Question

The electric force is exerted by a 40-nC charged particle located at the origin of a Cartesian coordinate system on a 10-nC charged particle located at (2.0 m, 2.0 m ).
Required:
a. Determine the direction of the force.
b. Determine the magnitude of the force.

Answer 1

Answer:

450 × 10 ⁻⁹ N

Explanation:

From the given:

The direction of force is;

$tan \ \theta = \dfrac{2}{2}$

$\theta =tan ^{-1} (1)$

$\theta = 45^0$

Hence, the angle is 45° counterclockwise from + x-axis

However;

the magnitude of the force is:

$F = \dfrac{kq_1q_2}{r^2}$

$F = \dfrac{(9*10^9 \ Nm^2/C^2 )*( 40 * 10^{-9} \ C ) * (10*10^{-9} \ C)}{(\sqrt{8})^2}$

$F = 450 \times 10^{-9} \ N$