Home
Class 12
PHYSICS
n conducting plates are placed face to f...

`n` conducting plates are placed face to face. Distance between two consecutive plates is `d`.
Area of plates is `A,(A)/(2),(A)/(4),(A)/(8)....((1)/(2^(n-1)))A`
A dielectric slab of dielectric constant `k` is inserted between the first and second plates and the assembly is charged by a battery of `emf xi`. Find the charge stored in the assembly.
.

A

`(epsilon_(0) A xi)/(2d[(1)/(k)+2^(n-1)-2])`

B

`(epsilon_(0) A xi)/(2d[(1)/(k)+2^(n-1)-1])`

C

`(epsilon_(0) A xi)/(d[(1)/(k)+2^(n-1)-2])`

D

`(epsilon_(0) A xi)/(3d[(1)/(k)+2^(n-1)-2])`

Text Solution

Verified by Experts

The correct Answer is:
A
`(n-1)` capacitors are in series, So,
`(1)/(C)=(1)/((epsilon_(0)A//2)/(d))+(1)/((epsilon_(0) A//4)/(d))+...+(1)/((epsilon_(0) A//2^(n-1))/(d)s)`
`C_(eq)=(epsilon_(0)A)/(2d[(1)/(k)+2^(n-1)-2])`
` :. q=C_(eq) xi rArr 1=(epsilon_(0) A xi)/(2d[(1)/(k)+2^(n-1)-2])`
`V_(1) = 30 V and V_(2) = 30 V` (Conceptual).
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

Find the charge on each plate given area of each plate is A and separation between two consecutive plates is d.

Introduction of a dielectric slab of dielectric constant K between the plates |When the battery is disconnected |When the battery remains connected |Questions |Force between capacitor plates

Knowledge Check

  • The separation between the plates of a parallel plate capacitor is d and the area of each plate is A. iff a dielectric slab of thickness x and dielectric constant K is introduced between the plates, then the capacitance will be

    A
    `(epsi_(0)A)/(d-x(1-(1)/(K)))`
    B
    `(epsi_(0)A)/(d+x(1-(1)/(K)))`
    C
    `(epsi_(0)A)/(d+x(1+(1)/(K)))`
    D
    `(epsi_(0)A)/(d-x(1+(2)/(K)))`

  • A parallel plate capacitor carries a harge Q. If a dielectric slab with dielectric constant K=2 is dipped between the plates, then

    A
    the stored energy remains unchanged.
    B
    the stored energy is increased by a factor of 2.
    C
    the stored energy is reduced to half its previous value.
    D
    none of the above is correct.

  • 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

    • Similar Questions

    Explore conceptually related problems

    A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q' .

    Two dielectric slabs of dielectric constants K_1 and K_2 are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.

    A parallel-plate capacitor has plate area A , the distance between its two plates x and y is d and a dielectric slab of dielectric constant epsilon_(r) is filled between the regions. It is connected to a cell of emf E . If the dielectric slab is removed, then

    Suppose n conducting face to face, and the distance between two successive plates is d. Each plate is half of the area of the previous one. If area of first plate is A . (i) If the area of the first plate is A, find the equyivalent capacitance of the system. (ii) If a cell of emf V is connected as shown in and a dielectric of dielectric constant k is inserted between the first and second plates, dind the charge on the nth plate.

    The potential enery of a charged parallel plate capacitor is U_(0) . If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

    Home
    Profile