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. `(17 rhor_(0))/(54in_(0))` left
B. `(rhor_(0))/(6in_(0))` left
C. `(17 rhor_(0))/(54in_(0))` right
D. `(rho r_(0))/(6in_(0))` right

Answer

A

Explanation

Electric field on surface of a uniformly charged sphere is given by `(Q)/(4piepsilon_(0)R^(2))=(rhoR)/(3epsilon_(0))`
Electric field at outside point is given by
`E=(Q)/(4piepsilonr^(2))=(rhoR^(3))/(3 epsilon_(0)r^(2))`
`E_(B)= E_("whole sphere")-E_("cavity")`
`E_(B)= (rhor_(0))/(3epsilon_(0))-(rho(r_(0)/(2))^(3))/(3epsilon_(0)((3r_(0))/(2))^(2))=(16rhor_(0))/(54epsilon_(0))`