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) \[\dfrac{C}{2}\]
Answer
274.8k+ views
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.
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.
Recently Updated Pages
Wheatstone Bridge – Principle, Formula, Diagram & Applications

Mass vs Weight: Key Differences Explained for Students

Circuit Switching vs Packet Switching: Key Differences Explained

Uniform Acceleration Explained: Formula, Examples & Graphs

Young’s Double Slit Experiment Derivation Explained

Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Understanding Electromagnetic Waves and Their Importance

Understanding Combined Translation and Rotational Motion

