Explanation:
It is known that inside a sphere with uniform volume charge density the field will be radial and has a magnitude E that can be expressed as follows.
E = [tex]\frac{q}{4 \p \epsilon_{o}R^{3}}r[/tex]
V = [tex]-\int_{0}^{r}E. dr[/tex]
= [tex]-\int_{0}^{r}\frac{q}{4 \p \epsilon_{o}R^{3}}r dr[/tex]
= [tex](\frac{q}{4 \p \epsilon_{o}R^{3}})(\frac{r^{2}}{2})[/tex]
= [tex]\frac{-qr^{2}}{8 \pi \epsilon_{o}R^{3}}[/tex]
At r = 1.45 cm = [tex]1.45 \times 10^{-2}[/tex] (as 1 m = 100 cm)
V = [tex]\frac{2.50 \times 10^{-15} \times 1.45 \times 10^{-2}}{8 \times 3.1416 \times 8.85 \times 10^{-12} \times 2.36 \times 10^{-2}}[/tex]
= [tex]6.905 \times 10^{-6}[/tex] mV
Thus, we can conclude that value of V at radial distance r = 1.45 cm is [tex]6.905 \times 10^{-6}[/tex] mV.
