Answer:

Step-by-step explanation:
give data:
inside diameter = 5.0 cm
charge q = 0.25 nC
Outside diameter = 15 cm
potential V at inside sphere is =

potential V at outside sphere is =

k is constant whose value is =

then potential difference between two point is
![\Delta V = kq \left [(1)/(R)-(1)/(r) \right ]](https://img.qammunity.org/2020/formulas/physics/college/843e9vgzm2cwuy4zi5ptdw4qj5s7ewljhe.png)
![\Delta V = 9*10^(9)*0.25*10^(-9) \left [(1)/(0.05)-(1)/(0.15) \right ]](https://img.qammunity.org/2020/formulas/physics/college/9u2brme6f0fgb276eez688mio7w2gl74o2.png)
