A small electric dipole of dipole moment  $\overrightarrow{P}$ is placed on the x-axis at the point $\left(1,0\right).$ The dipole moment vector forms an angle of ${30}^{\circ }$ with the x-axis. A non uniform electric field has been applied in the region given by the vector  $\overrightarrow{E}=x^2 \widehat{i}+y^2 \widehat{j}.$ The electric force acting on the dipole is most appropriately given by 

We will first calculate the $x$ component of force on the dipole. $x$ component of field at a point having $x$ co-ordinate $x$ is

$E_x=x^2$

Field at  $\left(x+\Delta x\right)$ is $\left(E_x+\Delta E_x\right)$ where

$\Delta E_x=\left(\frac{d{E}_x}{dx}\right)\Delta x=2x\left(d\cos 30°\right)=2xd\cos 30°$

$\therefore F_x=q\left(E_x+\Delta E_x\right)-q{E}_x=q\Delta E$$=2qxd\cos 30°=\left[2P\cos 30°\right]x$

At $x=1$  $F_x=2P\cos 30°$

Similarly, $F_y=\left[2P\sin 30°\right]y$ due to $E_y$

At $y=0,F_y=0.$ (According to the given function, $E_x$ does not give $F_y$ and $E_y$ does not give $F_x$ )