The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner D of the square, is :

Solution

Physics sir · 3 steps

Combine the two q/2 fields along diagonal BD; the field of q from B points the same way, so add them.

1. Side charges. A and C are each a distance a from D, so each gives E₁ = 1/4πε₀(q/2)/a². The field from A points along AD and the field from C points along CD, at 90° to each other. Their resultant is √2 E₁ = 1/4πε₀q/(√2 a²), pointing along the diagonal from B towards D.
2. Field from B. B is diagonally opposite D, a distance a√2 away, so r² = 2a². Its field is E_B = 1/4πε₀q/2a², pointing away from B along BD. That is the same direction as the resultant from A and C.

   E = 1/4πε₀q/(√2 a²) + 1/4πε₀q/2a² = q/(4πε₀ a²)(1/√2 + 1/2)

3. So. All the contributions point along BD, so the magnitudes add directly. This gives option (a).

The idea to keep. At a square's corner, first combine the two equal side-charge fields into one diagonal field. The diagonal charge then adds or subtracts along that same line, with r² = 2a².

Answer from the NTA final answer key. Solution by PrepWiser.