Chapter 18, Problem 1SP

(a)

To determine

The direction in which the electron beam deflect as it passes between the plates bearing the potential difference.

Answer to Problem 1SP

The electrons will be deflected in the upward direction, which is toward the positive plate.

Explanation of Solution

The charge of electrons is negative. The electric field is directed from positive charge to the negative charge.

Any particle with negative charge, in an electric field experiences a force in the direction opposite to the direction of the electric field. This will result the particle to move in a direction opposite to the electric field.

In the given situation, the potential difference is set up between the two plates such that the positive plate is in the upper side and negative plate is in the lower side. Thus, the electric field will be directing in the downward direction.

Since the electrons are accelerated between these plates, the force on the electrons will be in the upward direction and hence they will get deflected in the upward direction.

Conclusion:

Therefore, the electrons will be deflected in the upward direction, which is toward the positive plate.

(b)

To determine

The electric field in the region between the plates.

Answer to Problem 1SP

The electric field in the region between the plates is $15000\text{N/C}$.

Explanation of Solution

Given info: The separation between the plates is $2\text{cm}$ and the potential difference is $300\text{V}$.

Write the expression between the potential difference in terms of electric field.

$\Delta V=Ed$

Here,

$\Delta V$ is the potential difference

$E$ is the electric field

$d$ is the distance of separation between the charge sources

Solve for $E$.

$E=\frac{\Delta V}{d}$

Substitute $300\text{V}$ for $\Delta V$ and $2\text{cm}$ for $d$ to find the electric field $E$.

$\begin{array}{c}E=\frac{300\text{V}}{2\text{cm}}\\ =\frac{300\text{V}}{2\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\\ =15,000\text{N/C}\end{array}$

Conclusion:

Therefore, the electric field in the region between the plates is $15,000\text{N/C}$.

(c)

To determine

The magnitude of force exerted on an individual electron by the electric field between the plates.

Answer to Problem 1SP

The magnitude of force exerted on an individual electron by the electric field between the plates is $2.4\times {10}^{-15}\text{N}$.

Explanation of Solution

Given info: The electric field is $15,000\text{N/C}$.

Electric field is the force exerted on a unit charge.

Write the expression for the electric field.

$E=\frac{F}{q}$

Here,

$F$ is the magnitude of force

$q$ is the magnitude of charge

Solve for $F$.

$F=qE$

The magnitude of charge of electron is $1.6\times {10}^{-19}\text{C}$.

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$ and $15,000\text{N/C}$ for $E$ to find the force $F$.

$\begin{array}{c}F=\left(1.6\times {10}^{-19}\text{C}\right)\left(15,000\text{N/C}\right)\\ =2.4\times {10}^{-15}\text{N}\end{array}$

Conclusion:

Therefore, the magnitude of force exerted on an individual electron by the electric field between the plates is $2.4\times {10}^{-15}\text{N}$.

(d)

To determine

The magnitude and direction of the acceleration of the electron.

Answer to Problem 1SP

The electrons will have acceleration of $2.64\times {10}^{15}{\text{m/s}}^{\text{2}}$ which is directed in the upward direction.

Explanation of Solution

Given info: The force on the electron is $2.4\times {10}^{-15}\text{N}$.

Write the expression for the force on an object in terms of accelerations.

$F=ma$

Here,

$m$ is the mass of the object

$a$ is the acceleration of the object

Solve for $a$.

$a=\frac{F}{m}$

The mass of electron is $9.1\times {10}^{-31}\text{kg}$.

Substitute $2.4\times {10}^{-15}\text{N}$ for $F$ and $9.1\times {10}^{-31}\text{kg}$ for $m$ to find the acceleration of the electron.

$\begin{array}{c}a=\frac{2.4\times {10}^{-15}\text{N}}{9.1\times {10}^{-31}\text{kg}}\\ =2.64\times {10}^{15}{\text{m/s}}^{\text{2}}\end{array}$

Since the force is acting on the upward direction, the acceleration will also be in the same direction. Thus, the electron will be accelerated in the upward direction.

Conclusion:

Therefore, the electrons will have acceleration of $2.64\times {10}^{15}{\text{m/s}}^{\text{2}}$ which is directed in the upward direction.

(e)

To determine

The path followed by the electron when it passes through the potential difference.

Answer to Problem 1SP

The electron will follow a parabolic path towards the positively charged plate when it is accelerated between the plates.

Explanation of Solution

The trajectory of motion of a charged particle in an electric field is similar to the motion of a body in a constant gravitational field.

When the electron is accelerated between the charged plated beating a potential difference, the motion of the electron is deviated due to the force exerted by the electric field on the electron.

Since the electron experiences a force in the upward direction, offered by the electric field, it will follow a parabolic path towards the positively charged plate.

Conclusion:

Therefore, the electron will follow a parabolic path towards the positively charged plate when it is accelerated between the plates.