Question

Determine the magnitude of the acceleration experienced by an electron in an electric field of 825N/C .

Express your answer to three significant figures and include the appropriate units.

How does the direction of the acceleration depend on the direction of the field at that point?

along to the field
	perpendicular to the field
	opposite to the field

Answer 1

Concepts and reason

The concept required to solve the given problem is electrical force on a charge which is placed in an electric field.

Firstly, write the expression of the electric force on a charge which is placed in an electric field and determine its value. Then, write force equation in terms of mass and acceleration and solve acceleration.

Finally, find the direction of force acting on the electron.

Fundamentals

The electrical force acting on a charged particle which is placed in an electric field can be calculated using the following formula:

$F = qE$

Here, q is the charge of the particle and E is the electric field experienced by the charge.

From the Newton’s second law of motion, the net force acting on a particle to accelerate it in a region of space is as follows:

$F = ma$

Here, m is the mass of the particle and a is the acceleration of the particle.

(a)

The electrical force acting on a particle is given as follows:

$F = qE$

Substitute -$1.6 \times {10^{ - 19}}{\rm{ C}}$ for q and 825 N/C for E in the above equation.

$\begin{array}{c}\\F = \left( { - 1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {825{\rm{ N/C}}} \right)\\\\ = - 1.32 \times {10^{ - 16}}{\rm{ N}}\\\end{array}$

Here, the negative sign indicates that the direction of force on the electron is opposite to the direction of field.

The net force acting on electron to accelerate it in a region of electric field is as follows:

$F = ma$

Rearrange the above expression for a.

$a = \frac{F}{m}$

Substitute $- 1.32 \times {10^{ - 16}}{\rm{ N}}$for F and $9.1 \times {10^{ - 31}}{\rm{ kg}}$ for m in the above expression.

$\begin{array}{c}\\a = \frac{{ - 1.32 \times {{10}^{ - 16}}{\rm{ N}}}}{{9.1 \times {{10}^{ - 31}}{\rm{ kg}}}}\\\\ = - 1.45 \times {10^{14}}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Here, the negative sign indicates that the direction of the acceleration of the electron is opposite to the direction of field.

Thus, the magnitude of the acceleration of the electron is$1.45 \times {10^{14}}{\rm{ m/}}{{\rm{s}}^2}$.

(b)

The electric force acting on the electron in the electric field region is as follows:

$F = - 1.32 \times {10^{ - 16}}{\rm{ N}}$

It has a negative sign which means that its direction is opposite to that of the direction of the electric field.

Due to this, the acceleration also has the negative direction. Thus, the electron is accelerating in a direction opposite to the direction of the electric field.

Ans: Part a

The magnitude of the acceleration of the electron is$1.45 \times {10^{14}}{\rm{ m/}}{{\rm{s}}^2}$.