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.

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field experience by three charges are, a(kq/a(square) and kq/2a(square)

but kq/usare),kq/usare) are added as vectors by 90 degree angle so answer is 5kq/a(square)

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