Answer:
[tex]\Delta V = 30V[/tex]
Explanation:
give data:
inside diameter = 5.0 cm
charge q = 0.25 nC
Outside diameter = 15 cm
potential V at inside sphere is = [tex]V=\frac{kq}{R}[/tex]
potential V at outside sphere is = [tex]V=\frac{kq}{r}[/tex]
k is constant whose value is = [tex]8.99 *10^{9}Nm^{2}/c^2[/tex]
then potential difference between two point is
[tex]\Delta V = kq \left [\frac{1}{R}-\frac{1}{r} \right ][/tex]
[tex]\Delta V = 9*10^{9}*0.25*10^{-9} \left [\frac{1}{0.05}-\frac{1}{0.15} \right ][/tex]
[tex]\Delta V = 30V[/tex]