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(B) The electric field lines are closer together near the $+q$ charge,which means the magnitude of the electric field is stronger at the position of t

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towards the right as its potential energy will decrease.

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(B) The electric field lines are closer together near the $+q$ charge,which means the magnitude of the electric field is stronger at the position of the $+q$ charge than at the position of the $-q$ charge. Let $E_1$ be the field at $+q$ and $E_2$ be the field at $-q$. Thus,$|E_1| > |E_2|$.The force on the positive charge is $F_+ = qE_1$ (towards the right) and the force on the negative charge is $F_- = qE_2$ (towards the left).Since $|E_1| > |E_2|$,the net force $F_{net} = q(E_1 - E_2)$ acts towards the right.$A$ physical system naturally moves in a direction that decreases its potential energy. Therefore,the dipole will move towards the right as its potential energy decreases.

$A$ dipole is placed in an electric field as shown. In which direction will it move?

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