In: Physics
If the radius of the equipotential surface of point charge is 15.2 m at a potential of 2.30 kV, what is the magnitude of the point charge creating the potential?
Note that
V = kq / r
Which implies that
q = V r / k
where
k = 8.99E+09 N m^2/C^2
V = the electric potential = 2300
N/C
r = the distance from the source charge
15.2 m
Thus,
q = 3.89E-06 C
[ANSWER]
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