Question

A parallel plate condenser with oil between the plates (dielectric constant of oil \[K=2\]) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes
\[\begin{align}
  & \text{A) }\sqrt{2}C \\
 & \text{B) 2C} \\
 & \text{C) }\dfrac{C}{\sqrt{2}} \\
 & \text{D) }\dfrac{C}{2} \\
\end{align}\]

Answer 1

Hint: We need to understand how the capacitance of the parallel plate capacitor can vary depending on the dielectric constant of the medium used between the plates of the capacitor. The relation between these will help us find the required solution.

Complete answer:
We know that the capacitance is the ability of two metals to hold the electric field between them without conducting. The capacitance for this reason is dependent on the insulating medium which is present between the parallel plates, the distance between the plates and the area of the plates used. The capacitance for a parallel plate capacitor with a dielectric medium can be given as –
\[C=\dfrac{\varepsilon A}{d}\]
Where, C is the capacitance of the capacitor, \[\varepsilon \] is the permeability of the medium, A is the surface area of the plates and d is the distance between the plates.
Now, we are given that the oil used as the medium in the capacitor has a dielectric constant of 2. We can find the capacitance as –
\[\begin{align}
  & K=2 \\
 & \Rightarrow K=\dfrac{\varepsilon }{{{\varepsilon }_{0}}}=2 \\
 & \Rightarrow \varepsilon =2{{\varepsilon }_{0}} \\
 & \text{and,} \\
 & C=\dfrac{\varepsilon A}{d} \\
 & \therefore C=\dfrac{2{{\varepsilon }_{0}}A}{d} \\
\end{align}\]
Now, when the oil is removed, the medium becomes the air, for which the dielectric constant is 1 and therefore, the capacitance can be given as –
\[\begin{align}
  & {{K}_{air}}=1 \\
 & \Rightarrow {{K}_{air}}=\dfrac{\varepsilon }{{{\varepsilon }_{0}}}=1 \\
 & \Rightarrow \varepsilon ={{\varepsilon }_{0}} \\
 & \text{and,} \\
 & {{C}_{air}}=\dfrac{\varepsilon A}{d} \\
 & \therefore {{C}_{air}}=\dfrac{{{\varepsilon }_{0}}A}{d}=\dfrac{C}{2} \\
\end{align}\]
The capacitance of the parallel plate condenser reduces to half when the oil is removed from the capacitor and air becomes the dielectric medium of the capacitor.
This the required capacitance relative to the first case.

The correct answer is option D.

Note:
The dielectric medium present in a parallel plate capacitor increases the ability of storing the electric charges on the plates. The greater insulating the medium, the higher capacity is possessed by the capacitor. Or it can be explained as the absence of conduction between the plates.