Electric FieldsElectric Fields

The soluion is already clear enough.

Use the basic equation for electric field E due to a point charge of magnitude Q,
E = kQ/r^2
where k is a constant (= 9 x 10^9 F/m)
r is the distance from the point charge

Hence, electric field at P due to charge at X
Ex = kQ/a^2 in the direction XP
Electric field at P due to charge at Z
Ez = kQ/a^2 in the direction ZP
Resultant field (Exz) of Ex and Ez is
Exz = square-root[Ex^2 + Ez^2] = square-root[2(kQ/a^2)^2]
i.e. Exz = (kQ/a^2).square-root[2]
The field direction is along the diagonal YP point away from Y

Electric field at P due to charge at Y
Ey = k(-Q)/(YP)^2
Becaue (YP)^2 = (YX)^2 + (XP)^2 = a^2 + a^2 = 2a^2
hence, Ey = -kQ/(2a^2)
The direction of Ey is along the diagonal PY pointing towards Y

Therefore, resultant field of Ey and Exy
= -kQ/(2a^2) + (kQ/a^2).square-root[2]
= (kQ/a^2).[square-root(2) - (1/2)]
Since the resultant field is +ve, the same sign as Exz, the field direction is along the diagonal YP pointing away from Y.