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Q19.2-23PE

Expert-verified
Found in: Page 696

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# An electron is to be accelerated in a uniform electric field having a strength of $${\bf{2}}.{\bf{00}} \times {\bf{1}}{{\bf{0}}^6}{\rm{ }}{\bf{V}}/{\bf{m}}$$. (a) What energy in $$keV$$ is given to the electron if it is accelerated through $${\bf{0}}.{\bf{400}}{\rm{ }}{\bf{m}}$$? (b) Over what distance would it have to be accelerated to increase its energy by $${\bf{50}}.{\bf{0}}{\rm{ }}{\bf{GeV}}$$?

1. Value of energy given to electron is $${U_e} = 800{\rm{ }}keV$$.
2. Distance through which electron is accelerated is $$d = 25.0 \times {10^3}\;m$$.
See the step by step solution

## Step 1: Information Given

• Electric field strength of electron is: $${\bf{2}}.{\bf{00}} \times {\bf{1}}{{\bf{0}}^6}{\rm{ }}{\bf{V}}/{\bf{m}}$$
• Distance the electron is accelerated to: $${\bf{0}}.{\bf{400}}{\rm{ }}{\bf{m}}$$
• Energy to be increased is: $${\bf{50}}.{\bf{0}}{\rm{ }}{\bf{GeV}}$$

## Step 2: Definition of electric potential energy and electron volt.

Electron Volt: is the energy given to a fundamental charge accelerated through a potential difference of $$1\;V$$. In equation form,

$$1{\rm{ }}eV = \left( {1.60 \times {{10}^{ - 19}}{\rm{ }}C} \right)(1\;V) = 1.60 \times {10^{ - 19}}\;J$$

Electric Potential Energy: If a particle with charge $$q$$ is positioned at a location where a charged object's electric potential is $$V$$, the particle-object system's electric potential energy is $${U_e} = qV$$.

And the potential difference between two points separated by a distance $$d$$ in a uniform electric field of magnitude $$E$$ is $$\Delta V = Ed$$.

## Step 3: Calculation of the energy given to the electron in Kev.

(a)

The potential difference between the initial and final positions of the electron is found from equation:

$$\Delta V = - Ed$$

Entering the values for $$E$$ and $$d$$:

$$\begin{array}{c}\Delta V = - \left( {2.00 \times {{10}^6}\;V/m} \right)(0.0400\;m)\\\Delta V = - 8.00 \times {10^5}\;V\end{array}$$

The electric potential energy acquired by the electron after moving a distance $$d$$ is found from equation below:

$${U_e} = q\Delta V$$

Where$$q = - e$$ is the electric charge of the electron:

$${U_e} = - e\Delta V$$

Substitute numerical values:

$${U_e} = 1.28 \times {10^{ - 13}}\;J$$

We convert the electric potential energy of the electron from Joules to $$keV$$ using the conversion factor in equation:

$$\begin{array}{c}{U_e} = \left( {1.28 \times {{10}^{ - 13}}\;J} \right)\left( {\frac{{1{\rm{ }}eV}}{{1.60 \times {{10}^{ - 19}}\;J}}} \right)\left( {\frac{{1{\rm{ }}keV}}{{1000{\rm{ }}eV}}} \right)\\{U_e} = 800{\rm{ }}keV\end{array}$$

Hence, the energy value is obtained as $${U_e} = 800{\rm{ }}keV$$.

## Step 4: Calculation of the distance required to accelerate to increase electron’s energy.

(b)

The new required electric potential energy of the electron is:

$$\begin{array}{c}U_e^' = 800{\rm{ }}keV + 50.0{\rm{ }}GeV\\U_e^' = (800{\rm{ }}keV)\left( {\frac{{1000{\rm{ }}eV}}{{1{\rm{ }}keV}}} \right) + (50.0{\rm{ }}GeV)\left( {\frac{{{{10}^9}{\rm{ }}eV}}{{1{\rm{ }}GeV}}} \right)\\U_e^' = 5.00008 \times {10^{10}}{\rm{ }}eV\end{array}$$

Convert the new electric potential energy back to joules using the conversion factor in equation below:

$$\begin{array}{c}U_e^' = \left( {5.00008 \times {{10}^{10}}{\rm{ }}eV} \right)\left( {\frac{{1.6 \times {{10}^{ - 19}}\;J}}{{1{\rm{ }}eV}}} \right)\\U_e^' = 8.000128 \times {10^{ - 9}}\;J\end{array}$$[sg1]

The potential difference between the initial position of the electron and the position at which it acquires an electric potential energy $$U_e^'$$ is found from equation below:

$$\Delta V = \frac{{U_e^'}}{{ - e}}$$

Entering numerical values, it gives

$$\begin{array}{c}\Delta V = \frac{{8.000128 \times {{10}^{ - 9}}\;J}}{{ - 1.60 \times {{10}^{ - 19}}{\rm{ }}C}}\\\Delta V = - 5.00008 \times {10^{10}}\;V\end{array}$$

The distance through which the electron must be accelerated to increase its energy by $$50.0{\rm{ }}GeV$$ is then found by solving equation below for $$d$$:

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

Substituting numerical values, it gives:

$$\begin{array}{c}d = - \frac{{ - 5.00008 \times {{10}^{10}}\;V}}{{2.00 \times {{10}^6}\;V/m}}\\d = 25.0 \times {10^3}\;m\end{array}$$

Therefore, distance value is obtained as $$d = 25.0 \times {10^3}\;m$$.