Home
Class 12
PHYSICS
A dielectric slab of thickness d is inse...

A dielectric slab of thickness `d` is inserted in a parallel plate capacitor whose negative plate is at `x=0` and positive plate is at `x = 3d`. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from `0` to `3d(1998)`.

A

The magnitude of the electric field remains the same

B

the direction of the electric field remains the same

C

The electrical potential increases continuously

D

The electric potentical increases at first, then decreases and again increases

Text Solution

Verified by Experts

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

Topper's Solved these Questions

  • CAPACITANCE

    RESNICK AND HALLIDAY|Exercise PRACTICE QUESTIONS (LINKED COMPREHENSION)|8 Videos
  • CAPACITANCE

    RESNICK AND HALLIDAY|Exercise PRACTICE QUESTION MATRIX MATCH|8 Videos
  • CAPACITANCE

    RESNICK AND HALLIDAY|Exercise PRACTICE QUESTIONS (SINGLE CORRECT CHOICE TYPE)|23 Videos
  • ALL ABOUT ATOMS

    RESNICK AND HALLIDAY|Exercise CHECKPOINT|1 Videos
  • CENTER OF MASS

    RESNICK AND HALLIDAY|Exercise Practice Questions (Integer )|4 Videos

Similar Questions

Explore conceptually related problems

A dielectric slab of thicknesss d is inserted in a parallel - plate capacitor whose negative plate is at x = 0 and positive plate is x = 3d. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d,

A dielectric slab of thickness 't' is kept between the plates of a parallel plate capacitor separated by a distance 'd' (t lt d) . Derive the expression for the capacity of the capacitor .

Knowledge Check

  • A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance

    A
    becomes zero
    B
    remains the same
    C
    decreases
    D
    increases
  • A positive charge q is given to each plate of a parallel plate air capacitor having plate area A and plate separation d, then

    A
    since both the plates are identially charged, therefore, capacitance become equal to zero
    B
    energy stored in the space between the capacitor plates is equal to `(q^(2))/(epsilon_(0)A^(2))`
    C
    no charge appears on inner surface of the plates
    D
    potential difference between the plates is equal to `(2qd)/(epsilon_(0)A)`
  • A parallel plate capacitor is charged by a battery. The battery is removed and a thick glass slab is inserted between the plates. Now,

    A
    The capacity of the capacitor is increased
    B
    The electrical energy stored in the capacitor is decreased
    C
    The potential across the plate is decreased
    D
    The electric field between the plates is decreased.
  • Similar Questions

    Explore conceptually related problems

    A conducting slab of thickness ‘t’ is introduced between the plates of a parallel plate capacitor, separated by a distance d ( t lt d ) . Derive an expression for the capacitance of the capacitor. What will be its capacitance when t = d ?

    A parallel plate capacitor having plate area A and separation between the plates d is charged to potential V .The energy stored in capacitor is ....???

    A conducting slab of thickness 't' is introduced without touching between the plates of a parallel plate capacitor , separated by a distance 'd' ( t lt d) . Derive an expression for the capacitance of the capacitor .

    A slab of copper of thickness b is inserted in between the plates of a parallel plate capacitor as shown in figure. The separation of the plates is d. If b = d//2 , then the ratio of capacities of the capacitor after and before inserting the slab will be

    A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in figure. The separation of the plate is d . If b=d//2 , then the ratio of capacities of the capacitor after and before inserting that slab will be