Chapter 23, Problem 38P

(a)

To determine

The electric field on the axis of the disk at $5.00\text{cm}$ from the center of the disk.

Answer to Problem 38P

The electric field on the axis of the disk at $5.00\text{cm}$ from the center of the disk is $383.03\text{MN}/\text{C}$.

Explanation of Solution

Write the expression for the electric field on the axis of the disk at $5.00\text{cm}$ from the center of the disk.

    $E_{\text{x}}=2\pi k_{\text{e}}\sigma \left(1-\frac{x}{{\left(x^2+R^2\right)}^{\frac{1}{2}}}\right)$                                                                                    (I)

Here, $\sigma$ is the surface charge density, $R$ is the radius and $k_{\text{e}}$ is the Coulomb’s constant.

Write the constant value for the Coulomb’s constant $k$.

    $k_{\text{e}}=8.9876\times {\text{10}}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$

Conclusion:

Convert the radius of the disk from $\text{cm}$ to $\text{m}$.

    $\begin{array}{c}R=35.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.35\text{m}\end{array}$

Convert the distance from the disk from $\text{cm}$ to $\text{m}$.

    $\begin{array}{c}x=5.00\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.05\text{m}\end{array}$

Substitute $8.9876\times {\text{10}}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, $0.35\text{m}$ for $R$ and $0.05\text{m}$ for $x$ in equation (I), to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}=0.05\text{m}}=2\pi \left(8.9876\times {\text{10}}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}\right)\left(1-\frac{0.05\text{m}}{{\left({\left(0.05\text{m}\right)}^2+{\left(0.35\text{m}\right)}^2\right)}^{\frac{1}{2}}}\right)\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{0.05}{{\left(0.125\right)}^{\frac{1}{2}}}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{0.05}{0.3535}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-0.1414\right)\text{N}/\text{C}\end{array}$

Further simplify the above equation to find $E_{\text{x}=0.05\text{m}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{0}\text{.05}\text{m}}=\left(446.1189\times {10}^6\right)\times 0.8586\text{N}/\text{C}\\ =383.03\times {10}^6\text{N}/\text{C}\left(\frac{\text{1}\text{MN}/\text{C}}{{\text{10}}^{\text{6}}\text{N}/\text{C}}\right)\\ =383.03\text{MN}/\text{C}\end{array}$

Therefore, the electric field on the axis of the disk at $5.00\text{cm}$ from the center of the disk is $383.03\text{MN}/\text{C}$.

(b)

To determine

The electric field on the axis of the disk at $10.0\text{cm}$ from the center of the disk.

Answer to Problem 38P

The electric field on the axis of the disk at $10.0\text{cm}$ from the center of the disk is $323.5\text{MN}/\text{C}$.

Explanation of Solution

Conclusion:

Convert the distance from the disk from $\text{cm}$ to $\text{m}$.

    $\begin{array}{c}x=10.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.10\text{m}\end{array}$

Substitute $8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, $0.35\text{m}$ for $R$ and $0.10\text{m}$ for $x$ in equation (I), to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{0}\text{.10}\text{m}}=2\pi \left(8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}\right)\left(1-\frac{0.10\text{m}}{{\left({\left(0.10\text{m}\right)}^2+{\left(0.35\text{m}\right)}^2\right)}^{\frac{1}{2}}}\right)\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{0.10}{{\left(0.1325\right)}^{\frac{1}{2}}}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{0.10}{0.36400}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-0.2747\right)\text{N}/\text{C}\end{array}$

Further simplify the above equation to find $E_{\text{x}=\text{0}\text{.10}\text{m}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{0}\text{.10}\text{m}}=\left(446.1189\times {10}^6\right)\times 0.72527\text{N}/\text{C}\\ =323.5\times {10}^6\text{N}/\text{C}\left(\frac{\text{1}\text{MN}/\text{C}}{{\text{10}}^{\text{6}}\text{N}/\text{C}}\right)\\ =323.5\text{MN}/\text{C}\end{array}$

Therefore, the electric field on the axis of the disk at $10.0\text{cm}$ from the center of the disk is $323.5\text{MN}/\text{C}$.

(c)

To determine

The electric field on the axis of the disk at $50.0\text{cm}$ from the center of the disk.

Answer to Problem 38P

The electric field on the axis of the disk at $50.0\text{cm}$ from the center of the disk is $80.44\text{MN}/\text{C}$.

Explanation of Solution

Conclusion:

Convert the distance of the disk from $\text{cm}$ to $\text{m}$.

    $\begin{array}{c}x=50.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.50\text{m}\end{array}$

Substitute $8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, $0.35\text{m}$ for $R$ and $0.50\text{m}$ for $x$ in equation (I), to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{0}\text{.50}\text{m}}=2\pi \left(8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}\right)\left(1-\frac{0.50\text{m}}{{\left({\left(0.50\text{m}\right)}^2+{\left(0.35\text{m}\right)}^2\right)}^{\frac{1}{2}}}\right)\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{0.50}{{\left(0.3725\right)}^{\frac{1}{2}}}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{0.50}{0.610}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-0.8196\right)\text{N}/\text{C}\end{array}$

Further simplify the above equation to find $E_{\text{x}=\text{0}\text{.50}\text{m}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{0}\text{.50}\text{m}}=\left(446.1189\times {10}^6\right)\times 0.18032\text{N}/\text{C}\\ =80.44\times {10}^6\text{N}/\text{C}\left(\frac{\text{1}\text{MN}/\text{C}}{{\text{10}}^{\text{6}}\text{N}/\text{C}}\right)\\ =80.44\text{MN}/\text{C}\end{array}$

Therefore, the electric field on the axis of the disk at $50.0\text{cm}$ from the center of the disk is $80.44\text{MN}/\text{C}$.

(d)

To determine

The electric field on the axis of the disk at $200\text{cm}$ from the center of the disk.

Answer to Problem 38P

The electric field on the axis of the disk at $200\text{cm}$ from the center of the disk is $6.88\text{MN}/\text{C}$.

Explanation of Solution

Conclusion:

Convert the distance of the disk from $\text{cm}$ to $\text{m}$.

    $\begin{array}{c}x=200\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =2.0\text{m}\end{array}$

Substitute $8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, $0.35\text{m}$ for $R$ and $2.0\text{m}$ for $x$ in equation (I) to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}=2.0\text{m}}=2\pi \left(8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(7.90\times {10}^{-3}\text{C}/{\text{m}}^{\text{2}}\right)\left(1-\frac{2\text{m}}{{\left({\left(2\text{m}\right)}^2+{\left(0.35\text{m}\right)}^2\right)}^{\frac{1}{2}}}\right)\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{2}{{\left(4.1225\right)}^{\frac{1}{2}}}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-\frac{2}{2.030}\right)\text{N}/\text{C}\\ =\left(446.1189\times {10}^6\right)\left(1-0.9852\right)\text{N}/\text{C}\end{array}$

Further simplify the above equation to find $E_{\text{x}=\text{0}\text{.50}\text{m}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{2}\text{.0}\text{m}}=\left(446.1189\times {10}^6\right)\times 0.01477\text{N}/\text{C}\\ =6.88\times {10}^6\text{N}/\text{C}\left(\frac{\text{1}\text{MN}/\text{C}}{{\text{10}}^{\text{6}}\text{N}/\text{C}}\right)\\ =6.88\text{MN}/\text{C}\end{array}$

Therefore, the electric field on the axis of the disk at $200\text{cm}$ from the center of the disk is $6.88\text{MN}/\text{C}$.