A semi-infinite insulating rod has linear charge density `lambda`. The electric field atthe pointPshown in figure-1.418 is:

Options

A. `(2 lambda^(2))/((4 pi epsi_(0)r)^(2))` at `45^(@)` with AB
B. `(sqrt(2 lambda)^(2))/(4 pi epsi_(0)r^(2)` at `45^(@)` with AB
C. `(sqrt2 lambda)/(4 pi epsi_(0)r)` at `45^(@)` with AB
D. `(sqrt2 lambda)/(4 pi epsi_(0)r)` at `135^(@)` with AB