q, 2q, 3q and 4q charges are placed at the four corners A, B, C and D of a square. The field at the centre P of the square has the direction along

Let each side of the square be a.
Then distance of the centre of the square from each vertices will be a2
From the above figure,
Electric field at the center due to charge q
${\mathrm{E}}_{\mathrm{A}}=\frac{\mathrm{kq}}{{\left(\frac{\mathrm{a}}{\overline{2}}\right)}^2}=\frac{2\mathrm{kq}}{{\mathrm{a}}^2}$
Electric field at the center due to charge 2q
${\mathrm{E}}_{\mathrm{B}}=\frac{2\mathrm{kq}}{{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}^2}=2\left(\frac{2\mathrm{kq}}{{\mathrm{a}}^2}\right)$
Electric field at the center due to charge 3q
${\mathrm{E}}_{\mathrm{C}}=3\left(\frac{2\mathrm{ka}}{{\mathrm{a}}^2}\right)$
Electric field at the center due to charge 4q
${\mathrm{E}}_{\mathrm{D}}=4\left(\frac{2\mathrm{kg}}{{\mathrm{a}}^2}\right)$
Now,
Resultant of Ec and EA will be in the direction of EC because its value is greater

The resultant of EB and ED is in the direction of ED because its value is greater

Now from the figure, we can say that the results of electric field in the direction of Ec and ED will be along CB So, the resultant is along CB.