Answer:
(a) V = 65.625 Volts
(b) V = 131.25 Volts
(c) V = 131.25 Volts
Step-by-step explanation:
Recall that:
1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:
![V=k(Q)/(R)](https://img.qammunity.org/2021/formulas/physics/college/xvwxz72jwsfhsk781zjzbzxku5mz8fdgez.png)
2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:
![V=k(Q)/(r)](https://img.qammunity.org/2021/formulas/physics/college/p2ji9ys1zhwvmwqfqfdmunqq9lb29gczbe.png)
where k is the Coulomb constant (
), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.
Then, at a distance of:
(a) 48 cm = 0.48 m, the electric potential is:
![V=k(Q)/(r)=9\,\,10^9 \,(3.5\,\,10^(-9))/(0.48) =65.625\,\,V](https://img.qammunity.org/2021/formulas/physics/college/2rjpvs51wb2gefwpnh5koh1f85j9c325zx.png)
(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:
![V=k(Q)/(r)=9\,\,10^9 \,(3.5\,\,10^(-9))/(0.24) =131.25\,\,V](https://img.qammunity.org/2021/formulas/physics/college/cz9p1e461i81sbh4nvxprqwvjimksdpa4t.png)
(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:
![V=k(Q)/(R)=9\,\,10^9 \,(3.5\,\,10^(-9))/(0.24) =131.25\,\,V](https://img.qammunity.org/2021/formulas/physics/college/9ii0bh321t2sp21v46sekv7srusq87mmvb.png)