Answer to Question #91603 in Electricity and Magnetism for Priyanchal Kumar

Question

4 point charge of +Q each are located at corners of square, what point charge to be kept at center of the square so that the resulted force acting on any charge which are located at the corners of square is 0?

Answer 1

Consider 5 charges: Q in the points (l,0), (0,l), (-l, 0), (0, -l) and q in (0,0) The force between two charges q1 and q2 is given by

$$\vec F = \frac{kq_1 q_2}{r^3} \vec{r}$$

where r is the distance between them. Due to the symmetry it is sufficient to consider only one charge, for example the one in the point (l,0). We have 4 forces acting on it:

$$F_1 = \frac{kQ^2}{(2l)^3} (2l, 0)$$ $$F_2 = \frac{kQ^2}{(2l)^3} (0, 2l)$$ $$F_3 = \frac{kQ^2}{(2\sqrt{2}l)^3} (2l, 2l)$$ $$F_4 = \frac{kQq}{(\sqrt{2}l)^3} (l, l)$$

where we taking into account that the diagonal of the square is

$$d = \sqrt{2}a$$

The total force is

$$F = \frac{kQ}{l^2} (\frac{Q}{4} +\frac{Q}{8\sqrt{2}} + \frac{q}{2\sqrt{2}} , \frac{Q}{4} +\frac{Q}{8\sqrt{2}} + \frac{q}{2\sqrt{2}})$$

Hence F = 0 if

$$\frac{Q}{4} +\frac{Q}{8\sqrt{2}} + \frac{q}{2\sqrt{2}} = 0$$ $$q = - Q \frac{2\sqrt{2}+1}{4}$$