Answer:
V = 346.15 Volts
Explanation:
Given that,
Diameter of the sphere, d = 0.390 cm
Radius, r = 0.195 cm
Charge, [tex]q=75\ pC =75\times 10^{12}\ C[/tex]
The electric potential near its surface is given by :
[tex]V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 75\times 10^{-12}}{0.195\times 10^{-2}}\\\\V=346.15\ V[/tex]
So, the potential near its surface is equal to 346.15 V.
