Chapter 19, Problem 45P

(a)

To determine

The electric field at $0\text{cm}$ from the centre of the sphere.

Answer to Problem 45P

The electric field at $0\text{cm}$ from the centre of the sphere is $0\text{N}/\text{C}$.

Explanation of Solution

Given info: The radius of solid sphere is $40.0\text{cm}$ and the total positive charge on the sphere is $26.0\text{μC}$.

The diagram for the given condition is shown below.

Figure (1)

The charge enclosed by the Gaussian surface is,

$\begin{array}{c}{q}_{\text{en}}=\frac{q}{V}\left(\frac{4}{3}\text{π}{r}^3\right)\\ =\frac{q}{\frac{4}{3}\text{π}{R}^3}\left(\frac{4}{3}\text{π}{r}^3\right)\\ =\frac{q{r}^3}{R^3}\end{array}$

Here,

$V$ is the volume of the outer sphere.

$r$ is the radius of the inner sphere.

$R$ is the radius of the outer sphere.

The area of the sphere is,

$A=4\text{π}{r}^2$

The Gauss law is,

$\begin{array}{c}{\displaystyle \oint \overrightarrow{\text{E}}\cdot \text{d}\overrightarrow{\text{A}}}=\frac{q_{\text{en}}}{{\epsilon }_0}\\ EA=\frac{q_{\text{en}}}{{\epsilon }_0}\end{array}$

Here,

${\epsilon }_0$ is the permittivity of the electrical field.

$E$ is the electric field.

Substitute $4\text{π}{r}^2$ for $A$ and $\frac{q{r}^3}{R^3}$ for $q_{\text{en}}$ in above equation.

$\begin{array}{c}E\left(4\text{π}{r}^2\right)=\frac{\frac{q{r}^3}{R^3}}{{\epsilon }_0}\\ E=\frac{1}{4\text{π}{\epsilon }_0}\left(\frac{qr}{R^3}\right)\\ =K\left(\frac{qr}{R^3}\right)\end{array}$ (1)

Here,

$K$ is the coulomb constant.

Substitute $40.0\text{cm}$ for $R$, $26.0\text{μC}$ for $q$ and $0$ for $r$ in equation (1) to find $E$.

$\begin{array}{c}E=K\left(\frac{\left(26.0\text{μC}\right)\left(0\right)}{{\left(40.0\text{cm}\right)}^3}\right)\\ =0\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field at $0\text{cm}$ from the centre of the sphere is $0\text{N}/\text{C}$.

(b)

To determine

The electric field at $10.0\text{cm}$ from the centre of the sphere.

Answer to Problem 45P

The electric field at $10.0\text{cm}$ from the centre of the sphere is $3.65\times {10}^5\text{N}/\text{C}$.

Explanation of Solution

Given info: The radius of solid sphere is $40.0\text{cm}$ and the total positive charge on the sphere is $26.0\text{μC}$.

Recall the equation (1).

$E=K\left(\frac{qr}{R^3}\right)$

Substitute $40.0\text{cm}$ for $R$, $26.0\text{μC}$ for $q$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $K$ and $10.0\text{cm}$ for $r$ in above equation to find $E$.

$\begin{array}{c}E=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\frac{\left(26.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(10.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{{\left(40.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^3}\right)\\ =3.65\times {10}^5\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field at $10.0\text{cm}$ from the centre of the sphere is $3.65\times {10}^5\text{N}/\text{C}$.

(c)

To determine

The electric field at $40.0\text{cm}$ from the centre of the sphere.

Answer to Problem 45P

The electric field at $40.0\text{cm}$ from the centre of the sphere is $1.46\times {10}^6\text{N}/\text{C}$.

Explanation of Solution

Given info: The radius of solid sphere is $40.0\text{cm}$ and the total positive charge on the sphere is $26.0\text{μC}$.

Recall the equation (1).

$E=K\left(\frac{qr}{R^3}\right)$

Substitute $40.0\text{cm}$ for $R$, $26.0\text{μC}$ for $q$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $K$ and $40.0\text{cm}$ for $r$ in above equation to find $E$.

$\begin{array}{c}E=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\frac{\left(26.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(40.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{{\left(40.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^3}\right)\\ =1.46\times {10}^6\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field at $40.0\text{cm}$ from the centre of the sphere is $1.46\times {10}^6\text{N}/\text{C}$.

(d)

To determine

The electric field at $60.0\text{cm}$ from the centre of the sphere.

Answer to Problem 45P

The electric field at $60.0\text{cm}$ from the centre of the sphere is $6.49\times {10}^5\text{N}/\text{C}$.

Explanation of Solution

Given info: The radius of solid sphere is $40.0\text{cm}$ and the total positive charge on the sphere is $26.0\text{μC}$.

The distance $60.0\text{cm}$ is outside the solid sphere because the radius of the sphere is $40.0\text{cm}$. It means $r>R$.

Then,

$\begin{array}{c}{\displaystyle \oint \overrightarrow{\text{E}}\cdot \text{d}\overrightarrow{\text{A}}}=\frac{q_{\text{en}}}{{\epsilon }_0}\\ E\left(4\text{π}{r}^2\right)=\frac{q}{{\epsilon }_0}\\ E=\frac{1}{4\text{π}{\epsilon }_0}\left(\frac{q}{r^2}\right)\\ =K\left(\frac{q}{r^2}\right)\end{array}$

Substitute $26.0\text{μC}$ for $q$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $K$ and $60.0\text{cm}$ for $r$ in above equation to find $E$.

$\begin{array}{c}E=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\frac{\left(26.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}{{\left(60.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\right)\\ =6.49\times {10}^5\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field at $60.0\text{cm}$ from the centre of the sphere is $6.49\times {10}^5\text{N}/\text{C}$.