The figure shows electric field lines in whic | vedclass

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(C) The electric field lines are closer together on the left side and further apart on the right side,indicating that the electric field intensity $E$

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The dipole will experience a force towards left.

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(C) The electric field lines are closer together on the left side and further apart on the right side,indicating that the electric field intensity $E$ is stronger on the left and weaker on the right. The force on a charge $q$ is given by $F = qE$. The dipole consists of a negative charge $-q$ on the left and a positive charge $+q$ on the right. The force on the negative charge $-q$ is towards the left (opposite to the field direction) and has a magnitude $F_{-q} = qE_{left}$. The force on the positive charge $+q$ is towards the right (in the field direction) and has a magnitude $F_{+q} = qE_{right}$. Since the electric field is stronger on the left $(E_{left} > E_{right})$,the magnitude of the force on the negative charge is greater than the magnitude of the force on the positive charge $(F_{-q} > F_{+q})$. Therefore,the net force on the dipole is directed towards the left.

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

The figure shows electric field lines in which an electric dipole $p$ is placed as shown. Which of the following statements is correct?

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