# Couple acting on quadrupole due to a point charge

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

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