Chapter 22, Problem 12P

(a)

To determine

To find the electric field strength at a distance $0.010\text{cm}$ and compare the values in both approximate and exact case.

Explanation of Solution

Given:

Radius of a disk is $2.5\text{cm}$ .

Surface charge density of the disk is $3.6{\text{μC/m}}^{\text{2}}$ .

Distance of a point on axis is $0.010\text{cm}$ .

Formula used:

Write the expression of the electric field magnitude on the axis of a uniformly charged disk.

  $\left|E\right|=\frac{\sigma }{2{\epsilon }_0}\left[1-{\left(1+\frac{R^2}{z^2}\right)}^{-\frac{1}{2}}\right]$

Here, $E$ is the electric field, ${\epsilon }_0$ is the electric permittivity, $R$ is radius of the disk, $z$ is the distance.

Simplify above equation.

  $\left|E\right|=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{z}{\sqrt{{\left(z\right)}^2+{\left(R\right)}^2}}\right]$ ....... (1)

Write the expression of the electric field magnitude when $\left(\left|z\right|>>R\right)$ .

  $E=\frac{Q}{4\pi {\epsilon }_0{z}^2}$ ....... (2)

Here, $Q$ is the total charge.

Write the expression of the electric field magnitude when $\left(\left|z\right|<<R\right)$ .

  $E=\frac{\sigma }{2{\epsilon }_0}$

  $$ ....... (3)

Write the expression for surface charge density.

  $\sigma =\frac{Q}{A}$

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

Substitute $\pi R^2$ for $A$ and rearrange the expression for charge.

  $Q=\pi R^2\sigma$

Here, $R$ is the radius of the disk.

Substitute $\pi {\text{R}}^2\sigma$ for $Q$ in equation (2).

  $E=\left(\frac{\sigma R^2}{4{\epsilon }_0{z}^2}\right)$ ....... (4)

Calculation:

Approximate value of electric field is calculated below.

Substitute $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$ and $3.6{\text{μC/m}}^{\text{2}}$ for $\sigma$ in equation (3)

  $\begin{array}{c}\left|E\right|=\frac{\left(3.5{\text{μC/m}}^{\text{2}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)}\\ =197740.11 \text{N/C }\\ \approx \text{2}\times {\text{10}}^5\text{N/C }\end{array}$

The exact value of the electric field is calculated below.

Substitute $\text{3}\text{.6}\text{μC}/{\text{m}}^2$ for $\sigma$ , $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$ , $2.5\text{cm}$ for R and $0.010\text{cm}$ for $z$ in equation (1).

  $\begin{array}{c}\left|E\right|=\frac{\left(3.5{\text{μC/m}}^{\text{2}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)}\left[1-\frac{0.010\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{\sqrt{{\left(0.010\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2+{\left(\text{2}\text{.5}\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}}\right]\\ =196949.15 \text{N/C }\\ \approx \text{2}\times {\text{10}}^5\text{ N/C }\end{array}$

Conclusion:

Thus, near the disk the approximate result and the exact result are near about same. There isa parity between two values.

(b)

To determine

To find the electric field strength at a distance $0.040\text{cm}$ and compare the values in both approximate and exact case.

Explanation of Solution

Given:

Radius of a disk is $2.5\text{cm}$ .

Surface charge density of the disk is $3.6{\text{μC/m}}^{\text{2}}$ .

Distance of a point on axis is $0.040\text{cm}$ .

Formula used:

Write the expression of the electric field magnitude on the axis of a uniformly charged disk.

  $\left|E\right|=\frac{\sigma }{2{\epsilon }_0}\left[1-{\left(1+\frac{R^2}{z^2}\right)}^{-\frac{1}{2}}\right]$

Simplify above equation.

  $\left|E\right|=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{z}{\sqrt{{\left(z\right)}^2+{\left(R\right)}^2}}\right]$

Here, $E$ is the electric field, ${\epsilon }_0$ is the electric permittivity, $R$ is radius of the disk, $z$ is the distance.

Write the expression of the electric field magnitude when $\left(\left|z\right|>>R\right)$ .

  $E=\frac{Q}{4\pi {\epsilon }_0{z}^2}$

Here, $Q$ is the total charge.

Write the expression of the electric field magnitude when $\left(\left|z\right|<<R\right)$ .

  $E=\frac{\sigma }{2{\epsilon }_0}$

Write the expression for surface charge density.

  $\sigma =\frac{Q}{A}$

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

Substitute $A=\pi R^2$ and rearrange the expression for charge.

  $Q=\pi R^2\sigma$

Here, $R$ is the radius of the disk.

