Home
Class 12
PHYSICS
A parallel plate air capacitor of capaci...

A parallel plate air capacitor of capacitance `C` is connected to a cell of `emF V` and then disconnected from it. A dielectric slab of dielectric constant `K`, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?

A

The potential difference between the plates decreases K times

B

The energy stored in the capacitor decreases K times

C

The change in energy `(1)/(2) C_(0)V^(2) (K-1)`

D

The change in energy `(1)/(2) C_(0)V^(2) ((1)/(K) -1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NARAYNA|Exercise C.U.Q (Potential and Potential Difference)|30 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NARAYNA|Exercise C.U.Q (Potential energy of system of charges )|6 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NARAYNA|Exercise Evaluate yourself-9|4 Videos

  • ELECTROMAGNETIC WAVES

    NARAYNA|Exercise EXERCISE -4|15 Videos

  • ELECTROSTATICS AND GAUSS LAW

    NARAYNA|Exercise Intergers type question|11 Videos

Similar Questions

Explore conceptually related problems

A paralle plate air capacitor has capacitance C. Now half of the space is fillel with a dielectric material with dielectric constant K as shown in figure . The new capacitance is C . Then

A parallel plate air capacitor has capacitance C. Half of space between the plates is filled with dielectric of dielectric constant K as shown in figure . The new capacitance is C . Then

Knowledge Check

  • A parallel plate capacitor of capacitance C_(0) is charged with a charge Q_(0) to a potential difference V_(0) and the battery is then disconnected Now a dielectric slab of dielectric constant k is inserted between the plates ofcapacitor. The dimensions of the slab are such that it completely fills the space between the plates, then :

    A
    Charge on the' plates remains the same
    B
    Charge on the plates decreases to `Q_(0)//K`
    C
    Charge on _the plates increases to `KQ_(0)`
    D
    Potential difference between plates remains the same

  • A parallel capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E . If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become.

    A
    6E,6C
    B
    E,C
    C
    `E//6,6C`
    D
    E,6C

  • A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K= (5)/(3) is inserted between the plates, the magnitude of the induced charge will be :

    A
    `0.9 n C`
    B
    `1.2 n C`
    C
    `0.3 n C`
    D
    `2.4 n C`

    • Similar Questions

    Explore conceptually related problems

    A parallel plate capacitor having capacitance C farad is connected with a battery of emf V volts. Keeping the capacitor cannected with the battery, a dielectric slab of dielectric constant K is inserted between the plates. The dimensions of the slab are such that it fills the space between the capacitor plates. Then,

    A parallel -plate capacitor, whose plates are kept horizotnal is charged from a cell and then isolated from it. A dielectric slab which can just fit in the gap between the plates is now inserted to fill exactly half of the gap and then left alone. Neglect gravity and friction. The slab will

    A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

    A parallel plate capacitor having capacitance C Farad is connected with a bettery of emf V volts. Keeping the capacitor connected with the bettery, a dieletric slab of dielectric consutant K is inserted between the plates. The dimensions of the slab are such that it fills the space between the capacitor plates, consider the following sttements (i) Potential difference between the capacitor plated remains the same (ii) The capacitance increases by a factor K (iii) The energy stored increases by a factor K Then the correct statements are

    A parallel -plate air capacitor has capacity C . A dielectric slab of dielectric constant K , whose thickness is half of the air gap between the plates, is now inserted between the plates. The capacity of the capacitor will now be

    Home
    Profile