Three charges each of +4\, are placed at the corners A, B, C of a...

Let the center of the square be O. The distance from each corner to the center is $r = \dfrac{a}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}$ m.

The electric field at O due to the charge at A is equal and opposite to the electric field at O due to the charge at C. Hence, they cancel each other out.

The net electric field at O is solely due to the charge at B. Since the charge at B is positive, the electric field points away from B, i.e., towards D.

The magnitude of the electric field is given by $E = \dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r^2}$.

Substituting the values, we get $E = \dfrac{9 \times 10^9 \times 4 \times 10^{-6}}{(1/\sqrt{2})^2}$.

$E = \dfrac{36 \times 10^3}{1/2} = 72 \times 10^3 = 7.2 \times 10^4$ N/C.

The direction is towards point D.

Answer: $7.2 \times 10^4$ N/C towards point D