The individual charges onYou would expect that the two plates areforce would not all separated bydepend on whether or not the capacitor was connected to a distanceconstant voltage source $d$ which$V$.
If the capacitor $C = \dfrac {\epsilon_o A}{d}$ is connected to a constant voltage source the the energy stored in the capacitor is $U = \frac 1 2 CV^2 = \dfrac 1 2 \dfrac{\epsilon_o A}{d}V^2$.
The energy stored in the capacitor changes by $\Delta U$ if the separation of the plates $d$ increases by an amount $\Delta d$.
$\Delta U = - \dfrac 1 2 \left( \dfrac {\epsilon_o A}{d}\right)V \dfrac V d \Delta d = - \dfrac 1 2 Q E \Delta d $
A decrease in stored energy is because the capacitance has decreased at constant voltage.
$Q=CV= \dfrac {\epsilon_o A}{d} V\Rightarrow \Delta Q = - \dfrac {\epsilon A}{d^2} V \Delta d$
Thus the charge on the capacitor decreases and alsofor the forces between individual charges are not all perpendicularchange $\Delta Q$ to flow through the plates.voltage source the amount of work which must be done is $\Delta Q V $
$\Delta Q V = \left( \dfrac {\epsilon_o A}{d}\right)V \dfrac V d \Delta d = Q E \Delta d $
The electric fieldnet amount of work done by the external force $E$$F$ is $F \Delta d$ and this is the magnitude of the attractive force between the plates.
$F\Delta x = Q E \Delta d - \dfrac 1 2 Q E \Delta d \Rightarrow F = \frac 1 2 QE$
Note that in the constant voltage case there is contributed equally bya decrease in the twoenergy stored in the capacitor but at the same time work must be done in moving charge through the constant voltage source.
In the constant charge case increasing the separation of the plates increases the potential difference across the plates so each plate contributes an electric field $\frac E2$ which produces a force onmore energy is stored in the chargecapacitor and work must be done to do this.
For the constant change case $Q$ on$Q=CV = \dfrac {\epsilon_o A}{d}V$ so in the other plate equal toderivation the electric field $ Q \frac E2$$E$ is constant.