Answer to Question #138544 in Electric Circuits for vuty

Question

Three charged particles are arranged on corners of a square. (i) What is the direction of the electric field at the upper right corner?

Answer 1

Solution

For 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

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