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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 # An electron has an initial velocity of $$5.00 \times {10^6}{\rm{ m}}/{\rm{s}}$$ in a uniform $$2.00 \times {10^5}{\rm{ N}}/{\rm{C}}$$ strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron’s velocity when it returns to its starting point?

(a) The electric field is in the same direction as the initial velocity of the electron.

(b) The electron will travel a distance of $$3.55 \times {10^{ - 4}}{\rm{ m}}$$ before coming to rest.

(c) The time taken by the electron to come to rest is $$1.43 \times {10^{ - 10}}{\rm{ s}}$$.

(d) The velocity of electron when it returns to its start point is $$5.00 \times {10^6}{\rm{ m}}/{\rm{s}}$$.

See the step by step solution

## Step 1: Acceleration of the electron in an electric field

The force acting on the electron due to electric field is,

$$F = qE$$

Here, $$q$$ is the magnitude of the charge, and $$E$$ is the magnetic field.

The force acting on electron causes the acceleration of the electron. The force is,

$$F = ma$$

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

Since, the electrostatic force is responsible for the acceleration of the electron. Therefore,

$$ma = qE$$

Rearranging the above expression to get an expression for the acceleration of the electron.

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

## Step 2: Calculating the acceleration of the electron

The acceleration of the electron can be calculated using equation (1.3).

Substitute $$- 1.6 \times {10^{ - 19}}{\rm{ C}}$$ for $$q$$, $$2.00 \times {10^5}{\rm{ N}}/{\rm{C}}$$ for $$E$$, and $$9.1 \times {10^{ - 31}}{\rm{ kg}}$$ for $$m$$ into equation (1.3).

$$\begin{array}{c}a = \frac{{\left( { - 1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right) \times \left( {2.00 \times {{10}^5}{\rm{ N}}/{\rm{C}}} \right)}}{{\left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)}}\\ = - 3.52 \times {10^{16}}{\rm{ m}}/{{\rm{s}}^2}\end{array}$$

## Step 3: (a) Direction of the electric field

The electric field accelerates the electron in the opposite direction to its initial velocity.

Hence, the electric field is in the direction of the initial velocity.

## Step 4: (b) Distance traveled by the electron before coming to rest

The third equation of the motion is given as,

$${v^2} = {u^2} + 2as$$

Here, $$v$$ is the final velocity of the electron ($$v = 0$$ as the electron comes to rest), $$u$$ is the initial velocity of the electron $$\left( {u = 5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)$$, $$a$$ is the acceleration of the electron in the electric field $$\left( {a = - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)$$, and $$s$$ is the distance traveled by the electron before coming to rest.

The expression for the distance traveled by the electron before coming to the rest is,

$$s = \frac{{{v^2} - {u^2}}}{{2a}}$$

Substituting all known values,

$$\begin{array}{c}s = \frac{{{{\left( 0 \right)}^2} - {{\left( {5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( { - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 3.55 \times {10^{ - 4}}{\rm{ m}}\end{array}$$

Hence, the electron will travel a distance of $$3.55 \times {10^{ - 4}}{\rm{ m}}$$ before coming to rest.

## Step 5: (c) Time taken by the electron to come to rest

The first equation of the motion is,

$$v = u + at$$

Here, $$v$$ is the final velocity of the electron ($$v = 0$$ as the electron comes to rest), $$u$$ is the initial velocity of the electron $$\left( {u = 5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)$$, $$a$$ is the acceleration of the electron in the electric field $$\left( {a = - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)$$, and $$t$$ is the time taken by the electron to come to rest.

The expression for the time taken by the electron to come to rest is,

$$t = \frac{{v - u}}{a}$$

Substituting all known values,

$$\begin{array}{c}t = \frac{{\left( 0 \right) - \left( {5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)}}{{\left( { - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 1.43 \times {10^{ - 10}}{\rm{ s}}\end{array}$$

Hence, the time taken by the electron to come to rest is $$1.43 \times {10^{ - 10}}{\rm{ s}}$$.

## Step 6: (d) Velocity of electron when it returns to its stating position

After stopping the direction of the motion of the electron reverses. Thus, the acceleration becomes negative of the initial acceleration.

The third equation of motion is,

$${v'^2} = {u'^2} - 2as$$

Here, $$v'$$ is the final velocity of the electron or the velocity at the starting point, $$u'$$ is the initial velocity of the election in the reverse motion ($$u' = 0$$ as the electron starts from rest while staring the reverse motion), $$\left( {u = 5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)$$, $$a$$ is the acceleration of the electron in the electric field $$\left( {a = - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)$$, and $$s$$ is the distance traveled by the electron before reaching the starting point $$\left( {s = 3.55 \times {{10}^{ - 4}}{\rm{ m}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{{v'}^2} = {\left( 0 \right)^2} - 2 \times \left( { - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {3.55 \times {{10}^{ - 4}}{\rm{ m}}} \right)\\{{v'}^2} = 2.5 \times {10^{13}}{\rm{ }}{{\rm{m}}^2}/{{\rm{s}}^2}\\v' = 5.00 \times {10^6}{\rm{ m}}/{\rm{s}}\end{array}$$

Since, this velocity is opposite of the initial velocity.

Hence, the velocity of electron when it returns to its start point is $$5.00 \times {10^6}{\rm{ m}}/{\rm{s}}$$. ### Want to see more solutions like these? 