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

450 × 10 ⁻⁹ N

Step-by-step explanation:

From the given:

The direction of force is;

`tan \ \theta = (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 = (kq_1q_2)/(r^2)`

`F = ((9*10^9 \ Nm^2/C^2 )*( 40 * 10^(-9) \ C ) * (10*10^(-9) \ C))/((√(8))^2)`

`F = 450 * 10^(-9) \ N`