JEE Main 201815 Apr, Evening shiftPhysicsMediumElectrostatics167× since 2002A solid ball of radius R has a charge density ρ\rhoρ given by ρ\rhoρ = ρ\rhoρ_o (1 −-− \raise0.5exr/\lower0.25exR{\raise0.5ex\hbox{$\scriptstyle r$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle R$}}\raise0.5exr/\lower0.25exR) for 0 ≤\le≤ r ≤\le≤ R. The electric field outside the ball is :AOption A. ρoR3∈or2{{{\rho _o}{R^3}} \over {{ \in _o}{r^2}}}∈or2ρoR3BOption B. ρoR312∈or2{{{\rho _o}{R^3}} \over {12{ \in _o}{r^2}}}12∈or2ρoR3COption C. 4ρoR33∈or2{{4{\rho _o}{R^3}} \over {3{ \in _o}{r^2}}}3∈or24ρoR3DOption D. 3ρoR34∈or2{{3{\rho _o}{R^3}} \over {4{ \in _o}{r^2}}}4∈or23ρoR3CheckSkip · show solution
