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Expert-verified Found in: Page 666 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # A simple and common technique for accelerating electrons is shown in Figure 18.55, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is $$2.50 \times {10^4}{\rm{ N/C}}$$. (b) Explain why the electron will not be pulled back to the positive plate once it moves through the hole. Figure 18.55 Parallel conducting plates with opposite charges on them create a relatively uniform electric field used to accelerate electrons to the right. Those that go through the hole can be used to make a TV or computer screen glow or to produce X-rays.

(a) The acceleration of the electron is $$4.39 \times {10^{15}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}$$.

(b) The electron will not be pulled back to the positive plate once it moves through the hole because there is no electric field outside the plate.

See the step by step solution

## Step 1: Coulomb force

The force experienced by the charge when it is placed is placed in an electric field created by some other charge is known as Coulomb force or electrostatic force. The expression for the Coulomb force is,

$$F = qE$$

Here, $$q$$ is the magnitude of charge on electron, and $$E$$ is the electric field.

## Step 2: Acceleration of the electron

The force acting on the particle is,

$$F = ma$$

Here, $$m$$ is the mass of the electron, and $$a$$ is the acceleration of the electron.

Comparing equations (1.1) and (1.2) we get,

$$ma = qE$$

Therefore, the expression for the acceleration of the electron is,

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

Substitute $$1.6 \times {10^{ - 19}}{\rm{ }}C$$ for $$q$$, $$2.50 \times {10^4}{\rm{ N/C}}$$ for $$E$$, and $$9.1 \times {10^{ - 31}}{\rm{ kg}}$$ for $$m$$.

$$\begin{array}{c}a = \frac{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ }}C} \right) \times \left( {2.50 \times {{10}^4}{\rm{ N/C}}} \right)}}{{\left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)}}\\ = 4.39 \times {10^{15}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\end{array}$$

Hence, the acceleration of the electron is $$4.39 \times {10^{15}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}$$.

## Step 3: Reason the electron will not be pulled back to the positive place

Once the electron passes the hole, there will be no electric field exist outside the plates.

Hence, the electron will not be pulled back to the positive plate once it moves through the hole because there is no electric field outside the plate. ### Want to see more solutions like these? 