Chapter 18, Problem 1SP

An electron beam in a cathode-ray tube passes between two parallel plates that have a voltage difference of 500 V across them and are separated by a distance of 4 cm. as shown in the following diagram.

a.    In what direction will the electron beam deflect as it passes between these plates? Explain.

b.    Using the expression for a uniform field. ΔV = Ed, find the value of the electric field in the region between the plates.

c.    What is the magnitude of the force exerted on individual electrons by this field? (F = qE, q = 1.6 × 10^–19C)

d.    What are the magnitude and direction of the acceleration of an electron? (m = 9.1 × 10^–31 kg)

e.    What type of path will the electron follow as it passes through the region between the plates? Explain.

To determine

The direction in which the electron beam will deflect as is passes between the plates.

Answer to Problem 1SP

The electron beam will deflect upward.

Explanation of Solution

Electron is the first discovered fundamental particles. It was discovered by J. J. Thomson in 1897. The mass of electron is $9.1\times {10}^{-31}\text{ kg}$.

Electrons are negatively charged particles. The magnitude of the charge of electron is $1.60\times {10}^{-19}\text{ C}$. In the given situation the upward plate has positive charge and the downward plate has negative charge.

One of the fundamental property of the electric charges is that the like charges repel whereas the unlike charges attract. Thus it will be attracted towards the upward plate which has positive charge. In another way, the direction of the electric field is from positive to negative and the force on a negative charge is opposite to the direction of electric field. This implies the force on the electron is towards the positive plate.

Conclusion:

Thus the electron beam will deflect upward.

To determine

The value of the electric field in the region between the plates.

Answer to Problem 1SP

The value of the electric field in the region between the plates is $12.5\text{ kN/C}$.

Explanation of Solution

Given Info: The distance between the plates is $4\text{ cm}$ and the voltage difference between the plates is $500\text{ V}$.

Write the equation for the electric field between the plates.

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

Here,

$E$ is the electric field between the plates

$\Delta V$ is the voltage difference across the plates

$d$ is the distance between the plates

Substitute $500\text{ V}$ for $\Delta V$ and $4\text{ cm}$ for $d$ in the above equation to find $E$.

$\begin{array}{c}E=\frac{500\text{ V}}{4\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =\frac{500\text{ V}}{0.04\text{ m}}\\ =12500\text{ N/C}\\ =12.5\text{ kN/C}\end{array}$

Conclusion:

Thus the value of the electric field in the region between the plates is $12.5\text{ kN/C}$.

To determine

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

Answer to Problem 1SP

The magnitude of the force exerted on individual electron by the electric field is $2.0\times {10}^{-15}\text{ N}$.

Explanation of Solution

Write the equation for the magnitude of the force on the electron.

$F=eE$

Here,

$F$ is the magnitude of the force on the electron

$e$ is the charge of the electron

The magnitude of $e$ is $1.60\times {10}^{-19}\text{ C}$.

Substitute $1.60\times {10}^{-19}\text{ C}$ for $e$ and $12500\text{ N}$ for $E$ in the above equation to find $F$.

$\begin{array}{c}F=\left(1.60\times {10}^{-19}\text{ C}\right)\left(12500\text{ N/C}\right)\\ =2.0\times {10}^{-15}\text{ N}\end{array}$

Conclusion:

Thus the magnitude of the force exerted on individual electron by the electric field is $2.0\times {10}^{-15}\text{ N}$.

To determine

The magnitude and direction of the acceleration of an electron.

Answer to Problem 1SP

The magnitude of acceleration of an electron is $2.20\times {10}^{15}{\text{ m/s}}^2$ and its direction is towards upward.

Explanation of Solution

Write the equation for the force on a body.

$F=ma$

Here,

$m$ is the mass of the particle

$a$ is the acceleration of the particle

Rewrite the above equation for $a$.

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

Substitute $2.0\times {10}^{-15}\text{ N}$ for $F$ and $9.1\times {10}^{-31}\text{ kg}$ for $m$ in the above equation to find $a$.

$\begin{array}{c}a=\frac{2.0\times {10}^{-15}\text{ N}}{9.1\times {10}^{-31}\text{ kg}}\\ =2.20\times {10}^{15}{\text{ m/s}}^2\end{array}$

The direction of acceleration will be in the direction force on the particle. In part (b) it is found that the force on the particle is toward upward so that the direction of acceleration will also be upward.

Conclusion:

Thus the magnitude of acceleration of an electron is $2.20\times {10}^{15}{\text{ m/s}}^2$ and its direction is towards upward.

To determine

The type of path the electron will follow as it passes through the region between the plates.

Answer to Problem 1SP

The type of path the electron will follow as it passes through the region between the plates will be parabolic and it will be towards the positively charged plate.

Explanation of Solution

The path followed by a particle under the action of given forces is called the trajectory of the particle. The trajectory may have different shapes. It can be expressed by a particular equation.

In the given situation, the electrons are in a uniform electric field between the plates. This implies the electron will have constant acceleration. This is similar to the situation where a body is under the influence of gravitational force where the constant acceleration is equal to the acceleration due to gravity.

A body falling under gravitational force will follow a parabolic path. Similarly the electron moving with the constant acceleration will also follow a parabolic path and it will be directed towards the positive plate situated upward.

Conclusion:

Thus the type of path the electron will follow as it passes through the region between the plates will be parabolic and it will be towards the positively charged plate.