Chapter 16, Problem 16E

(a)

To determine

The magnitude and direction of electric field between the plates.

Answer to Problem 16E

The magnitude of electric field between the plates is $\underset{\_}{1.13\times {10}^7\text{N}\cdot {\text{C}}^{-1}}$ and it is directed from positive plate to negative plate.

Explanation of Solution

Given that the side length of the square plates is $0.1\text{m}$, the magnitude of charge on the plates is $1.0\times {10}^{-6}\text{C}$, and the separation between the plates is $1.0\times {10}^{-2}\text{m}$.

Write the expression for the magnitude of electric field between two oppositely charged plates.

  $E=4\pi k\sigma$        (I)

Here, $E$ is the electric field, $\sigma$ is the surface charge density, and $k$ is a constant ($k=9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$).

The surface charge density is defined as the charge per unit area. Write the expression for the surface charge density.

  $\sigma =\frac{Q}{A}$        (II)

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

Write the expression for the area of a square plate.

  $A=a^2$        (III)

Here, $a$ is the side length of the plate.

Use equation (II) in (I).

  $E=\frac{4\pi kQ}{A}$        (IV)

Use equation (III) in (IV).

  $E=\frac{4\pi kQ}{a^2}$        (V)

Conclusion:

Substitute $9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}$ for $k$, $1.0\times {10}^{-6}\text{C}$ for $Q$, and $0.1\text{m}$ for $a$ in equation (V) to find $E$.

  $\begin{array}{c}E=\frac{4\pi \left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2\cdot {\text{C}}^{-2}\right)\left(1.0\times {10}^{-6}\text{C}\right)}{{\left(0.1\text{m}\right)}^2}\\ =1.13\times {10}^7\text{N}\cdot {\text{C}}^{-1}\end{array}$

The total electric field between two oppositely charged plates are always directed from positive plate to negative plate (refer Figure 16.15 d).

Therefore, the magnitude of electric field between the plates is $\underset{\_}{1.13\times {10}^7\text{N}\cdot {\text{C}}^{-1}}$ and it is directed from positive plate to negative plate.

(b)

To determine

The magnitude and direction of the force on an electron placed in between the plates.

Answer to Problem 16E

The magnitude of the force on the electron is $\underset{\_}{1.81\times {10}^{-12}\text{N}}$ and it is directed opposite to the direction of electric field.

Explanation of Solution

It is obtained that the electric field is $1.13\times {10}^7\text{N}\cdot {\text{C}}^{-1}$ directed from positive plate to negative plate.

Write the expression for the electrostatic force on a charged particle in an electric field.

  $F=qE$        (VI)

Here, $F$ is the force, and $q$ is the charge.

The charge of electron is $-e$. Replace $q$ by $-e$ in equation (VI).

  $F=-eE$        (VII)

Conclusion:

Substitute $1.13\times {10}^7\text{N}\cdot {\text{C}}^{-1}$ for $E$, and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (VII) to find $F$.

  $\begin{array}{c}F=-\left(1.6\times {10}^{-19}\text{C}\right)\left(1.13\times {10}^7\text{N}\cdot {\text{C}}^{-1}\right)\\ =-1.81\times {10}^{-12}\text{N}\end{array}$

The negative sign indicate that the direction of force is opposite to the direction of electric field.

Therefore, the magnitude of the force on the electron is $\underset{\_}{1.81\times {10}^{-12}\text{N}}$ and it is directed opposite to the direction of electric field.

(c)

To determine

The work needed to move an electron from positive plate to negative plate.

Answer to Problem 16E

The work needed to move an electron from positive plate to negative plate is $\underset{\_}{1.81\times {10}^{-14}\text{J}}$.

Explanation of Solution

It is obtained that the magnitude of force on the electron is $1.81\times {10}^{-12}\text{N}$. The separation between the plates is $1.0\times {10}^{-2}\text{m}$.

Work has to be done to move an electron from positive plate to negative plate against the electric field. The amount of work required can be computed using the equation,

  $W=Fd$        (VIII)

Here, $W$ is the work done, $d$ is the distance moved (separation between the plates).

Conclusion:

Substitute $1.81\times {10}^{-12}\text{N}$ for $F$, and $1.0\times {10}^{-2}\text{m}$ for $d$ in equation (VIII) to find $W$.

  $\begin{array}{c}W=\left(1.81\times {10}^{-12}\text{N}\right)\left(1.0\times {10}^{-2}\text{m}\right)\\ =1.81\times {10}^{-14}\text{J}\end{array}$

Therefore, the work needed to move an electron from positive plate to negative plate is $\underset{\_}{1.81\times {10}^{-14}\text{J}}$.