A positively charge sphere of radius `r_0` carries a volume charge density `rho`. A spherical cavity of radius `r_0//2` is then scooped out and left empty. `C_1` is the center of the sphere and `C_2` that of the cavity. What is the direction and magnitude of the electric field at point B?

Options

A. `(17rhor_0)/(54epsilon_0)`left
B. `(rhor_0)/(6epsilon_0)` left
C. `(17rhor_0)/(54epsilon_0)`right
D. `(rhor_0)/(6epsilon_0)`right

Answer

A

Explanation

a. Electric field on surface of a uniformly charged sphere is given
by
`Q/(4piepsilon_0R^2) = (rhoR)/(3epsilon_0)`
Electric field at outside point is given by
`E = Q/(4piepsilon_0r^2) = (rhoR^3)/(3epsilon_0r^2)`
`E_B = E_(whol e sphere) - E_(cavity)`
`= (rhor_0)/(3epsilon_0) - (rho(r_0/2)^3)/(3epsilon_0(3r_0/2)^2) = (17rhor_0)/(54epsilon_0)`.