The distance between charges +q and -q is 2l | vedclass

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(C) The system consists of two dipoles. The first dipole has charges $+q$ and $-q$ separated by $2l$,so its dipole moment is $p_1 = q(2l) = 2ql$ direc

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) The system consists of two dipoles. The first dipole has charges $+q$ and $-q$ separated by $2l$,so its dipole moment is $p_1 = q(2l) = 2ql$ directed from $-q$ to $+q$ (towards the right).The second dipole has charges $+2q$ and $-2q$ separated by $4l$,so its dipole moment is $p_2 = 2q(4l) = 8ql$ directed from $-2q$ to $+2q$ (towards the left).The net dipole moment is $p_{	ext{net}} = p_2 - p_1 = 8ql - 2ql = 6ql$ directed towards the left.The angle between the position vector $\vec{r}$ and the net dipole moment $\vec{p}_{	ext{net}}$ is $	heta = 180^{\circ} - 60^{\circ} = 120^{\circ}$.The electrostatic potential due to a dipole is given by $V = \frac{kp_{	ext{net}} \cos 	heta}{r^2}$.Substituting the values: $V = \frac{(9 	imes 10^9)(6ql) \cos(120^{\circ})}{r^2}$.Since $\cos(120^{\circ}) = -0.5$,we get $V = \frac{(9 	imes 10^9)(6ql)(-0.5)}{r^2} = -27 \left[ \frac{ql}{r^2} ight] 	imes 10^9 	ext{ V}$.Comparing this with the given expression $-\alpha \left[ \frac{ql}{r^2} ight] 	imes 10^9 	ext{ V}$,we find $\alpha = 27$.

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