Question

Find the magnitude and direction of an electric field that exerts a 1.38 * 10^-18N northward force on a proton.

Answer 1

Given:

Force = 1.38 x 10⁻¹⁸ N northward.

Let's find the magnitude and direction of the electric field

To find the magnitude, apply the formula:

$E=\frac{F}{q}$

Where:

E is the electric field

F is the force = 1.38 x 10⁻¹⁸ N

q is the charge of proton = -1.60 x 10⁻¹⁹ C

Input the values in the equation and solve for E:

$\begin{gathered} E=\frac{1.38\times10^{-18}}{-1.60\times10^{-19}} \\ \\ E=-8.625\text{ N/C} \end{gathered}$

The magnitude of the electric field is -8.625 N/C.

Since the direction of the force is Northward and the polarity of the charge is negative, the direction of the electric field will be opposite the direction of the force.

Therefore, the direction of the electric field is downward.

ANSWER:

8.625 N/C Southward