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Answer: \(F = \tfrac{1}{2}QE\), because the energy stored increases by \(\tfrac{1}{2}QE\,\Delta x\) when the plates are pulled apart by \(\Delta x\); the ½ appears because each plate feels only the field of the other plate, which is E/2.
Derivation (energy method). Let the capacitor have plate area A, separation d and charge ±Q (isolated, so Q stays constant).
\[C = \frac{\varepsilon_0 A}{d},\quad U = \frac{Q^{2}}{2C} = \frac{Q^{2}d}{2\varepsilon_0 A}\]
Pull one plate away slowly by a small distance \(\Delta x\) against the attractive force F. The work done is \(F\Delta x\), and it equals the increase in stored energy:
\[F\,\Delta x = \frac{Q^{2}\,\Delta x}{2\varepsilon_0 A}\]
\[F = \frac{Q^{2}}{2\varepsilon_0 A} = \frac{1}{2}Q\left(\frac{Q}{\varepsilon_0 A}\right)\]
The field between the plates is \(E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}\). Hence
\[F = \frac{1}{2}QE.\]
Origin of the factor ½: the field E between the plates is produced by both plates together, each contributing \(\dfrac{\sigma}{2\varepsilon_0} = \dfrac{E}{2}\). A plate cannot exert a force on itself, so the force on one plate comes only from the field of the other plate, \(E/2\). So \(F = Q \times \dfrac{E}{2}\). Equivalently, the field is E just outside the plate and zero inside the conductor, and the charge on the surface feels the average value E/2.
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