Question

(II) An electron is released from rest in a uniform electric field and accelerates to the north at a rate of\({\bf{105 m/}}{{\bf{s}}^{\bf{2}}}\).Find the magnitude and direction of the electric field.

Short answer

The magnitude of the electric field is \(5.97 \times {10^{ - 10}}{\rm{ N/C}}\). The direction of the electric field is in the south.

Step-by-step solution

Understanding the electric field strength

The electric field strength is the ratio of the force acting on the electron to the electron's charge. Due to this force, the electron starts accelerating.It is a type of vector quantity. The direction of the force acting on a charged particle in an electric field depends on the type of charge.

Identification of given data

The given data can be listed below as:

• The acceleration of the electron is\(a = 105{\rm{ m/}}{{\rm{s}}^2}\).
• The charge on the electron is\(q = 1.6 \times {10^{ - 19}}{\rm{ C}}\).
• The mass of the electron is\(m = 9.1 \times {10^{ - 31}}{\rm{ kg}}\).

Determination of the magnitude and the direction of the electric field

The magnitude of the force on an electron can be expressed as:

\(F = qE\) … (i)

Here, E is the electric field strength.

From Newton’s second law, the magnitude of the force can be expressed as:

\(F = ma\) … (ii)

Here, ais the acceleration of the electron.

Equate equations (i) and (ii), and evaluate the expression of the electric field.

\(\begin{aligned}{c}qE = ma\\E = \frac{{ma}}{q}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}E = \frac{{9.1 \times {{10}^{ - 31}}{\rm{ kg}} \times 105{\rm{ m/}}{{\rm{s}}^2}}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\ = \frac{{{\rm{9}}{\rm{.56}} \times {\rm{1}}{{\rm{0}}^{ - 29}}{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ = 5.97 \times {10^{ - 10}}{\rm{ N/C}}\end{aligned}\)

The electron has a negative charge, and it is accelerating in the north direction. Due to its negative charge, the electric field strength is directed opposite to the electron’s motion. The direction opposite to the north is south.

Thus, the magnitude of the electric field is \(5.97 \times {10^{ - 10}}{\rm{ N/C}}\), and it is directed toward south.