Question

A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potential is 150 V. What is the radius of the sphere

Answer 1

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, $V_s$ = 450 V

potential at radial distance, $V_r$ = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

$V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}$

$450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m$

Therefore, the radius of the sphere is 4.05 m