Question

If the radius of the equipotential surface of the point charge is 14.3 m at a potential of 2.20 kV, what will be the magnitude of the point charge that generates the potential?

Answer 1

The magnitude of the point charge is 3.496 x 10⁻⁶ C

Step-by-step explanation:

Given;

radius of the surface, r = 14.3 m

magnitude of the potential, V = 2.2 kV = 2,200 V

The magnitude of the point charge is calculated as follows;

`V = ((1)/(4\pi \epsilon _0) )((Q)/(r) )\\\\V = (KQ)/(r) \\\\Q = (Vr)/(K) \\\\Q = (2,200 * 14.3)/(9* 10^9) \\\\Q = 3.496 * 10^(-6) \ C\\\\Q = 3.496 \ \mu C`

Therefore, the magnitude of the point charge is 3.496 μC