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;
[tex]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[/tex]
Therefore, the magnitude of the point charge is 3.496 μC
