Chapter 23, Problem 5P

Example 23.3 derives the exact expression for the electric field at a point on the axis of a uniformly charged disk. Consider a disk of radius R = 3.00 cm having a uniformly distributed charge of +5.20 μC. (a) Using the result of Example 23.3, compute the electric field at a point on the axis and 3.00 mm from the center. (b) What If? Explain how the answer to part (a) compares with the field computed from the near-field approximation E = σ/2ϵ₀. (We derived this expression in Example 23.3.) (c) Using the result of Example 23.3, compute the electric field at a point on the axis and 30.0 cm from the center of the disk. (d) What If? Explain how the answer to part (c) compares with the electric field obtained by treating the disk as a +5.20-μC charged particle at a distance of 30.0 cm.

To determine

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

Answer to Problem 5P

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

Explanation of Solution

Given info: The value of the charge is $+5.20\text{μC}$, radius of the disk is $3.00\text{cm}$, distance from center of the disk is $3.00\text{mm}$.

The formula to calculate the surface charge density is,

$\sigma =\frac{q}{A}$ (1)

The formula to calculate the area is,

$A=\pi R^2$ (2)

Substitute $\pi R^2$ for $A$ from equation (2) in equation (1).

$\sigma =\frac{q}{\pi R^2}$ (3)

Substitute $+5.20\text{μC}$ for the value of $q$, $3.00\text{cm}$ for $R$ in equation (3) to calculate $\sigma$.

$\begin{array}{c}=\frac{5.20\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}}{3.14\times {\left(3.00\text{cm×}\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1}\text{cm}}\right)}^2}\\ =0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}\end{array}$

The formula to calculate the electric field at a point from the center of a uniformly charged disk is,

$E=\frac{\sigma }{2\epsilon }\left(1-\frac{x}{\sqrt{R^2+x^2}}\right)$ (4)

Here,

$E$ is the electric field at that point.

$\sigma$ is the surface charge density.

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

$x$ is the distance of the point from the center of the disk.

$R$ is the radius of the disk.

Substitute $0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_o$, $3.00\text{cm}$ for $R$, $3.00\text{mm}$ for $x$ in equation (4).

$\begin{array}{c}E=\frac{0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}}{2\times 8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}}\times \left(1-\frac{3\text{mm×}\frac{{\text{10}}^{-\text{3}}\text{m}}{\text{1}\text{mm}}}{\sqrt{\left({\left(3\text{mm×}\frac{{\text{10}}^{-3}\text{m}}{\text{1}\text{mm}}\right)}^{-2}+{\left(3\text{cm×}\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)}^{-2}\right)}}\right)\\ =9.35\times {10}^7\text{N}/\text{C}\end{array}$

Conclusion:

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

To determine

The percentage change when part (a) is compared with the field computed from the near field approximation.

Answer to Problem 5P

The magnitude of electric field when computed from near field approximation is $1.04\times {10}^8\text{N}/\text{C}$ and this value is $11\%$ higher.

Explanation of Solution

The formula to calculate the electric field at a point from the center of a uniformly charged disk is,

$E=\frac{\sigma }{2\epsilon }$ (5)

Here,

$E$ is the electric field at that point.

$\sigma$ is the surface charge density.

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

Substitute $0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_o$, in equation (5) as,

$\begin{array}{c}E=\frac{0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}}{2\times 8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}{\text{.m}}^{\text{2}}}\\ =1.04\times {10}^8\text{N}/\text{C}\end{array}$

The value of electric field for uniformly charged disk for a point $3\text{mm}$ from the center is $9.35\times {10}^7\text{N}/\text{C}$.

The value of the electric field computed from near field approximation is $1.04\times {10}^8\text{N}/\text{C}$.

The formula to calculate the percentage change with respect to the field computed from near field approximation is,

$\begin{array}{c}\%\text{ change}=\frac{\left(1.04\times {10}^8-9.35\times {10}^7\right)}{1.04\times {10}^8}\times 100\\ =\frac{1.05\times {10}^7}{1.04\times {10}^8}\times 100\\ \approx 11\%\text{ higher}\end{array}$

Conclusion:

Therefore, the magnitude of electric field when computed from near field approximation is $1.04\times {10}^8\text{N}/\text{C}$ and this value is $11\%$ higher.

