Answer :
Explanation:
The given data is as follows.
Potential (V) = [tex]5.75 \times 10^{2} V[/tex]
Distance (r) = 15.6 m
Let us assume that q is the magnitude of charge. Formula to calculate charge will be as follows.
V = [tex]\frac{1}{4 \pi \epsilon_{o}} \times \frac{q}{r}[/tex]
[tex]5.75 \times 10^{2} V = 9 \times 10^{9} \times \frac{q}{15.6 m}[/tex]
q = [tex]\frac{89.7 \times 10^{2}}{9 \times 10^{9}}[/tex]
= [tex]9.96 \times 10^{-7}[/tex] C
As the potential due to charge is positive. Therefore, q is positive.
Thus, we can conclude that the sign of charge is positive and magnitude of the charge is [tex]9.96 \times 10^{-7}[/tex] C.