Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
[tex]E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}[/tex]
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
[tex]E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}[/tex]
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
[tex]V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r<R)=- \int\limits^r_\infty{E.dl\\\\= -\int\limits^R_\infty{\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } -\int\limits^r_R{\frac{1}{4\pi(e_{0)} }\frac{Q}{R^2}\frac{r}{R} dr\\[/tex]
=[tex]\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ][/tex]
∴NOTE: Graph is attached
