What are the magnitude and direction of the electric field at a distance of 1.50 m from a 50.0-nC charge?

Question

Quizplus · Accepted Answer

The electric field (E) due to a point charge is given by E = k * |q| / r^2, where k is Coulomb's constant (8.99 x 10^9 N m²/C²), q is the charge (50.0 x 10^-9 C), and r is the distance (1.50 m). Calculating this gives approximately 200 N/C, and the field direction is away from the positive charge.

What Are the Magnitude and Direction of the Electric Field