homework and exercises - Attractive force between capacitor plates

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2~ ?$$

Answer

The energy of the capacitor is $U= \frac{\epsilon_0}{2} S\,\mathrm d E^2$ where $S$ is the area of a plate. If we increase of $\Delta d$ the distance of, say, the right plate from the left one, keeping fixed the charge $Q$ on each plate, $E$ does not change and we find a variation of energy $$\Delta U = \frac{\epsilon_0}{2} S E^2 \Delta d = \frac{\epsilon_0}{2} SE \: E \Delta d= \frac{1}{2} Q \: E \Delta d\:.$$ This variation of energy, up to a sign, is due to the electric work $F \Delta d$, in turn, it is due to the force $F$ the left plate applies on the charges of the right plate. Therefore $$F= \frac{1}{2} QE \:.$$