Chapter 22, Problem 76P

(a)

To determine

The magnitude and the direction of the electric field for a non-conducting spherical shell concentric with a solid sphere.

Explanation of Solution

Given:

The diameter of the sphere is $1.20\text{m}$ .

The volume charge density of the sphere is $+5.00{\text{μC/m}}^{\text{3}}$ .

The diameter of the shell is $2.40\text{m}$ .

The surface charge density of the sphere is $-1.50{\text{μC/m}}^2$ .

Formula used:

Write the expression of the electric field at any point for a non-conducting sphere.

  ${\overrightarrow{E}}_{\text{sphere}}=k\frac{Q}{r^2}\widehat{r}$ …… (1)

Here, ${\overrightarrow{E}}_{\text{sphere}}$ is the electric field, $k$ is Coulomb’s constant, $Q$ is the total charge, $r$ is the distance of any point and $\widehat{r}$ is the unit vector along $r$ .

Write the expression for charge for a sphere.

  $Q=\rho V$

Here, $\rho$ is the volume charge density and $V$ is the volume of the sphere.

Substitute $\frac{4\pi }{3}{a}^3$ for $V$ in the above expression.

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

Here, $a$ is the radius of the sphere.

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

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\frac{\rho a^3}{r^2}\widehat{r}$ …… (2)

Write the above expression when $\left(r\le a\right)$ .

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\rho r\widehat{r}$ …… (3)

Simplify the above equation.

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\rho r\widehat{r}$

Write the expression for the electric field at any point due to a spherical shell.

  ${\overrightarrow{E}}_{\text{shell}}=k\frac{q}{r^2}\widehat{r}$ …… (4)

Here, ${\overrightarrow{E}}_{\text{shell}}$ is the electric field, $k$ is Coulomb’s constant, $q$ is the total charge, $r$ is the distance of any point and $\widehat{r}$ is the unit vector along $r$ .

Write the expression charge of a spherical shell.

  $q=\rho A$

Here, $\sigma$ is the surface charge density and $A$ is the surface area of the sphere.

Substitute $4\pi a^2$ for $A$ in the above equation.

  $q=\rho \left(4\pi a^2\right)$

Here, $a$ is the radius of the sphere.

Substitute $\rho \left(4\pi a^2\right)$ for $q$ in equation (4).

  ${\overrightarrow{E}}_{\text{shell}}=k\frac{\rho \left(4\pi a^2\right)}{r^2}\widehat{r}$ …… (5)

Write the expression for the resultant electric field at any point in space due to a spherical shell and a solid sphere.

  $\overrightarrow{E}={\overrightarrow{E}}_{\text{shell}}+{\overrightarrow{E}}_{\text{sphere}}$ …… (6)

Calculation:

The electric field at point $x=4.50\text{m , }y=0$ for the sphere is calculated below.

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$ , $+5.00{\text{μC/m}}^{\text{3}}$ for $\rho$ , $\left(4.50\text{m}-4.00\text{m}\right)$ for $r$ and $\widehat{i}$ for $\widehat{r}$ in equation (3).

  $\begin{array}{c}{\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(+5.00{\text{μC/m}}^{\text{3}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(4.50\text{m}-4.00\text{m}\right)\widehat{i}\\ =94122.11\text{N/C}\widehat{i}\\ \approx \text{94}\text{kN/C}\widehat{i}\end{array}$

The direction of ${\overrightarrow{E}}_{\text{sphere}}$ is calculated below.

  $\theta =0°$

  $\left(4.50\text{m ,}0\right)$ is inside the spherical shell.

The electric field at point $x=4.50\text{m , }y=0$ for the spherical shell is calculated below.

  ${\overrightarrow{E}}_{\text{shell}}=0$

The electric field at point $\left(4.50\text{m ,}0\right)$ is calculated below.

Substitute $\text{94}\text{kN/C}\widehat{i}$ for ${\overrightarrow{E}}_{\text{sphere}}$ and $0$ for ${\overrightarrow{E}}_{\text{shell}}$ in equation (6).

