Home
Class 12
PHYSICS
Statement I: A dielectric slab is insert...

Statement I: A dielectric slab is inserted between the plates of an isolated charged capacitor. The charge on the capacitor will remain the same.
Statement II: Charge on an isolated system is conserved.

A

Statement I is true, statement II is ture, statement II is a correct explanation for statement I.

B

Statement I is true, statement II is ture, statement II is not a correct explanation for statement I.

C

Statement I is true, statement II is false.

D

Statement I is false, statement II is true.

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • CAPACITANCE AND CAPACITOR

    CHHAYA PUBLICATION|Exercise MULTIPLE CORRECT ANSWER TYPE|4 Videos

  • CAPACITANCE AND CAPACITOR

    CHHAYA PUBLICATION|Exercise MATRIX MATCH TYPE|5 Videos

  • CAPACITANCE AND CAPACITOR

    CHHAYA PUBLICATION|Exercise HOTS NUMERICAL PROBLEMS|19 Videos

  • ATOMIC NUCLEUS

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|14 Videos

  • COMMUNICATION STSTEM

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|28 Videos

Similar Questions

Explore conceptually related problems

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

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'

A glass slab if introduced between the plates of a parallel plate capacitor. Does the capacitances of the capacitor increase, decrease or remain unchanged?

The electrostatic force between the metal plates of an isotated parallel plate capacitor C having a charge Q and area A, is

Two plates of an isolated charged capacitor are connected by a copper wire. What will happen to the energy stored in the capacitor?

The distance between the two plates of an isolated charged parallel plate air capacitor is increased. The potential difference between the plates will……………………..

A parallel plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant k is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively the charge on each plate, the electric field between the plates and the work done in the process of inserting the slab, then

If the distance between the plates of an isolated parallel plate 200 mu F capacitor, charged to 400 V, is halved, calulate the change in the energy stored in it.