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Q19.2-23PE

Expert-verifiedFound in: Page 696

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 \({\bf{50}}.{\bf{0}}{\rm{ }}{\bf{GeV}}\)?**

- Value of energy given to electron is \({U_e} = 800{\rm{ }}keV\).
- Distance through which electron is accelerated is \(d = 25.0 \times {10^3}\;m\).

- 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: \({\bf{50}}.{\bf{0}}{\rm{ }}{\bf{GeV}}\)

**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 \(\Delta V = Ed\).

(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 \({U_e} = 800{\rm{ }}keV\).

(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 \(d = 25.0 \times {10^3}\;m\).

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