Chapter 23, Problem 41P

(a)

To determine

The electric field at a point on the axis and $3.00\text{mm}$ from the center of the disk.

Answer to Problem 41P

The electric field at a point on the axis and $3.00\text{mm}$ from the center of the disk is $9.35\times {10}^7\text{N}/\text{C}$.

Explanation of Solution

Write the expression for the surface charge density.

    $\sigma =\frac{Q}{\pi R^2}$                                                                                                                    (I)

Here, $\sigma$ is the surface charge density, $R$ is the radius of the disk and $Q$ is the charge on the disk.

Write the expression for the electric field at a point along the axis of the circular disk carrying charge.

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

Here, $E_{\text{x}}$ is the electric field at a point along the axis, $k_{\text{e}}$ is the Coulomb’s constant, $x$ is the distance on the axis from the center of the circular disk and $R$ is the radius of the disk.

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

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

Conclusion:

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

    $\begin{array}{c}R=\left(3.0\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.03\text{m}\end{array}$

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

    $\begin{array}{c}x=\left(3.0\text{mm}\right)\left(\frac{1\text{mm}}{1000\text{m}}\right)\\ =0.003\text{m}\end{array}$

Substitute $5.2\times {10}^{-6}\text{C}$ for $Q$ and $0.03\text{m}$ for $R$ in equation (I) to calculate $\sigma$.

    $\begin{array}{c}\sigma =\frac{5.2\times {10}^{-6}\text{C}}{\pi {\left(0.03\text{m}\right)}^2}\\ =1.839\times {10}^{-3}\text{C}/{\text{m}}^2\end{array}$

Substitute $8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $1.839\times {10}^{-3}\text{C}/{\text{m}}^2$ for $\sigma$, $0.03\text{m}$. for $R$ and $0.003\text{m}$ for $x$ in equation (II) to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}=0.003\text{m}}=\left[\begin{array}{l}2\pi \left(8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.839\times {10}^{-3}\text{C}/{\text{m}}^2\right)\\ \left(1-\frac{0.003\text{m}}{{\left({\left(0.003\text{m}\right)}^2+{\left(0.03\text{m}\right)}^2\right)}^{\frac{1}{2}}}\right)\end{array}\right]\\ =\left(103.8150\times {10}^6\right)\left(1-\frac{0.003}{{\left(9.04\times {10}^{-4}\right)}^{\frac{1}{2}}}\right)\text{N}/\text{C}\\ =\left(103.8150\times {10}^6\right)\left(1-\frac{0.003}{0.030}\right)\text{N}/\text{C}\\ =\left(103.8150\times {10}^6\right)\left(1-0.1\right)\text{N}/\text{C}\end{array}$

Here, $E_{\text{x}=0.003\text{m}}$ is the electric field at $0.003\text{m}$ along the axis from the center of the circular disk.

Simplify the above equation to find $E_{\text{x}=0.003\text{m}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{0}\text{.003}\text{m}}=\left(103.8150\times {10}^6\right)\times 0.9\text{N}/\text{C}\\ =9.35\times {10}^7\text{N}/\text{C}\end{array}$

Therefore, the electric field at a point on the axis and $3.00\text{mm}$ from the center of the disk is $9.35\times {10}^7\text{N}/\text{C}$.

(b)

To determine

The comparison of the answer from part (a) with the field computed from the near-filed approximation $E={\sigma /2\epsilon }_0$.

Answer to Problem 41P

The value of the field computed from the near field approximation is $1.04\times {10}^8\text{N}/\text{C}$ and is greater than the part (a) by approximately $11\%$.

Explanation of Solution

Write the expression for the magnitude of the electric field

    $E=2\pi k_{\text{e}}\sigma$                                                                                                               (III)

Here, $\sigma$ is the surface charge density and ${\epsilon }_0$ is the permittivity of space.

Write the expression for the change in the field computed from the near-filed approximation.

    $\Delta E=\left(\frac{E-E_{\text{x}=\text{0}\text{.003}\text{m}}}{E_{\text{x}=\text{0}\text{.003}\text{m}}}\right)\times 100\%$                                                                                 (IV)

Here, $\Delta E$ is the change in the electric field.

Conclusion:

Substitute $1.839\times {10}^{-3}\text{C}/{\text{m}}^2$ for $\sigma$ and $8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$. in equation (III) to calculate $E$.

    $\begin{array}{c}E=2\pi \times 8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 1.839\times {10}^{-3}\text{C}/{\text{m}}^2\\ =1.04\times {10}^8\text{N}/\text{C}\end{array}$

Substitute $9.35\times {10}^7\text{N}/\text{C}$ for $E_{\text{x}=\text{0}\text{.003}\text{m}}$ and $1.04\times {10}^8\text{N}/\text{C}$ for $E$ in equation (IV) to find $\Delta E$.