  $\overrightarrow{E}=\text{94}\text{kN/C}\widehat{i}$

Conclusion:

Thus, the electric magnitude and the direction of electric field at point $\left(4.50\text{m ,}0\right)$ is $\text{94}\text{kN/C}$ and $0°$ respectively.

(b)

To determine

The magnitude and the direction of the electric field for a non-conducting spherical shell concentric with a solid sphere.

Explanation of Solution

Given:

The diameter of the sphere is $1.20\text{m}$ .

The volume charge density of the sphere is $+5.00{\text{μC/m}}^{\text{3}}$ .

The diameter of the shell is $2.40\text{m}$ .

The surface charge density of the sphere is $-1.50{\text{μC/m}}^2$ .

Formula used:

Write the expression of the electric field at any point for a non-conducting sphere.

  ${\overrightarrow{E}}_{\text{sphere}}=k\frac{Q}{r^2}\widehat{r}$

Here, ${\overrightarrow{E}}_{\text{sphere}}$ is the electric field, $k$ is Coulomb’s constant, $Q$ is the total charge, $r$ is the distance of any point and $\widehat{r}$ is the unit vector along $r$ .

Write the expression for charge for a sphere.

  $Q=\rho V$

Here, $\rho$ is the volume charge density and $V$ is the volume of the sphere.

Substitute $\frac{4\pi }{3}{a}^3$ for $V$ in the above expression.

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

Here, $a$ is the radius of the sphere.

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

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\frac{\rho a^3}{r^2}\widehat{r}$

Write the above expression when $\left(r\le a\right)$ .

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\rho r\widehat{r}$

Simplify the above equation.

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\rho r\widehat{r}$

Write the expression for the electric field at any point due to a spherical shell.

  ${\overrightarrow{E}}_{\text{shell}}=k\frac{q}{r^2}\widehat{r}$

Here, ${\overrightarrow{E}}_{\text{shell}}$ is the electric field, $k$ is Coulomb’s constant, $q$ is the total charge, $r$ is the distance of any point and $\widehat{r}$ is the unit vector along $r$ .

Write the expression charge of a spherical shell.

  $q=\rho A$

Here, $\sigma$ is the surface charge density and $A$ is the surface area of the sphere.

Substitute $4\pi a^2$ for $A$ in the above equation.

  $q=\rho \left(4\pi a^2\right)$

Here, $a$ is the radius of the sphere.

Substitute $\rho \left(4\pi a^2\right)$ for $q$ in equation (4).

  ${\overrightarrow{E}}_{\text{shell}}=k\frac{\rho \left(4\pi a^2\right)}{r^2}\widehat{r}$

Resultant electric field at any point in space due to a spherical shell and a solid sphere.

  $\overrightarrow{E}={\overrightarrow{E}}_{\text{shell}}+{\overrightarrow{E}}_{\text{sphere}}$

Calculation:

The electric field at point $x=4.0\text{m , }y=1.10\text{m}$ for the sphere is calculated below.

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$ , $+5.00{\text{μC/m}}^{\text{3}}$ for $\rho$ , $0.600\text{m}$ for $a$ , $1.10\text{m}$ for $r$ , $0.600\text{m}$ for $a$ and $\widehat{j}$ for $\widehat{r}$ in equation (2).

  $\begin{array}{c}{\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(+5.00{\text{μC/m}}^{\text{3}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right){\left(0.600\right)}^3}{{\left(1.10\text{m}\right)}^2}\widehat{j}\\ =33603.92\text{N/C}\widehat{j}\\ =33.6\text{kN/C}\widehat{j}\end{array}$

The direction of ${\overrightarrow{E}}_{\text{sphere}}$ at point $\left(4.0\text{m ,}1.10\text{m}\right)$ is calculated below.

