Three charges each equal to q are placed at the three corners of a square of side a. Find the electric field at the fourth corner.
Electric field at D due to the charge at A is, EA = kq/a2 along AD
Electric field at D due to the charge at C is, EC = kq/a2 along CE
Electric field at D due to the charge at B is, EB = kq/(2a2) along BE
Since, EA and EC are equal in magnitude and are perpendicular to each other so their resultant will be along the diagonal BD.
Resultant of EA and EC is, EAC = [(EA)2 + (EC)2]1/2 = √2 kq/a2 along BD.
Now, EAC and EB are in the same direction so they can be directly added. Thus, the resultant electric field at D is,
E = EAC + EB = √2 kq/a2 + kq/(2a2) = (2√2 + 1)kq/(2a2) along BD.