91Ó°ÊÓ

The force acting on the electron due to electric field is,

\(F = qE\)

Here,\(q\)is the magnitude of the charge, and\(E\)is the magnetic field.

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

\(ma = qE\)

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

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

The acceleration of the electron can be calculated using equation (1.3).

Substitute\( - 1.6 \times {10^{ - 19}}{\rm{ C}}\)for\(q\),\(2.00 \times {10^5}{\rm{ N}}/{\rm{C}}\)for\(E\), and\(9.1 \times {10^{ - 31}}{\rm{ kg}}\)for\(m\)into equation (1.3).

\(\begin{array}{c}a = \frac{{\left( { - 1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right) \times \left( {2.00 \times {{10}^5}{\rm{ N}}/{\rm{C}}} \right)}}{{\left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)}}\\ = - 3.52 \times {10^{16}}{\rm{ m}}/{{\rm{s}}^2}\end{array}\)

The electric field accelerates the electron in the opposite direction to its initial velocity.

Hence, the electric field is in the direction of the initial velocity.

The third equation of the motion is given as,

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

Here,\(v\)is the final velocity of the electron (\(v = 0\)as the electron comes to rest),\(u\)is the initial velocity of the electron\(\left( {u = 5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)\),\(a\)is the acceleration of the electron in the electric field\(\left( {a = - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)\), and\(s\)is the distance traveled by the electron before coming to rest.

The expression for the distance traveled by the electron before coming to the rest is,

\(s = \frac{{{v^2} - {u^2}}}{{2a}}\)

Substituting all known values,

\(\begin{array}{c}s = \frac{{{{\left( 0 \right)}^2} - {{\left( {5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( { - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 3.55 \times {10^{ - 4}}{\rm{ m}}\end{array}\)

Hence, the electron will travel a distance of \(3.55 \times {10^{ - 4}}{\rm{ m}}\) before coming to rest.

The first equation of the motion is,

\(v = u + at\)

Here,\(v\)is the final velocity of the electron (\(v = 0\)as the electron comes to rest),\(u\)is the initial velocity of the electron\(\left( {u = 5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)\),\(a\)is the acceleration of the electron in the electric field\(\left( {a = - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)\), and\(t\)is the time taken by the electron to come to rest.

The expression for the time taken by the electron to come to rest is,

\(t = \frac{{v - u}}{a}\)

Substituting all known values,

\(\begin{array}{c}t = \frac{{\left( 0 \right) - \left( {5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)}}{{\left( { - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 1.43 \times {10^{ - 10}}{\rm{ s}}\end{array}\)

Hence, the time taken by the electron to come to rest is \(1.43 \times {10^{ - 10}}{\rm{ s}}\).

After stopping the direction of the motion of the electron reverses. Thus, the acceleration becomes negative of the initial acceleration.

The third equation of motion is,

\({v'^2} = {u'^2} - 2as\)

Here,\(v'\)is the final velocity of the electron or the velocity at the starting point,\(u'\)is the initial velocity of the election in the reverse motion (\(u' = 0\)as the electron starts from rest while staring the reverse motion),\(\left( {u = 5.00 \times {{10}^6}{\rm{ m}}/{\rm{s}}} \right)\),\(a\)is the acceleration of the electron in the electric field\(\left( {a = - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right)\), and\(s\)is the distance traveled by the electron before reaching the starting point\(\left( {s = 3.55 \times {{10}^{ - 4}}{\rm{ m}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{{v'}^2} = {\left( 0 \right)^2} - 2 \times \left( { - 3.52 \times {{10}^{16}}{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {3.55 \times {{10}^{ - 4}}{\rm{ m}}} \right)\\{{v'}^2} = 2.5 \times {10^{13}}{\rm{ }}{{\rm{m}}^2}/{{\rm{s}}^2}\\v' = 5.00 \times {10^6}{\rm{ m}}/{\rm{s}}\end{array}\)

Since, this velocity is opposite of the initial velocity.

Hence, the velocity of electron when it returns to its start point is\(5.00 \times {10^6}{\rm{ m}}/{\rm{s}}\).