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

Answer:

The sign of the charge is negative

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

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;

$V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\where;\\\\k \ is \ coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2\\\\q = \frac{-1.5 \times 2\times 10^{-3}}{9\times 10^9 } \\\\q = -3.33 \times 10^{-13} \ C\\\\Magnitude \ of \ the\ charge, q = 3.33 \times 10^{-13} \ C$