    $\begin{array}{c}\Delta E=\left(\frac{1.04\times {10}^8\text{N}/\text{C}-9.35\times {10}^7\text{N}/\text{C}}{9.35\times {10}^7\text{N}/\text{C}}\right)\times 100\%\\ =11.23\%\end{array}$

Therefore, the value of the field computed from the near field approximation is $1.04\times {10}^8\text{N}/\text{C}$ and is greater than the part (a) by $11.23\%$.

(c)

To determine

The electric field at a point on the axis of the disk and $30\text{cm}$ from the center of the disk.

Answer to Problem 41P

The electric field at a point on the axis and $30\text{cm}$ from the center of the disk is $5.15\times {10}^5\text{N}/\text{C}$.

Explanation of Solution

Write the expression for the electric field at a point on the axis of the circular disk carrying charge.

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

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

Conclusion:

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

    $\begin{array}{c}x=\left(30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.3\text{m}\end{array}$

Substitute $8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $1.839\times {10}^{-3}\text{C}/{\text{m}}^2$ for $\sigma$, $0.03\text{m}$ for $R$ and $0.3\text{m}$ for $x$ in equation (V), to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}=0.3\text{m}}=\left[\begin{array}{c}2\pi \left(8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.839\times {10}^{-3}\text{C}/{\text{m}}^2\right)\\ \left(1-\frac{0.3\text{m}}{{\left({\left(0.3\text{m}\right)}^2+{\left(0.03\text{m}\right)}^2\right)}^{\frac{1}{2}}}\right)\end{array}\right]\\ =\left(103.8497\times {10}^6\right)\left(1-\frac{0.3}{{\left(0.0909\right)}^{\frac{1}{2}}}\right)\text{N}/\text{C}\\ =\left(103.8497\times {10}^6\right)\left(1-\frac{0.3}{0.301496}\right)\text{N}/\text{C}\\ =\left(103.8497\times {10}^6\right)\left(1-0.99503\right)\text{N}/\text{C}\end{array}$

Here, $E_{\text{x}=0.3\text{m}}$ is the electric field at $0.3\text{m}$ along the axis from the center of the circular disk.

Simplify the above equation to find $E_{\text{x}=0.3\text{m}}$.

    $\begin{array}{c}{E}_{\text{x}=\text{0}\text{.3}\text{m}}=\left(103.8150\times {10}^6\right)\times \left(4.97\times {10}^{-3}\right)\text{N}/\text{C}\\ =5.15\times {10}^5\text{N}/\text{C}\end{array}$

Therefore, the electric field at a point on the axis and $30\text{cm}$ from the center of the disk is $5.15\times {10}^5\text{N}/\text{C}$.

(d)

To determine

The comparison of the answer from part (c) with the electric field obtained by treating the disk as a $+5.20\text{-}\mu \text{C}$ charged particle at a distance of $30.0\text{cm}$.

Answer to Problem 41P

The value of the field computed is $5.19\times {10}^5\text{N}/\text{C}$ and is greater than part (c) by approximately $0.7\%$.

Explanation of Solution

Write the expression for the electric field at a distance of $30\text{cm}$, if the disk is treated as a point charge of $5.2\mu \text{C}$.

    $E_{\text{x}}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{Q}{x^2}\right)$                                                                                                      (VI)

Conclusion:

Substitute $8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $\frac{1}{4\pi {\epsilon }_0}$, $5.2\times {10}^{-6}\text{C}$ for $Q$ and $0.3\text{m}$ for $x$ in equation (VI), to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}=0.3\text{m}}=\left(8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\frac{5.2\times {10}^{-6}\text{C}}{{\left(0.\text{3}\text{m}\right)}^2}\right)\\ =\left(8.9876\times {\text{10}}^9\right)\left(57.778\times {10}^{-6}\right)\text{N}/\text{C}\\ =5.19\times {10}^5\text{N}/\text{C}\end{array}$

Substitute $5.15\times {10}^5\text{N}/\text{C}$ for $E_{\text{x}=0.3\text{m}}$ and $5.19\times {10}^5\text{N}/\text{C}$ for $E$ in equation (IV) to find $\Delta E$.

    $\begin{array}{c}\Delta E=\left(\frac{5.19\times {10}^5\text{N}/\text{C}-5.15\times {10}^5\text{N}/\text{C}}{5.15\times {10}^5\text{N}/\text{C}}\right)\times 100\%\\ =0.7\%\end{array}$

Therefore, the value of the field computed is $5.19\times {10}^5\text{N}/\text{C}$ and is greater than the part (c) by $0.7\%$.