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

Answer:

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

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;

$V = (\frac{1}{4\pi \epsilon _0} )(\frac{Q}{r} )\\\\V = \frac{KQ}{r} \\\\Q = \frac{Vr}{K} \\\\Q = \frac{2,200 \times 14.3}{9\times 10^9} \\\\Q = 3.496 \times 10^{-6} \ C\\\\Q = 3.496 \ \mu C$

Therefore, the magnitude of the point charge is 3.496 μC