A parallel plate capacitor is made of two circular plates separated by a distance 5mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is `3xx10^4V//m` the charge density of the positive plate will be close to:
A
`3 xx 10^(4) C//m^(2)`
B
`6 xx 10^(4) C//m^(2)`
C
`6 xx 10^(-7) C//m^(2)`
D
`3 xx 10^(-7) C//m^(2)`
Text Solution
AI Generated Solution
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
Consider a parallel plate capcaitor with plates 20 cm by 20 cm and separated by 2 mm. The dielectric constant of the material between the plates is 5. The plates are connected to a voltage source of 500 V. The energy density of the field between the plates will be close to
A parallel plate capacitor with square plates is filed with four dielectrics of dielectric constants K_(1),K_(2),K_(3),K_(4), arranged as shown in the figure. The effective dielectric constant K will be :
A parallel plate capacitor is formed by two plates, each of area 100 cm^2 , separated by a distance of 1 mm. A dielectric of dielectric constant 5.0 and dielectric strength 1.9 X 10 ^7 Vm^(-1) is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing any dielectric breakdown.
A parallel plate capacitor, with plate area A and plate separation d, is filled with a dielectric slabe as shown. What is the capacitance of the arrangement ?
An uncharged parallel plate capacitor is connected to a battery. The electric field between the plates is 10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entries space. The electric field between the plates now is
A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected ? Justify your answer.
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 (without dielectric) is charged by a battery and kept connected to the battery. A dielectric salb of dielectric constant 'k' is inserted between the plates fully occupying the space between the plates. The energy density of electric field between the plates will:
A parallel plate capacitor consists of two circular plates of radius R= 0.1 m. they are separated by a distance d=0.5 mm , if electric field between the capacitor plates changes as (dE)/(dt) = 5 xx 10^(13) (V)/(mxxs) . Find displacement current between the plates.
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) .
JEE MAINS PREVIOUS YEAR ENGLISH-JEE MAINS-Chemistry