  $\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{1.10}{4.0-4.0}\right)\\ ={\text{tan}}^{-1}\left(0\right)\\ =90°\end{array}$

  $\left(4.0\text{m ,}1.10\text{m}\right)$ is inside the spherical shell.

The electric field at point $x=4.0\text{m , }y=1.10\text{m}$ for the spherical shell is calculated below.

  ${\overrightarrow{E}}_{\text{shell}}=0$

The electric field at point $\left(4.50\text{m ,}0\right)$ is calculated below.

Substitute $33.6\text{kN/C}\widehat{j}$ for ${\overrightarrow{E}}_{\text{sphere}}$ and $0$ for ${\overrightarrow{E}}_{\text{shell}}$ in equation (6).

  $\overrightarrow{E}=33.6\text{kN/C}\widehat{j}$

Conclusion:

Thus, the electric magnitude and the direction of electric field at point $\left(4.0\text{m ,}1.10\text{m}\right)$ is $33.6\text{kN/C}$ and $90°$ respectively.

(c)

To determine

The magnitude and the direction of the electric field for a non-conducting spherical shell concentric with a solid sphere.

Explanation of Solution

Given:

The diameter of the sphere is $1.20\text{m}$ .

The volume charge density of the sphere is $+5.00{\text{μC/m}}^{\text{3}}$ .

The diameter of the shell is $2.40\text{m}$ .

The surface charge density of the sphere is $-1.50{\text{μC/m}}^2$ .

Formula used:

Write the expression of the electric field at any point for a non-conducting sphere.

  ${\overrightarrow{E}}_{\text{sphere}}=k\frac{Q}{r^2}\widehat{r}$

Here, ${\overrightarrow{E}}_{\text{sphere}}$ is the electric field, $k$ is Coulomb’s constant, $Q$ is the total charge, $r$ is the distance of any point and $\widehat{r}$ is the unit vector along $r$ .

Write the expression for charge for a sphere.

  $Q=\rho V$

Here, $\rho$ is the volume charge density and $V$ is the volume of the sphere.

Substitute $\frac{4\pi }{3}{a}^3$ for $V$ in the above expression.

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

Here, $a$ is the radius of the sphere.

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

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\frac{\rho a^3}{r^2}\widehat{r}$

Write the above expression when $\left(r\le a\right)$ .

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\rho r\widehat{r}$

Simplify the above equation.

  ${\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}k\rho r\widehat{r}$

Write the expression for the electric field at any point due to a spherical shell.

  ${\overrightarrow{E}}_{\text{shell}}=k\frac{q}{r^2}\widehat{r}$

Here, ${\overrightarrow{E}}_{\text{shell}}$ is the electric field, $k$ is Coulomb’s constant, $q$ is the total charge, $r$ is the distance of any point and $\widehat{r}$ is the unit vector along $r$ .

Write the expression charge of a spherical shell.

  $q=\rho A$

Here, $\sigma$ is the surface charge density and $A$ is the surface area of the sphere.

Substitute $4\pi a^2$ for $A$ in the above equation.

  $q=\rho \left(4\pi a^2\right)$

Here, $a$ is the radius of the sphere.

Substitute $\rho \left(4\pi a^2\right)$ for $q$ in equation (4).

  ${\overrightarrow{E}}_{\text{shell}}=k\frac{\rho \left(4\pi a^2\right)}{r^2}\widehat{r}$

Resultant electric field at any point in space due to a spherical shell and a solid sphere.

  $\overrightarrow{E}={\overrightarrow{E}}_{\text{shell}}+{\overrightarrow{E}}_{\text{sphere}}$

Calculation:

Write the expression for distance between the points $\left(4.0\text{m ,}0\right)$ and $\left(2.0\text{m ,}3.0\text{m}\right)$ .

  $\begin{array}{c}r=\sqrt{{\left(4.0\text{m}-2.0\text{m}\right)}^2+{\left(0\text{m}-3.0\text{m}\right)}^2}\\ =3.606\text{m}\end{array}$

Write the direction for $r$ .

