A parallel plate air capacitor has a capacitance $ 18\mu F $ . If the distance between the plates is trebled and a dielectric medium is introduced, the capacitance becomes $ 72\mu F $ . The dielectric constant of the medium is
(A) $ 4 $
(B) $ 9 $
(C) $ 12 $
(D) $ 2 $

Hint
First calculate the electric field between the two plates. Using that, we can find the potential difference between them. From here we can find the capacitance of the system applying a suitable formula.
 $ \Rightarrow E = \dfrac{\sigma }{{{\varepsilon _0}}} $ where $ \sigma $ is the charge per unit area on the plates and $ {\varepsilon _0} $ is the permittivity of free space.
 $ \Rightarrow Q = CV $ where $ Q $ is the charge on the positive conductor and $ C $ is called the capacitance.
 $ \Rightarrow \sigma = \dfrac{Q}{A} $ where $ A $ is the area of the capacitor plate.

Complete step by step answer
A capacitor is a system of conductors and dielectric that can store electric charge. It consists of two conductors containing equal and opposite charges and has a potential difference $ V $ between them.
The potential difference between the conductors is proportional to the charge on the capacitor and is given by the relation $ Q = CV $ where $ Q $ is the charge on the positive conductor and $ C $ is called the capacitance.
Now, for a parallel plate air capacitor, the electric field between the plates is $ E = \dfrac{\sigma }{{{\varepsilon _0}}} $ where $ \sigma $ is the charge per unit area on the plates and $ {\varepsilon _0} $ is the permittivity of free space.
Now, we know that the potential difference between the two plates is given by, $ V = E \times d $ where $ d $ is the distance between the two plates.
Thus, substituting the value of $ V $ in the equation $ Q = CV $ , we get,
 $ \Rightarrow Q = CEd $
Putting $ E = \dfrac{\sigma }{{{\varepsilon _0}}} $ and $ \sigma = \dfrac{Q}{A} $ where $ A $ is the area of the capacitor plate, we get
 $ \Rightarrow Q = C \times \dfrac{Q}{{A{\varepsilon _0}}} \times d $
 $ \Rightarrow C = \dfrac{{A{\varepsilon _0}}}{d} $
This value is given to be $ 18\mu F $ .
Now, we are told that a dielectric medium of constant $ K $ is introduced between the conducting plates and the separation is increased by three times. So substituting $ {\varepsilon _0} $ with $ \varepsilon = K{\varepsilon _0} $ and $ d $ with $ 3d $ in the capacitance equation we get,
 $ \Rightarrow C = \dfrac{{A\varepsilon }}{{3d}} $
 $ \Rightarrow C = \dfrac{{A \times K{\varepsilon _0}}}{{3d}} $
 $ \Rightarrow 72 = \dfrac{{K \times 18}}{3} $
Where this new capacitance is $ 72\mu F $
 $ \Rightarrow K = \dfrac{{72 \times 3}}{{18}} = 12 $
Therefore, the correct option is (C).

Note
To establish the capacitance of an isolated single conductor, we assume the conductor to be a part of a capacitor whose other conductor is at infinity. In this particular problem we deal with capacitance with uniform dielectric, but there are certain cases where a non-uniform dielectric is used which varies with distance. For such a case we take $ \dfrac{{dK}}{{dx}} = \alpha $ , where $ \alpha $ is a constant.