Question

(I) Determine the magnitude and direction of the electric force on an electron in a uniform electric field of strength $\mathbf{2460}\mathbf{N}\mathbf{/}\mathbf{C}$ that points due east.

Short answer

The magnitude of the force on the electron is $3.94\times {16}^{-16}\mathrm{N}$ due west.

Step-by-step solution

Understanding the electric field 

The electric field at any point in space is defined as the force per unit charge at that point.

The expression for electric field is given as:

$E=\frac{F}{q}$ … (i)

Here, F is the force and q is the charge.

Given data

The charge of an electron is,$q=1.6\times {16}^{-19}\mathrm{C}$.

The magnitude of the electric field is,$E=2460\mathrm{N}/\mathrm{C}$due east.

Determination of force on an electron in the electric field

From equation (i), the force on the electron is given as:

$F=qE$

Substitute the values in the above expression.

$\begin{array}{r}cF=\left(1.6\times {16}^{-19}\mathrm{C}\right)\times \left(2460\mathrm{N}/\mathrm{C}\right)\\ =3.94\times {16}^{-16}\mathrm{N}\end{array}$

As the charge of the electron is negative, then the force on the electron is due west.

Thus, themagnitude of the force on the electron is $3.94\times {16}^{-16}\mathrm{N}$ due west.