quadrupole and q'

I wish to find the couple acting on the quadrupole due to q', assuming r>>a. Here is my working:

Force acting on -2q:

$E_{at \ -2q} = \frac{q'}{4\pi\epsilon r^2} \\ F_1 = (-2q) E = \frac{-2qq'}{4\pi\epsilon r^2}$

i.e. $F_1 = \frac{2qq'}{4\pi\epsilon r^2}$ towards q'

Force acting on leftmost q:

distance q to q' $\approx r + a\cos\theta$

$E_{at \ q} = \frac{q'}{4\pi\epsilon(r+a\cos\theta)^2} \approx \frac{q'}{4\pi\epsilon r^2} \left(1-\frac{2a\cos\theta}{r}\right)$

$F_2 = qE = \frac{qq'}{4\pi\epsilon r^2}\left(1 - \frac{2a\cos\theta}{r}\right)$ away from q'

Taking moments about the rightmost q, the torque will act anticlockwise:

$\tau = F_1 a \sin\theta - F_2 2a \sin\theta = \frac{2qq'a^2 \sin{2\theta}}{4\pi\epsilon r^3}$ anticlockwise

but this is 2/3 of the answer I get using an energy method (which is the answer stated in the textbook). What am I misunderstanding about the problem when using this force method?

The correct answer should be: $\tau = \frac{3qq'a^2 \sin{2\theta}}{4\pi\epsilon r^3}$


2 Answers 2


You want the torque due to the difference in the forces on the two +q charges. The force on the -2q charge does not contribute to the torque.


In case anyone is interested here is the fix:

The angles of the two forces are slightly different, so the expression for the torque is actually:

$\tau = F_1 a\sin\theta - F_2 2a \sin(\theta + \delta\theta) \\ \sin(\theta + \delta\theta) = \sin\theta + \delta\theta\cos\theta \\ \delta\theta = \frac{a\sin\theta}{r} $

carrying on the calculation with the above expressions gets you the correct answer.


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