Chapter 22, Problem 71P

(a)

To determine

The electric field between the two parallel metal plates.

Explanation of Solution

Given:

Area of the two plates is $500{\text{cm}}^{\text{2}}$ .

Separation between the plates is $1.50\text{cm}$ .

The amount of charge that is transfer from left plate to the right is $+1.50\text{nC}$ .

Formula used:

Draw a diagram to show the surface charge density of the two metal parallel plates 1 and 2.

  

Write the expression of electric field due to a plane surface.

  $E=-\frac{\sigma }{2{\epsilon }_0}$

Here, $E$ is the electric field, ${\epsilon }_0$ is the electric permittivity in free space and $\sigma$ is the surface charge density.

Write the expression for the surface charge density due to the plane 1.

  ${\sigma }_1={\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$

Here, ${\sigma }_1$ is the surface charge density due to the plate 1, ${\sigma }_{\text{1L}}$ is the surface charge density in the left side of the plate 1 and ${\sigma }_{\text{1R}}$ is the surface charge density in the right side of the plate 1.

Write the expression for the surface charge density due to the plane 2.

  ${\sigma }_2={\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$

Here, ${\sigma }_2$ is the surface charge density due to the plate 2, ${\sigma }_{\text{2L}}$ is the surface charge density in the left side of the plate 2 and ${\sigma }_{\text{2R}}$ is the surface charge density in the right side of the plate 2.

Calculation:

The electric field at a distance $0.25\text{cm}$ from the plate on right between the two plates is calculated below.

  $\begin{array}{c}{E}_x=\frac{{\sigma }_{\text{1L}}}{2{\epsilon }_0}+\frac{{\sigma }_{\text{1R}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2L}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2R}}}{2{\epsilon }_0}\\ =\frac{{\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}}{2{\epsilon }_0}\end{array}$   ........ (1)

Here, $E_x$ is the total electric field between the metal plates due to plate 1 and 2.

Substitute ${\sigma }_1$ for ${\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$ and ${\sigma }_2$ for ${\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$ in equation (1).

  $E_x=\frac{{\sigma }_{\text{1}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2}}}{2{\epsilon }_0}$

Charge of the plates is equal but opposite sue to the transfer of charges from one plate to another.

Substitute $\left(-{\sigma }_2\right)$ for ${\sigma }_1$ in the above equation.

  $\begin{array}{c}{E}_x=\frac{-{\sigma }_2}{2{\epsilon }_0}-\frac{{\sigma }_2}{2{\epsilon }_0}\\ =\frac{-{\sigma }_2}{{\epsilon }_0}\end{array}$

Substitute $\frac{Q}{A}$ for ${\sigma }_2$ in the above equation.

  $E_x=-\frac{Q}{{\epsilon }_{0}A}$

Here, $Q$ is the total charge and $A$ is the area of the plates.

Substitute $8.854\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$ , $+1.50\text{nC}$ for $Q$ and $500{\text{cm}}^{\text{2}}$ for $A$ in the above equation.

  $\begin{array}{c}{E}_x=-\frac{\left(+1.50\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{\left(8.854\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(500{\text{cm}}^{\text{2}}\left(\frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^{\text{2}}}\right)\right)}\\ =-3389\text{N/C}\\ \approx -3.39\text{kN/C}\end{array}$

Conclusion:

Thus, the electric field is $-3.39\text{kN/C}$ at a distance $0.25\text{cm}$ from the plate on right between the two plates and the direction of the electric field is towards the left.

(b)

To determine

The electric field between the two parallel metal plates.

Explanation of Solution

Given:

Area of the two plates is $500{\text{cm}}^{\text{2}}$ .

Separation between the plates is $1.50\text{cm}$ .

The amount of charge that is transfer from left plate to the right is $+1.50\text{nC}$ .

Formula used:

Write the expression of electric field due to a plane surface.

  $E=-\frac{\sigma }{2{\epsilon }_0}$

Here, $E$ is the electric field, ${\epsilon }_0$ is the electric permittivity in free space and $\sigma$ is the surface charge density.

