91Ó°ÊÓ

When any charged particle is placed in an electric field, the field causes an acceleration in the particle.

The electrostatic field acting on the particle in the field is,

\({F_e} = qE\)

Here,\(q\)is the charge on the proton and\(E\)is the electric field.

The force acting on proton causes the acceleration of the proton. 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 proton. Therefore,

\(ma = qE\)

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

\(a = \frac{{qE}}{m}\)

The final velocity of the proton is 3.00% of the speed of the light. Thus,

\(\begin{array}{c}v = \left( {3.00\% } \right)c\\ = \left( {3.00 \times \frac{1}{{100}}} \right)c\\ = 0.03c\end{array}\)

Here,\(c\)is the speed of light.

Third equation of the motion is,

\({v^2} = {u^2} + 2as\)

Here,\(v\)is the final velocity of the proton,\(u\)is the initial velocity of the proton (\(u = 0\)as the proton starts from the rest),\(a\)is the acceleration of the proton, and\(s\)is the distance traveled by the proton.

Using equation (1.1) and (1.2),

\({\left( {0.03c} \right)^2} = {\left( 0 \right)^2} + 2 \times \left( {\frac{{qE}}{m}} \right)s\)

Rearranging the above equation in order to get an expression for the distance traveled by the proton.

\(s = \frac{{{{\left( {0.03c} \right)}^2}m}}{{2qE}}\)

Substitute\(1.67 \times {10^{ - 27}}{\rm{ kg}}\)for\(m\),\(3.00 \times {10^8}{\rm{ m}}/{\rm{s}}\)for\(c\),\(1.6 \times {10^{ - 19}}{\rm{ C}}\)for\(q\), and\(3.00 \times {10^6}{\rm{ N}}/{\rm{C}}\)for\(E\).

\(\begin{array}{c}s = \frac{{{{\left[ {0.03 \times {{\left( {3.00 \times {{10}^8}{\rm{ m}}/{\rm{s}}} \right)}^2}} \right]}^2} \times \left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)}}{{2 \times \left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {3.00 \times {{10}^6}{\rm{ N}}/{\rm{C}}} \right)}}\\ = 0.141{\rm{ m}}\end{array}\)

Hence, the distance traveled by a free proton is \(0.141{\rm{ m}}\).

While traveling this distance, the proton will collide with many air particles.

Hence, this is not practical in air.