Answer :
Answer:
The magnitude of charge is [tex]8.809\times 10^{-7}\ C[/tex] with a positive charge.
Explanation:
Given that,
The potential due to a point charge, [tex]V=5.05\times 10^2\ V[/tex]
Distance, d = 15.7 m
The potential due to a point charge is given by :
[tex]V=\dfrac{kq}{d}[/tex]
q is charge
k is electrostatic constant
[tex]q=\dfrac{Vd}{k}\\\\q=\dfrac{5.05\times 10^2\times 15.7}{9\times 10^9}\\\\q=8.809\times 10^{-7}\ C[/tex]
Here, the potential is positive, then the magnitude of charge is [tex]8.809\times 10^{-7}\ C[/tex] with a positive charge.