Home
Class 12
PHYSICS
A parallel plate capacitor is of area 6 ...

A parallel plate capacitor is of area 6 `cm^(2)` and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant `K_(1)=10`, `K_(2)=12 and K_(3) = 14`. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be :

A

4

B

14

C

12

D

36

Text Solution

Verified by Experts

Promotional Banner

Topper's Solved these Questions

  • JEE MAINS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise Chemistry|1 Videos

  • JEE MAIN

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|452 Videos

  • JEE MAINS 2020

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise PHYSICS|250 Videos

Similar Questions

Explore conceptually related problems

A parallel plate capacitor of plate area A and separation d filled with three dielectric materials as show in figure. The dielectric constants are K_1 , K_2 and K_3 respectively. The capacitance across AB will be :

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k_1 , k_2 and k_3 as shown. If a isngle dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectic constant k is given by

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials hving dielectric constants K_(1), K_(2), K_(3) and K_(4) as shown in the figure below. If a single dielectric material is to be used to have the same capacitance c in this capacitor, then its dielectric constant K is given by

In the figure a capacitor is filled with dielectrics K_(1) , K_(2) and K_(3) . The resultant capacitance is

A parallel plate capacitor has plate area A and plate separation d. The space betwwen the plates is filled up to a thickness x (ltd) with a dielectric constant K. Calculate the capacitance of the system.

A parallel plate capacitor of plate area A and separation d is filled with two materials each of thickness d/2 and dielectric constants in_1 and in_2 respectively. The equivalent capacitance will be

Air filled capacitor of capacitance 2 muF is filled with three dielectric material of dielectric constant K_(1) = 4, K _(2) = 4 and K_(3) = 6 as shown in the figure . The new capacitance of the capacitor is

Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K_(1), K_(2) and K_(3) . The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig. II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be ( E_(1) refers to capacitor (I) and E_(2) to capacitor (II)):

A capacitor of plate area A and separation d is filled with two dielectrics of dielectric constant K_(1) = 6 and K_(2) = 4 . New capacitance will be

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by K=K_(0)(1+alphax). . Calculate capacitance of system: (given alpha d ltlt 1) .