Electric Field

The electric field due to each charge at a point ((0, y)) on the y-axis must be resolved into components. The (+q) charge at ((d,0)) produces a field of magnitude (E_1 = kq/(y^2+d^2)) directed from ((d,0)) towards ((0,y)). Its x-component is (-E_1 \sin\theta) (negative x) and y-component is (+E_1 \cos\theta). The (-q) charge at ((-d,0)) produces a field of magnitude (E_2 = kq/(y^2+d^2)) directed from ((0,y)) towards ((-d,0)). Its x-component is also (-E_2 \sin\theta) (negative x) and y-component is (-E_2 \cos\theta). Since (E_1 = E_2), the y-components cancel and the x-components add. Therefore, the net electric field at every point on the y-axis is directed purely in the negative x-direction. Option 'Only at the origin' is incorrect because the field at the origin is also along (-x) but by no means exclusively. Option 'Only at \infty' is incorrect because the field vanishes at \infty. Option 'At no point' is incorrect because the symmetry argument clearly shows x-direction field everywhere on the y-axis. This JEE Advanced style problem tests vector component analysis. Plausibility check: the dipole field at midpoint of the axis is anti-parallel to the dipole moment, consistent with the (-x) direction here.