Substitute $\pi {\text{R}}^2\sigma$ for $Q$ in equation (2)

  $E=\left(\frac{\sigma R^2}{4{\epsilon }_0{z}^2}\right)$

Calculation:

Approximate value of electric field is calculated below.

Substitute $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$ and $3.6{\text{μC/m}}^{\text{2}}$ for $\sigma$ in equation (3)

  $\begin{array}{c}E=\frac{\left(3.5{\text{μC/m}}^{\text{2}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)}\\ =197740.11 \text{N/C }\\ \approx \text{2}\times {\text{10}}^5\text{N/C}\end{array}$

The exact value of the electric field is calculated below.

Substitute $\text{3}\text{.6}\text{μC}/{\text{m}}^2$ for $\sigma$ , $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$ , $2.5\text{cm}$ for R and $0.040\text{cm}$ for $z$ in equation (1).

  $\begin{array}{c}\left|E\right|=\frac{\left(3.5{\text{μC/m}}^{\text{2}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)}\left[1-\frac{0.040\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{\sqrt{{\left(0.040\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2+{\left(\text{2}\text{.5}\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}}\right]\\ =194576.67\text{N/C }\end{array}$

Conclusion:

Thus, for the in case of electric field at a distance $0.040\text{cm}$ approximate result is deflected from the exact value as the distance increases from the disk.

(c)

To determine

To find the electric field strength at a distance $5.0\text{m}$ and compare the values in both approximate and exact case.

Explanation of Solution

Given:

Radius of a disk is $2.5\text{cm}$ .

Surface charge density of the disk is $3.6{\text{μC/m}}^{\text{2}}$ .

Distance of a point on axis is $5.0\text{m}$ .

Formula used:

Write the expression of the electric field magnitude on the axis of a uniformly charged disk.

  $\left|E\right|=\frac{\sigma }{2{\epsilon }_0}\left[1-{\left(1+\frac{R^2}{z^2}\right)}^{-\frac{1}{2}}\right]$

Simplify above equation.

  $\left|E\right|=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{z}{\sqrt{{\left(z\right)}^2+{\left(R\right)}^2}}\right]$

Here, $E$ is the electric field, ${\epsilon }_0$ is the electric permittivity, $R$ is radius of the disk, $z$ is the distance.

Write the expression of the electric field magnitude when $\left(\left|z\right|>>R\right)$ .

  $E=\frac{Q}{4\pi {\epsilon }_0{z}^2}$

Here, $Q$ is the total charge.

Write the expression of the electric field magnitude when $\left(\left|z\right|<<R\right)$ .

  $E=\frac{\sigma }{2{\epsilon }_0}$

Write the expression for surface charge density.

  $\sigma =\frac{Q}{A}$

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

Substitute $A=\pi R^2$ and rearrange the expression for charge.

  $Q=\pi R^2\sigma$

Here, $R$ is the radius of the disk.

Substitute $\pi {\text{R}}^2\sigma$ for $Q$ in equation (2)

  $E=\left(\frac{\sigma R^2}{4{\epsilon }_0{z}^2}\right)$

Calculation:

Approximate value of electric field is calculated below.

Substitute $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$ , $3.6{\text{μC/m}}^{\text{2}}$ for $\sigma$ , $2.5\text{m}$ for $R$ and $5.0\text{m}$ for $z$ in equation (4)

  $\begin{array}{c}E=\frac{\left(3.5{\text{μC/m}}^{\text{2}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}{4\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)}\left(\frac{{\left(\text{2}\text{.5}\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{{\left(5.0m\right)}^2}\right)\\ =2.471\text{N/C}\\ \approx \text{2}\text{.5}\text{N/C}\end{array}$

The exact value of the electric field is calculated below.

Substitute $\text{3}\text{.6}\text{μC}/{\text{m}}^2$ for $\sigma$ , $8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}$ for ${\epsilon }_0$ , $2.5\text{cm}$ for R and $5.0\text{m}$ for $z$ in equation (1).

  $\begin{array}{c}\left|E\right|=\frac{\left(3.5{\text{μC/m}}^{\text{2}}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}{2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N-m}}^{\text{2}}\right)}\left[1-\frac{5.0\text{m}}{\sqrt{{\left(5.0\text{m}\right)}^2+{\left(\text{2}\text{.5}\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}}\right]\\ =2.471\text{N/C}\\ \approx \text{2}\text{.5}\text{N/C}\end{array}$

Conclusion:

Thus, for large distances approximate result has a good agreement with the exact result. Two values are equal.