To determine

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

Answer to Problem 5P

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

Explanation of Solution

Given info: The value of the charge is $+5.2\text{μC}$, radius of the disk is $3.00\text{cm}$, distance from center of the disk is $30\text{cm}$.

The formula to calculate the surface charge density is,

$\sigma =\frac{q}{A}$ (6)

The formula to calculate the area is,

$A=\pi \times R^2$ (7)

Substitute $A=\pi \times R^2$ from equation (2) in equation (1).

$\sigma =\frac{q}{\pi \times R^2}$ (8)

Substitute $+5.20\text{μC}$ for the value of $q$, $3.00\text{cm}$ for $R$ in equation (8) to calculate $\sigma$.

$\begin{array}{c}=\frac{5.20\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}}{3.14\times {\left(3.00\text{cm×}\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1}\text{cm}}\right)}^2}\\ =0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}\end{array}$

The formula to calculate the electric field at a point from the center of a uniformly charged disk is,

$E=\frac{\sigma }{2\epsilon }\left(1-\frac{x}{\sqrt{R^2+x^2}}\right)$ (9)

Here,

$E$ is the electric field at that point.

$\sigma$ is the surface charge density.

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

$x$ is the distance of the point from the center of the disk.

$R$ is the radius of the disk.

Substitute $0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_o$, $3.00\text{cm}$ for $R$, $30\text{cm}$ for $x$ in equation (9).

$\begin{array}{c}E=\frac{0.184005662\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}}{2\times 8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}}\times \left(1-\frac{30\text{cm×}\frac{{\text{10}}^{-2}\text{m}}{\text{1}\text{cm}}}{\sqrt{\left({\left(30\text{cm×}\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)}^{-2}+{\left(3\text{cm×}\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)}^{-2}\right)}}\right)\\ =5.15\times {10}^5\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the Electric field at a point on the axis and $3.00\text{mm}$ from the center is $5.15\times {10}^5\text{N}/\text{C}$ away from the center of the disk.

To determine

The percentage change when part (c) is compared with the field obtained by treating the disk as a $+5.20\text{μC}$ charged particle at a distance of $30\text{cm}$.

Answer to Problem 5P

The magnitude of electric field obtained by treating the disk as a $+5.20\text{μC}$ charged particle at a distance of $30\text{cm}$ is $5.19\times {10}^5\text{N}/\text{C}$ and this value is $0.7\%$ higher.

Explanation of Solution

The formula to calculate the electric field of a charged particle is,

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

Here

$k_{\text{e}}$ is the electrostatic constant.

$Q$ is the electric charge.

$R$ is the radius of the hollow sphere.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/\text{C}$ for $k_e$, $\text{5}\text{.20}\text{μC}$ for the value of $Q$, $30.0\text{cm}$ for $R$, in the above formula.

$\begin{array}{c}E=\frac{k_{\text{e}}Q}{R^2}\\ =\left(\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/\text{C}\right)\left(5.20\mu \text{C}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)}{\text{30}\text{cm×}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)}\right)\\ =5.19\times {10}^5\text{N}/\text{C}\end{array}$

the percentage change when part (c) is compared with the field obtained by treating the disk as a $+5.20\text{μC}$ charged particle.

The value of electric field for uniformly charged disk for a point $30\text{cm}$ from the center is $5.15\times {10}^5\text{N}/\text{C}$.

The value of the electric field computed from near field approximation is $5.19\times {10}^5\text{N}/\text{C}$.

The formula to calculate the percentage change when part (c) is compared with the field obtained by treating the disk as a $+5.20\text{μC}$ charged particle is,

$\begin{array}{c}\%\text{ change}=\frac{\left(5.19-5.15\right)\times {10}^5}{5.19\times {10}^5}\times 100\\ =\frac{0.04\times {10}^5}{5.19\times {10}^5}\times 100\\ =0.7\%\text{ higher}\end{array}$

Conclusion:

Therefore, the magnitude of electric field obtained by treating the disk as a $+5.20\text{μC}$ charged particle at a distance of $30\text{cm}$ is $5.19\times {10}^5\text{N}/\text{C}$ and this value is $0.7\%$ higher.