Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is a = $${{\sqrt 3 } \over 2}L$$.

A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field $$\overrightarrow E $$ through the shaded area is :

Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency : (m = mass of charged particle)

A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is $$-$$Q. The electric field at the centre of cube is :