Home
Class 12
PHYSICS
A parallel plate capacitor having plates...

A parallel plate capacitor having plates of area S and plate separation d, has capacitance `C_1` in air. When two dielectrics of different relative primitivities (`epsilon_1=2` and `epsilon_2=4`) are introduced between the two plates as shown in the figure, the capacitance becomes `C_2`. The ratio `C_2/C_1` is

A

(a) 6/5

B

(b) 5/3

C

(c) 7/5

D

(d) 7/3

Text Solution

Verified by Experts

The correct Answer is:
D


`C_(eq)=(C_1xxC_2)/(C_1+C_2)+C_3=((2epsilon_0s)/(d)xx(4epsilon_0s)/(d))/((6epsilon_0s)/(d))+(epsilon_0s)/(d)`
`=4/3(epsilon_0s)/(d)+(epsilon_0s)/(d)`
`:.` `C_(eq)=7/3(epsilon_0s)/(d)=7/3C_1` `[:' C_1=(epsilon_0s)/(d)]`
Promotional Banner

Topper's Solved these Questions

  • ELECTROSTATICS

    SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise Subjective Problems|2 Videos

  • ELECTROSTATICS

    SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise Comprehension Based Questions|2 Videos

  • ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT

    SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise JEE Main And Advanced|107 Videos

  • MODERN PHYSICS

    SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise MCQ (One Correct Answer|1 Videos

Similar Questions

Explore conceptually related problems

Three dielectrics of relative permittivities epsilon_(r_(1))=1,epsilon_(r_(2))=2 , and epsilon_(r_(3))=3 are introduced in a parallel plate capacitor of plate A and B. .

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be :

The capacitance of a parallel plate condenser is C_(0) If a dielectric of relative permittivity epsilon_(r) and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of (C)/(C_(0)) will be -

A parallel plate capacitor with vacuum between its plates has capacitance C . A slabe of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill 1//3^(rd) of the capacitor as shown in the figure. the new capacitance will be

A parallel plate capacitor with air between plates (dielectric constant K =2) has a capacitance C. If the air is removed, then capacitance of the capacitor is

A parallel-plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K_1 and K_2 of same area A//2 and thickness d//2 are inserted in the space between the plates. The capacitance of the capacitor will be given by :

A parallel plate capacitor of area A , plate separation d and capacitance C is filled with three different dielectric materials having dielectric constant K_(1),K_(2) and K_(3) as shown in fig. If a single dielectric material is to be used to have the same effective capacitance as the above combination then its dielectric constant K is given by :

Two parallel plate of area A and separated by two different dielectric as shown in the figure. The net capacitance is