Question

Four charged particles are at the corners of a square of side a. Determine (a) the electric field at the location of charge q and (b) the total electric force exerted on q.

Answer 1

Solutiona) this question consider side of square is "a" Charges are "q" Figure can be shown as

Electric field on fourth corner is given as

$E_1=\frac{kq}{a^2}$

$E_2=\frac{kq}{a^2}$

Resultant of both above field is

$E'=\frac{\sqrt{2}kq}{a^2}$

$E_3=\frac{kq}{2a^2}$

And finally the resultant electric field will be in direction of E₃ on fourth corner is given by

$E=\frac{\sqrt{2}kq}{a^2}+\frac{kq}{2a^2}$

$E=\frac{kq}{a^2}(\sqrt{2}+\frac{1}{2})$

Direction is resultant is 45° from x and y so vector is direction of resultant electric is given by

$E=\frac{kq}{a^2}(\sqrt{2}+\frac{1}{2}) \frac{1}{\sqrt2}(i+j)$

b) force on this charge is given by

$F=qE$

$F=\frac{kq^2}{a^2}(\sqrt{2}+\frac{1}{2})$