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**Hint:**The question can be answered with the help of the work done concepts and potential energy. Manipulating the capacitance formula and using it is also a key step in this solution. Capacitor has no electric effect inside it.

**Complete step by step answer:**

If we assume \[F\] to be the force applied on a parallel plate capacitor to separate the plates by a distance of $\Delta x$ then the work done that will be required to make that happen will be $F\Delta x$. Now using the conservation of energy theorem, the potential energy of the plates increases by $uA\Delta x$. Here $u$ is the energy density, $A$ is the area of each plate, $d$ is the distance between the plates and $V$ is the potential difference across the plates.

Therefore, according to the law of conservation of energy, the work done will be equal to the increase in the potential energy.

$F\Delta x=uA\Delta x$

Solving this we get

$F=uA=\left( \dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}} \right)A$,

where $E$ is the energy intensity

And the formula for energy intensity is

$E=\dfrac{V}{d}$

Therefore, the force equation becomes

$F=\dfrac{1}{2}{{\varepsilon }_{0}}\left( \dfrac{V}{d} \right)EA \\

\Rightarrow F=\dfrac{1}{2}\left( {{\varepsilon }_{0}}A\left( \dfrac{V}{d} \right) \right)E$

And we know that the formula of capacitance is $\dfrac{{{\varepsilon }_{0}}A}{d}$. Therefore, the formula of force becomes,

$F=\dfrac{1}{2}\left( CV \right)E$

And also the charge on the capacitor is given by $Q=CV$

Therefore, $F=\dfrac{1}{2}QE$, which is the solution to the first part of the question. The origin of the factor $\dfrac{1}{2}$ in the formula denotes the fact that the electric field is just present in the outside area and inside it is zero.

**Note:**This concept is very useful when we learn about the effect of force on the capacitors and how the capacitors of different types react to search force in return. The concept is useful in further understanding electric circuits.

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