Write the expression for the surface charge density due to the plane 1.

  ${\sigma }_1={\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$

Here, ${\sigma }_1$ is the surface charge density due to the plate 1, ${\sigma }_{\text{1L}}$ is the surface charge density in the left side of the plate 1 and ${\sigma }_{\text{1R}}$ is the surface charge density in the right side of the plate 1.

Write the expression for the surface charge density due to the plane 2.

  ${\sigma }_2={\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$

Here, ${\sigma }_2$ is the surface charge density due to the plate 2, ${\sigma }_{\text{2L}}$ is the surface charge density in the left side of the plate 2 and ${\sigma }_{\text{2R}}$ is the surface charge density in the right side of the plate 2.

Calculation:

The electric field at a distance $1.00\text{cm}$ from the plate on right between the two plates is calculated below.

  $\begin{array}{c}{E}_x=\frac{{\sigma }_{\text{1L}}}{2{\epsilon }_0}+\frac{{\sigma }_{\text{1R}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2L}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2R}}}{2{\epsilon }_0}\\ =\frac{{\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}}{2{\epsilon }_0}\end{array}$

Here, $E_x$ is the total electric field between the metal plates due to plate 1 and 2.

Substitute ${\sigma }_1$ for ${\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$ and ${\sigma }_2$ for ${\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$ in equation (1).

  $E_x=\frac{{\sigma }_{\text{1}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2}}}{2{\epsilon }_0}$

Charge of the plates is equal but opposite sue to the transfer of charges from one plate to another.

Substitute $\left(-{\sigma }_2\right)$ for ${\sigma }_1$ in the above equation.

  $\begin{array}{c}{E}_x=\frac{-{\sigma }_2}{2{\epsilon }_0}-\frac{{\sigma }_2}{2{\epsilon }_0}\\ =\frac{-{\sigma }_2}{{\epsilon }_0}\end{array}$

Substitute $\frac{Q}{A}$ for ${\sigma }_2$ in the above equation.

  $E_x=-\frac{Q}{{\epsilon }_{0}A}$

Here, $Q$ is the total charge and $A$ is the area of the plates.

Substitute $8.854\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$ , $+1.50\text{nC}$ for $Q$ and $500{\text{cm}}^{\text{2}}$ for $A$ in the above equation.

  $\begin{array}{c}{E}_x=-\frac{\left(+1.50\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{\left(8.854\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(500{\text{cm}}^{\text{2}}\left(\frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^{\text{2}}}\right)\right)}\\ =-3388\text{N/C}\\ \approx -3.39\text{kN/C}\end{array}$

Conclusion:

Thus, the electric field is $-3.39\text{kN/C}$ at a distance $1.00\text{cm}$ from the plate on right between the two plates and the direction of the electric field is towards the left.

(c)

To determine

The electric field to the left of the left plate.

Explanation of Solution

Given:

Area of the two plates is $500{\text{cm}}^{\text{2}}$ .

Separation between the plates is $1.50\text{cm}$ .

The amount of charge that is transfer from left plate to the right is $+1.50\text{nC}$ .

Formula used:

Write the expression of electric field due to a plane surface.

  $E=-\frac{\sigma }{2{\epsilon }_0}$

Here, $E$ is the electric field, ${\epsilon }_0$ is the electric permittivity in free space and $\sigma$ is the surface charge density.

Write the expression for the surface charge density due to the plane 1.

  ${\sigma }_1={\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$

Here, ${\sigma }_1$ is the surface charge density due to the plate 1, ${\sigma }_{\text{1L}}$ is the surface charge density in the left side of the plate 1 and ${\sigma }_{\text{1R}}$ is the surface charge density in the right side of the plate 1.

Write the expression for the surface charge density due to the plane 2.

  ${\sigma }_2={\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$

Here, ${\sigma }_2$ is the surface charge density due to the plate 2, ${\sigma }_{\text{2L}}$ is the surface charge density in the left side of the plate 2 and ${\sigma }_{\text{2R}}$ is the surface charge density in the right side of the plate 2.

Calculation:

The electric field to the left from the plate on left.

