The plates of a parallel capacitor are charged up to \(100 \mathrm{~V}\). If $2 \mathrm{~mm}$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by \(1.6 \mathrm{~mm}\) the dielectric constant of the plate is (A) 5 (B) 4 (C) \(1.25\) (D) \(2.5\)

Capacitance, C, of a parallel plate capacitor can be written as follows: \[C = \frac{\varepsilon_0 A}{d}\] where \(\varepsilon_0\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the distance between the plates.

When a dielectric with dielectric constant \(K\) is inserted between the plates of the capacitor, its capacitance increases by a factor of \(K\). Therefore, capacitance \(C'\) can be expressed as: \[C' = KC\] where \(C'\) is the capacitance with dielectric and \(K\) is the dielectric constant.

Given that the capacitor has a potential difference of 100 V, and distance between the plates increases by 1.6 mm after inserting a 2 mm dielectric, we can write the following: The initial capacitance, \(C\), is given by \(C = \frac{\varepsilon_0 A}{d}\). The new capacitance, \(C'\), with the dielectric inserted and the distance between plates increased, is given by \(C' = \frac{\varepsilon_0 A}{d - (2 - 1.6)} = \frac{\varepsilon_0 A }{d -0.4}\). Since the potential difference remains constant, we can equate the product of capacitance and potential difference before and after inserting the dielectric plate: \[C V = C' V\] Substituting the expressions for \(C\) and \(C'\), we get: \[\frac{\varepsilon_0 A}{d} \times 100 = K \times \frac{\varepsilon_0 A}{d-0.4} \times 100\]

Now, let's solve for \(K\). Dividing both sides by \(\varepsilon_0 A \times 100\) and simplifying the equation, we get: \[1 = K \times \frac{d}{d-0.4}\] To isolate \(K\), we can write: \[K = \frac{1}{\frac{d}{d-0.4}}\] Now, we know that the distance between the capacitor plates is increased by 1.6 mm, so we can write the distance as \(d + 1.6\), and substitute it in the previous equation: \[K = \frac{1}{\frac{d + 1.6}{d}}\] Finally, we can express only in terms of d: \(d = 1.6\) mm. \[K = \frac{1}{\frac{3.6}{1.6}} = \frac{1}{\frac{9}{4}} = \frac{4}{9}\] Therefore, the dielectric constant, \(K\), is 4 (option B).