A parallel-plate air capacitor has a capacity ' C ', distance with a separation ' x ', a dielectric material with a dielectric constant ' k ' between plates, and a potential difference with a ' V ' volt applied between plates. The force of attraction between the plates is

Step-by-step solution

The capacitance of the air capacitor is given by C = 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 = Q^2 2k ₀ A . Substituting Q = kCV and ₀ A = Cx , we get: F = (kCV)^2 2k(Cx) F = k^2C^2V^2 2kCx F = kCV^2 2x Answ