  $\begin{array}{c}{E}_x=-\frac{{\sigma }_{\text{1L}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{1R}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2L}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2R}}}{2{\epsilon }_0}\\ =-\frac{\left({\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}\right)}{2{\epsilon }_0}-\frac{\left({\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}\right)}{2{\epsilon }_0}\end{array}$   ........ (2)

Here, $E_x$ is the total electric field between the metal plates due to plate 1 and 2.

Substitute ${\sigma }_1$ for ${\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$ and ${\sigma }_2$ for ${\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$ in equation (2).

  $E_x=-\frac{{\sigma }_{\text{1}}}{2{\epsilon }_0}-\frac{{\sigma }_{\text{2}}}{2{\epsilon }_0}$

Charge of the plates is equal but opposite sue to the transfer of charges from one plate to another.

Substitute $\left(-{\sigma }_2\right)$ for ${\sigma }_1$ in the above equation.

  $\begin{array}{c}{E}_x=-\frac{-{\sigma }_2}{2{\epsilon }_0}-\frac{{\sigma }_2}{2{\epsilon }_0}\\ =0\end{array}$

Conclusion:

Thus, the electric field is $0$ to the left from the plate on left.

(d)

To determine

The electric field to the right of the right plate.

Explanation of Solution

Given:

Area of the two plates is $500{\text{cm}}^{\text{2}}$ .

Separation between the plates is $1.50\text{cm}$ .

The amount of charge that is transfer from left plate to the right is $+1.50\text{nC}$ .

Formula used:

Write the expression of electric field due to a plane surface.

  $E=-\frac{\sigma }{2{\epsilon }_0}$

Here, $E$ is the electric field, ${\epsilon }_0$ is the electric permittivity in free space and $\sigma$ is the surface charge density.

Write the expression for the surface charge density due to the plane 1.

  ${\sigma }_1={\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$

Here, ${\sigma }_1$ is the surface charge density due to the plate 1, ${\sigma }_{\text{1L}}$ is the surface charge density in the left side of the plate 1 and ${\sigma }_{\text{1R}}$ is the surface charge density in the right side of the plate 1.

Write the expression for the surface charge density due to the plane 2.

  ${\sigma }_2={\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$

Here, ${\sigma }_2$ is the surface charge density due to the plate 2, ${\sigma }_{\text{2L}}$ is the surface charge density in the left side of the plate 2 and ${\sigma }_{\text{2R}}$ is the surface charge density in the right side of the plate 2.

Calculation:

The electric field to the right from the plate on right.

  $\begin{array}{c}{E}_x=\frac{{\sigma }_{\text{1L}}}{2{\epsilon }_0}+\frac{{\sigma }_{\text{1R}}}{2{\epsilon }_0}+\frac{{\sigma }_{\text{2L}}}{2{\epsilon }_0}+\frac{{\sigma }_{\text{2R}}}{2{\epsilon }_0}\\ =\frac{\left({\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}\right)}{2{\epsilon }_0}+\frac{\left({\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}\right)}{2{\epsilon }_0}\end{array}$   ........ (3)

Here, $E_x$ is the total electric field between the metal plates due to plate 1 and 2.

Substitute ${\sigma }_1$ for ${\sigma }_{\text{1L}}+{\sigma }_{\text{1R}}$ and ${\sigma }_2$ for ${\sigma }_{\text{2L}}+{\sigma }_{\text{2R}}$ in equation (3).

  $E_x=\frac{{\sigma }_{\text{1}}}{2{\epsilon }_0}+\frac{{\sigma }_{\text{2}}}{2{\epsilon }_0}$

Charge of the plates is equal but opposite sue to the transfer of charges from one plate to another.

Substitute $\left(-{\sigma }_2\right)$ for ${\sigma }_1$ in the above equation.

  $\begin{array}{c}{E}_x=\frac{-{\sigma }_2}{2{\epsilon }_0}+\frac{{\sigma }_2}{2{\epsilon }_0}\\ =0\end{array}$

Conclusion:

Thus, the electric field is $0$ to the right from the plate on right.