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?

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 \(\Delta V = Ed\).

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:

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

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 \(5.12 \times {10^{ - 15}}\;J\).

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 \(1.60 \times {10^4}\;V\) .

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 \(8.00 \times {10^5}\;V/m\).