A parallel-plate air capacitor has a capacity ' C ', distance with...

The capacitance of the air capacitor is given by $C = \dfrac{\epsilon_0 A}{x}$, where $A$ is the area of the plates.

When a dielectric material of dielectric constant $k$ is introduced between the plates, the new capacitance becomes $C' = kC$.

The charge on the capacitor plates when a potential difference $V$ is applied is $Q = C'V = kCV$.

The magnitude of the force of attraction between the plates of the capacitor is given by $F = \dfrac{Q^2}{2k\epsilon_0 A}$.

Substituting $Q = kCV$ and $\epsilon_0 A = Cx$, we get:

$F = \dfrac{(kCV)^2}{2k(Cx)}$

$F = \dfrac{k^2C^2V^2}{2kCx}$

$F = \dfrac{kCV^2}{2x}$

Answer: $\dfrac{kCV^2}{2x}$