{2}dr\text{'}\right)$

Calculation:

Integrate the above expression from $r\text{'}=0$ to $r\text{'}=r$ ,

  $\begin{array}{c}V=\frac{3kQ}{R^{3}r}{\displaystyle \underset{0}{\overset{r}{\int }}r{\text{'}}^{2}dr\text{'}}\\ =\frac{3kQ}{R^{3}r}{\left[\frac{r{\text{'}}^3}{3}\right]}_0^r\\ =\frac{3kQ}{R^{3}r}\left[\frac{r^3}{3}-0\right]\\ =\frac{kQ}{R^3}{r}^2\end{array}$

Conclusion:

Therefore, the potential at $r$ due to all the charge in the region closer than $r$ from the centre of the sphere is $\frac{kQ}{R^3}{r}^2$ .

(f)

To determine

The total potential $V$ at $r$ by adding the equation $\frac{kQ}{R^3}{r}^2$ and $\frac{3kQ}{2R^3}\left(R^2-r^2\right)$ .

Answer to Problem 57P

The total potential $V$ at $r$ by adding the equation $\frac{kQ}{R^3}{r}^2$ and $\frac{3kQ}{2R^3}\left(R^2-r^2\right)$ is $\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$ .

Explanation of Solution

Formula used:

The expression for the potential at $r$ due to all the charge in the region closer than $r$ from the centre of the sphere is given as,

  $\frac{kQ}{R^3}{r}^2$

The expression for the potential at $r$ due to all the charge in the region farther than $r$ from the centre of the sphere is given as,

  $\frac{3kQ}{2R^3}\left(R^2-r^2\right)$

Calculation:

The total potential $V$ at $r$ is calculated as,

  $\begin{array}{c}V=\frac{kQ}{R^3}{r}^2+\frac{3kQ}{2R^3}\left(R^2-r^2\right)\\ =\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)\end{array}$

Conclusion:

Therefore, the total potential $V$ at $r$ by adding the equation $\frac{kQ}{R^3}{r}^2$ and $\frac{3kQ}{2R^3}\left(R^2-r^2\right)$ is $\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$ .