Question

(II) Determine the magnitude of the acceleration experienced by an electron in an electric field of\({\bf{756 N/C}}\).How does the direction of the acceleration depend on the direction of the field at that point?

Short answer

The magnitude of the acceleration of an electron is \(1.33 \times {10^{14}}{\rm{ m/}}{{\rm{s}}^2}\). The direction of acceleration is opposite to the direction of the electric field.

Step-by-step solution

Understanding the acceleration of an electron

When the electric force acts on an electron, the electron starts accelerating.The acceleration of an electron depends on the charge, electric field, and mass of an electron.The unit of acceleration of an electron is a meter per second square.

Identification of the given data

The given data is listed below:

• The magnitude of an electric field is\(E = 756{\rm{ N/C}}\).
• The charge on the electron is\(q = 1.6 \times {10^{ - 19}}{\rm{ C}}\).
• The mass of an electron is \(m = 9.1 \times {10^{ - 31}}{\rm{ kg}}\).

Determination of the magnitude and direction of acceleration

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

\(F = qE\) … (i)

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

\(F = ma\) … (ii)

Here, ais the acceleration of an electron.

Equate equations (i) and (ii). Then evaluate the expression of the acceleration.

\(\begin{aligned}{c}F = qE\\F = ma\\a = \frac{{qE}}{m}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}a = \frac{{1.6 \times {{10}^{ - 19}}{\rm{ C}} \times 756{\rm{ N/C}}\left( {\frac{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1{\rm{ N}}}}} \right)}}{{9.1 \times {{10}^{ - 31}}{\rm{ kg}}}}\\ = \frac{{1.21 \times {{10}^{ - 16}}{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{9.1 \times {{10}^{ - 31}}{\rm{ kg}}}}{\rm{ }}\\ = 1.33 \times {10^{14}}{\rm{ m/}}{{\rm{s}}^2}\end{aligned}\)

The charge of an electron is a negative sign. The acceleration is in the opposite direction of the field. The acceleration is dependent on the direction of the electric field.

Thus, the magnitude of the acceleration of an electron is \(1.33 \times {10^{14}}{\rm{ m/}}{{\rm{s}}^2}\). The direction of acceleration is opposite to the direction of the electric field.