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{0-3.0}{4.0-2.0}\right)\\ \approx -56.3°\end{array}$

When the above value is subtracted from $360°$ .

  $\theta =304°$

Write the expression for unit vector along $r$ .

  $\widehat{r}=cos\theta \widehat{i}+sin\theta \widehat{j}$

Substitute $123.7°$ for $\theta$ in the above equation.

  $\begin{array}{l}r=cos\left(123.7°\right)\widehat{i}+sin\left(123.7°\right)\widehat{j}\\ \approx -0.554\widehat{i}+0.832\widehat{j}\end{array}$

The electric field at point $x=2.0\text{m , }y=3.0\text{m}$ for the sphere is calculated below.

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$ , $+5.00{\text{μC/m}}^{\text{3}}$ for $\rho$ , $0.600\text{m}$ for $a$ , $3.606\text{m}$ for $r$ and $\left(-0.554\widehat{i}+0.832\widehat{j}\right)$ for $\widehat{r}$ in equation (3).

  $\begin{array}{c}{\overrightarrow{E}}_{\text{sphere}}=\frac{4\pi }{3}\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(+5.00{\text{μC/m}}^{\text{3}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right){\left(0.600\right)}^3}{{\left(3.606\text{m}\right)}^2}\left(-0.554\widehat{i}+0.832\widehat{j}\right)\\ =3.127\text{kN/C}\left(-0.554\widehat{i}+0.832\widehat{j}\right)\\ \approx -1.732\text{kN/C}\widehat{i}+2.601\text{kN/C}\widehat{j}\end{array}$

The electric field at point $x=2.0\text{m , }y=3.0\text{m}$ for the spherical shell is calculated below.

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$ , $+5.00{\text{μC/m}}^{\text{3}}$ for $\rho$ , $1.20\text{m}$ for $a$ , $3.606\text{m}$ for $r$ and $\left(-0.554\widehat{i}+0.832\widehat{j}\right)$ for $\widehat{r}$ in equation (5).

  $\begin{array}{c}{\overrightarrow{E}}_{\text{sphere}}=4\pi \left(8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(-1.50{\text{μC/m}}^2\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right){\left(1.20\right)}^2}{{\left(3.606\text{m}\right)}^2}\left(-0.554\widehat{i}+0.832\widehat{j}\right)\\ =-18.77\text{kN/C}\left(-0.554\widehat{i}+0.832\widehat{j}\right)\\ \approx 10.40\text{kN/C}\widehat{i}-15.61\text{kN/C}\widehat{j}\end{array}$

The electric field at point $\left(2.0\text{m ,}3.0\text{m}\right)$ is calculated below.

Substitute $\left(-1.732\text{kN/C}\widehat{i}+2.601\text{kN/C}\widehat{j}\right)$ for ${\overrightarrow{E}}_{\text{sphere}}$ and $\left(10.40\text{kN/C}\widehat{i}-15.61\text{kN/C}\widehat{j}\right)$ for ${\overrightarrow{E}}_{\text{shell}}$ in equation (6).

  $\begin{array}{c}\overrightarrow{E}=\left(-1.732\text{kN/C}\widehat{i}+2.601\text{kN/C}\widehat{j}\right)+\left(10.40\text{kN/C}\widehat{i}-15.61\text{kN/C}\widehat{j}\right)\\ =8.668\text{kN/C}\widehat{i}-13.01\text{kN/C}\widehat{j}\end{array}$

The magnitude of electric field at $\left(2.0\text{m ,}3.0\text{m}\right)$ is calculated below.

  $\begin{array}{c}E=\sqrt{{\left(-13.01\text{kN/C}\right)}^2+{\left(8.668\text{kN/C}\right)}^2}\\ \approx 15.62\text{kN/C}\end{array}$

Conclusion:

Thus, the electric magnitude and the direction of electric field at point $\left(2.0\text{m ,}3.0\text{m}\right)$ is $15.62\text{kN/C}$ and $304°$ respectively.