Question

What are the sign and magnitude in coulomb's of a point charge that produces a potential of -1.50 V at a distance of 2.00 mm

Answer 1

The sign of the charge is negative

The magnitude of the charge is 3.33 x 10⁻¹³ C

Step-by-step explanation:

Given;

potential difference, V = -1.5 V

distance of the point charge, r = 2 mm = 2 x 10⁻³ m

The magnitude of the charge is calculated as follows;

`V = (kq)/(r) \\\\q = (Vr)/(k) \\\\where;\\\\k \ is \ coulomb's \ constant = 9* 10^9 \ Nm^2/C^2\\\\q = (-1.5 * 2* 10^(-3))/(9* 10^9 ) \\\\q = -3.33 * 10^(-13) \ C\\\\Magnitude \ of \ the\ charge, q = 3.33 * 10^(-13) \ C`