Answer:
The radius of the sphere is 4.05 m
Explanation:
Given;
potential at surface, [tex]V_s[/tex] = 450 V
potential at radial distance, [tex]V_r[/tex] = 150
radial distance, l = 8.1 m
Apply Coulomb's law of electrostatic force;
[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]
[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]
Therefore, the radius of the sphere is 4.05 m
