A positively charged sphere of radius `r_(0)` carries a volume charge density `rho_(E)` (Figure). A spherical cavity of radius `r_(0)//2` is then scooped out and left empty, as shown. What is the direction and magnitude of the electric field at point B?

Options

A. `(17rhor_(0))/(54in_(0)) l eft`
B. `(rho_(0))/(6in_(0))l eft`
C. `(17rhor_(0))/(54in_(0))right`
D. `(rhor_(0))/(6in_(0)) right`

Answer

A

Explanation

Electric field on surface of a uniformly charged square is given by `Q/(4pivarepsilon_(0)R^(3)) = (rhoR)/(3varepsilon)_(0)`
Electric field at outside point is given by `E=Q/(4pivarepsilon_(0)r^(2))=(rhoR^(3))/(3varepsilon_(0)r^(2))`
`|vec(E)_(r)| = (rhor_(0))/(3varepsilon_(0)) - (rho(r_(0)/2)^(3))/(3varepsilon_(0)((3r_(0))/2)^(2))=(17rhor_(0))/(54varepsilon_(0))`