Question

The potential on the surface of a sphere (radius R) is given by heta%29" id="TexFormula1" title="V = V_0 cos(2 \theta)" alt="V = V_0 cos(2 \theta)" align="absmiddle" class="latex-formula">. (Assume $V(r = \infty) = 0$, as usual. Also, assume there is no charge inside or outside, it's ALL on the surface!) i) Find the potential inside and outside this sphere. ii) Find the charge density $\sigma(\theta)$ on the sphere.

Answer 1

σ = V/r

Explanation:

The potential on the surface of the sphere is V = V₀cos(2θ) . The potential inside the sphere is the same at all points , which is equal to V as above .

The potential outside the  sphere will decrease with  distance from the center . by the relation V = Q/4πε₀r  ,  where Q is the charge on the surface and r is the distance from the center of the sphere .

Thus the potential decreases inversely proportional to the distance from center of the sphere .

( ii ) The charge density means charge per unit area  . Thus σ = Q/4πε₀r²

Thus σ = V/r