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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:

(A) \[\sqrt 2 C\]
(B) \[2C\]
(C) \[\dfrac{C}{{\sqrt 2 }}\]
(D) \[\dfrac{C}{2}\]

Last updated date: 25th Jun 2024
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Hint It is given that a parallel plate condenser with a di-electric has a capacitance value C. Consider this as our initial case. Now it is said that the dielectric is removed, hence assume k value to be 1. Use capacitance formula to equate both the cases and hence find the capacitance of the second case.

Complete Step by Step Solution
Capacitance is defined as the electric property of the conductor which measures its ability to hold electric charge or store electric charge per unit area of the conductor. It is also given as charge stored per unit voltage passing through a given conductor in a closed circuit. Now, in our problem, it is given that a parallel plate condenser has oil as a dielectric between the surfaces has a capacitance of value C.
 Capacitance is directly proportional to the area of the plates and inversely proportional to the distance between the plates. Higher the di-electric constant value, higher the capacitance and higher the permittivity of di-electric, higher the capacitance value. Mathematically, we can write this as
\[ \Rightarrow C = \dfrac{{k{\varepsilon _0}A}}{d}\], where k is the dielectric constant and A is the area of the plates.
In our first scenario, the plates are said to have oil as dielectric with a dielectric constant value of 2. Now , substituting this in the equations above we get,
\[ \Rightarrow C = \dfrac{{2{\varepsilon _0}A}}{d}\]------(1)
 In the second case, the dielectric is said to be removed from the plates, hence the dielectric medium will be air with a dielectric constant value of 1. Now, the equation changes as,
\[ \Rightarrow {C_2} = \dfrac{{{\varepsilon _0}A}}{d}\]-------(2)
From (1) we can rewrite the common term and substitute it in (2). Now,
\[ \Rightarrow \dfrac{C}{2} = \dfrac{{{\varepsilon _0}A}}{d}\]----(3)
Substituting (3) in (2) we get,
\[ \Rightarrow {C_2} = \dfrac{C}{2}\]

Hence, Option (d) is the right answer for the given question.

Note Often we come across the term dielectric whenever we speak of capacitance. Dielectric is just an electrical insulator which has a capacity to get polarized once an electrical field is applied to it. It is said that dielectric will increase the capacitance value to a greater extent , permitting storage of charges through it.