A square of side d, made from a thin insulating plate, is uniformly charged and carries a total charge of Q. A point charge q is placed on the symmetrical normal axis of the square at a distance d/2 from the plate. How large is the force acting on the point charge ?

Explanation

According to Newton's third law, the insulating plate acts on the point charge with a force of the same magnitude (but opposite direction) as the point charge does on the plate. We calculate the magnitude of this latter force.
Divide the plate (notionally) inot small pieces, and denote the area of the `i^(th)` piece by `Delta A_(i)`. Because of the uniform charge distibution, the charge on this small piece is `Delta Q_(i) = (Q)/(d^(2)) Delta A_(i)`. and so the electric force acting on it is `Fi = Ei Qi`, where Ei is the magnitude of the electric field produced by the point charge q at the position of the small piece.
. The force acting on the insulating plate, as a whole, can be calculated as the vector sum of the forces acting on the individual pieces of the plate. Because of the axial symmetery, the net force is erpendicular to the plate, and so it is sufficient to sum the perpendicular components of the forces: :br" `F = underset(i)sum F_(i) cos theta_(i) = underset(i)sum E_(i) (q)/(d^(2)) DeltaA_(i) cos theta_(i) = (Q)/(d^(2)) underset(i) sum E_(i) Delta A_(i) cos theta_(i)`
Where `theta_(i)` is the angle between the normal to the plate and the line that connects the point charge to the its piece of it.
The Sum in the given expression is nothing other than the electric flux through the square sheet produced by the point charge `q : psi = (q)/(6_(epsi_(0)))`
Using this and our previous observations, we calculate the magnitude of the force acting on the point charge due to the presence of the charged insulating plates as `F = (Qq)/(6e_(0)d^(2))`