Chapter 24, Problem 24.54AP

A solid, insulating sphere of radius a has a uniform charge density throughout its volume and a total charge Q. Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are b and e as shown in Figure P24.45. We wish to understand completely the charges and electric fields at all locations. (a) Find the charge contained within a sphere of radius r < a. (b) From this value, find the magnitude of the electric field for r < a. (c) What charge is contained within a sphere of radius r when a < r < b? (d) From this value, find the magnitude of the electric field for r when a < r < b. (e) Now consider r when b < r < c. What is the magnitude of the electric field for this range of values of r? (f) From this value, what must be the charge on the inner surface of the hollow sphere? (g) From part (f), what must be the charge on the outer surface of the hollow sphere? (h) Consider the three spherical surfaces of radii a, b, and c. Which of these surfaces has the largest magnitude of surface charge density?

Figure P24.45 Problems 43 and 47.

To determine

The charge contained within the sphere of radius $r<a$.

Answer to Problem 24.54AP

The charge contained within the sphere of radius $r<a$ is $Q{\left(\frac{r}{a}\right)}^3$.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

Write the expression to calculate the charge density of the insulating sphere.

$\rho =\frac{Q}{\left(\frac{4}{3}\pi a^3\right)}$

Write the expression to calculate the charge on the outer sphere at radius $r<a$.

$q_{\text{in}}=\rho \left(\frac{4}{3}\pi r^3\right)$ (1)

Substitute $\frac{Q}{\left(\frac{4}{3}\pi a^3\right)}$ for $\rho$ in equation (1).

$\begin{array}{c}{q}_{\text{in}}=\frac{Q}{\left(\frac{4}{3}\pi a^3\right)}\left(\frac{4}{3}\pi r^3\right)\\ =Q{\left(\frac{r}{a}\right)}^3\end{array}$

Conclusion:

Therefore, the charge contained within the sphere of radius $r<a$ is $Q{\left(\frac{r}{a}\right)}^3$.

To determine

The magnitude of the electric field.

Answer to Problem 24.54AP

The magnitude of the electric field is $k_{\text{e}}\frac{Qr}{a^3}$.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

Write the expression to calculate the electric field.

$\begin{array}{l}E=\frac{q_{\text{in}}}{A\times {\epsilon }_0}\\ =\frac{q_{\text{in}}}{\left(4\pi r^2\right)\times {\epsilon }_0}\end{array}$

Here,

${\epsilon }_0$ is the permittivity of free space.

Substitute $Q{\left(\frac{r}{a}\right)}^3$ for $q_{\text{in}}$ in above equation.

$\begin{array}{c}E=\frac{Q{\left(\frac{r}{a}\right)}^3}{\left(4\pi r^2\right)\times {\epsilon }_0}\\ =\frac{1}{4\pi {\epsilon }_0}\frac{Qr}{a^3}\end{array}$

The expression of Coulomb’s law constant is,

$k_{\text{e}}=\frac{1}{4\pi {\epsilon }_0}$

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_0}$ in above equation to find $E$.

$E=k_{\text{e}}\frac{Qr}{a^3}$

Conclusion:

Therefore, the magnitude of the electric field is $k_{\text{e}}\frac{Qr}{a^3}$.

To determine

The charge contained within a sphere of radius $r$ when $a<r<b$.

Answer to Problem 24.54AP

The charge contained within a sphere of radius $r$ when $a<r<b$ is $Q$.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

The sphere of radius $r$ when $a<r<b$ is enclosed whole insulating sphere so the charge within the sphere is $Q$.

Conclusion:

Therefore, the charge contained within a sphere of radius $r$ when $a<r<b$ is $Q$.

To determine

The magnitude of the electric field for $r$ when $a<r<b$.

Answer to Problem 24.54AP

The magnitude of the electric field is $k_{\text{e}}\frac{Q}{r^2}$.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

Write the expression to calculate the electric filed due to sphere.

$\begin{array}{c}E=\frac{Q}{A\times {\epsilon }_0}\\ =\frac{Q}{\left(4\pi r^2\right)\times {\epsilon }_0}\\ =\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}\end{array}$

The expression of Coulomb’s law constant is,

$k_{\text{e}}=\frac{1}{4\pi {\epsilon }_0}$

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_0}$ in above equation to find $E$.

$E=k_{\text{e}}\frac{Q}{r^2}$

Thus, the magnitude of the electric field is $k_{\text{e}}\frac{Q}{r^2}$.

Conclusion:

Therefore, the magnitude of the electric field is $k_{\text{e}}\frac{Q}{r^2}$.

To determine

The magnitude of the electric field for $r$ when $b<r<c$.

Answer to Problem 24.54AP

The electric field for $r$ when $b<r<c$ is zero.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

The charge is exist only on the surface of the conductor due to which the electric field inside a conductor is zero.

Conclusion:

Therefore, the electric field for $r$ when $b<r<c$ is zero.

To determine

The charge on the inner surface of the sphere for $r$ when $b<r<c$.

Answer to Problem 24.54AP

The charge on the inner surface of the sphere for $r$ when $b<r<c$ is $-Q$.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

The charge on the inner sphere is $Q$ due to which the same charge is induce on the inner surface of the outer sphere but with the negative sign.

Conclusion:

Therefore, the charge on the inner surface of the sphere for $r$ when $b<r<c$ is $-Q$.

To determine

The charge on the outer surface of the sphere for $r$ when $b<r<c$.

Answer to Problem 24.54AP

The charge on the outer surface of the sphere for $r$ when $b<r<c$ is $+Q$.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

The total charge on the outer sphere is zero. That means the charge on the outer surface of the outer sphere is $+Q$.

Conclusion:

Therefore, the charge on the outer surface of the sphere for $r$ when $b<r<c$ is $+Q$.

To determine

The sphere which has the largest magnitude of surface charge density.

Answer to Problem 24.54AP

The sphere which has the largest magnitude of surface charge density is inner surface of radius $b$.

Explanation of Solution

Given info: The radius of the inner insulating sphere is $a$, the total charge on the sphere is $Q$. The inner and outer radius of uncharged concentric sphere is $b$ and $c$ respectively.

The surface charge density is inversely proportional to the surface area of the body.

$\rho \propto \frac{1}{A}$

Therefore, the sphere which has smallest surface area will have largest surface charge density.

Conclusion:

Therefore, the sphere which has the largest magnitude of surface charge density is inner surface of radius $b$.