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(D) The electric field $E$ increases along the positive $x-$ axis. Let $E_1$ be the field at the position of the negative charge $(-q)$ and $E_2$ be t

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Dipole moves along negative $x-$ axis and undergoes an anticlockwise rotation

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(D) The electric field $E$ increases along the positive $x-$ axis. Let $E_1$ be the field at the position of the negative charge $(-q)$ and $E_2$ be the field at the position of the positive charge $(+q)$. Since the field increases along the positive $x-$ axis,$E_2 > E_1$. The force on the positive charge is $F_2 = qE_2$ (directed along the positive $x-$ axis). The force on the negative charge is $F_1 = qE_1$ (directed along the negative $x-$ axis). Since $E_2 > E_1$,the net force $F_{net} = F_2 - F_1 = q(E_2 - E_1)$ is directed along the positive $x-$ axis. Wait,looking at the diagram,the negative charge is at a larger $x$ coordinate than the positive charge. Therefore,the field at the negative charge is stronger. Let $x_{-q} > x_{+q}$,so $E(-q) > E(+q)$. Force on $-q$ is $F_1 = qE(-q)$ (towards negative $x-$ axis). Force on $+q$ is $F_2 = qE(+q)$ (towards positive $x-$ axis). Since $E(-q) > E(+q)$,the net force is towards the negative $x-$ axis. The torque $	au = p 	imes E$ tends to rotate the dipole to align with the field. In this configuration,the torque causes an anticlockwise rotation.

There exists a non-uniform electric field along the $x-$ axis as shown in the figure below. The field increases at a uniform rate along the $+ve$ $x-$ axis. $A$ dipole is placed inside the field as shown. Which one of the following is correct for the dipole?

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