Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

A doubly charged ion is accelerated to an energy of \(32.0{\rm{ }}keV\) by the electric field between two parallel conducting plates separated by \(2.00{\rm{ }}cm\). What is the electric field strength between the plates?

Value of the electric strength between the plates is \(E = 8.00 \times {10^5}\;V/m\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of electric volt and electric potential energy.

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 2: Given data for the calculation of electric field strength.

As, the ion is doubly charged.

The electric potential energy of the ion is:

\(\begin{array}{c}{U_e} = (32.0{\rm{ }}keV)\left( {\frac{{1000{\rm{ }}eV}}{{1{\rm{ }}keV}}} \right)\\{U_e} = 32.0 \times {10^3}{\rm{ }}eV\end{array}\)

The distance between the two plates is:

\(d = (2.00\;cm)\left( {\frac{{1\;m}}{{100\;cm}}} \right)\)

The value of the fundamental charge is:

\(e = 1.60 \times {10^{ - 19}}{\rm{ }}C\)

Step 3: Converting electric potential energy from electron volt to Joule.

Convert the electric potential energy of the ion from electron volts to joules using the conversion factor in equation below:

\(\begin{array}{c}{U_e} = \left( {32.0 \times {{10}^3}{\rm{ }}eV} \right)\left( {\frac{{1.60 \times {{10}^{ - 19}}\;J}}{{1{\rm{ }}eV}}} \right)\\{U_e} = 5.12 \times {10^{ - 15}}\;J\end{array}\)

Hence, the potential energy from electron volt to joule is \(5.12 \times {10^{ - 15}}\;J\).

Step 4: Calculation for the potential difference between the plates.

The potential difference between the two plates is found by solving equation below for \(\Delta V\):

\(\Delta V = \frac{{{U_e}}}{q}\)

Where \(q = 2e\) is the charge of the ion:

\(\Delta V = \frac{{{U_e}}}{{2e}}\)

Entering the values for \({U_e}\) and \(e\), we obtain:

\(\begin{array}{c}\Delta V = \frac{{5.12 \times {{10}^{ - 15}}\;J}}{{2\left( {1.60 \times {{10}^{ - 19}}C} \right)}}\\\Delta V = 1.60 \times {10^4}\;V\end{array}\)

Hence, the potential difference between the plates is \(1.60 \times {10^4}\;V\) .

Step 5: Calculation for the electric field strength.

The electric field strength between the two plates is found by solving equation below for \(E\):

\(E = \frac{{\Delta V}}{d}\)

Substitute numerical values:

\(\begin{array}{c}E = \frac{{1.60 \times {{10}^4}\;V}}{{0.0200\;m}}\\E = 8.00 \times {10^5}\;V/m\end{array}\)

Therefore, electric field value is obtained as \(8.00 \times {10^5}\;V/m\).

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College Physics (Urone)

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Short Answer

A doubly charged ion is accelerated to an energy of \(32.0{\rm{ }}keV\) by the electric field between two parallel conducting plates separated by \(2.00{\rm{ }}cm\). What is the electric field strength between the plates?

Step by Step Solution

Step 1: Definition of electric volt and electric potential energy.

Step 2: Given data for the calculation of electric field strength.

Step 3: Converting electric potential energy from electron volt to Joule.

Step 4: Calculation for the potential difference between the plates.

Step 5: Calculation for the electric field strength.

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College Physics (Urone)

Answers without the blur.

Short Answer

A doubly charged ion is accelerated to an energy of \(32.0{\rm{ }}keV\) by the electric field between two parallel conducting plates separated by \(2.00{\rm{ }}cm\). What is the electric field strength between the plates?

Step by Step Solution

Step 1: Definition of electric volt and electric potential energy.

Step 2: Given data for the calculation of electric field strength.

Step 3: Converting electric potential energy from electron volt to Joule.

Step 4: Calculation for the potential difference between the plates.

Step 5: Calculation for the electric field strength.

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