Home
Class 12
PHYSICS
STATEMENT-1: A parallel plate capacitor ...

STATEMENT-1: A parallel plate capacitor is charged by a `d.c.` source supplying a constant voltage `V`. If the plates are kept connected to the source and the space between the plates is filled with a dielectric, the charge on the plates will increases.
STATEMENT-2: Additional charge will flow from teh source to the plates.

A

Statement -1 si Ture, Statement -2 is True, Statement -2 is a correct explanation for Statement -1.

B

Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for Statement-1.

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True.

Text Solution

Verified by Experts

The correct Answer is:
A

The correct choice is `(A)`. Since the source supplies a contant voltage `V`. The potential difference between the plates remains equal to `V` because the source is not disconnected. The capacitance `C` increases due to the introduction of the dielectric. since `Q = CV`, the charge `q` on the capacitor plates will increase.
Promotional Banner

Similar Questions

Explore conceptually related problems

A parallel plate capacitor is charged. If the plates are pulled apart

STATEMENT-1: A parallel plate capacitor is charged by a battery of voltage V . The battery is then disconnected. If the space between the plates is filled with a dielectric, the enegry stroed in the capacitor will decrease STATEMENT-2: The capacitance of a capacitor increases due to the introduciton of a dielectric between the plates.

A parallel plate capacitor with air between the plates has a capacitance C. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant 6, then the capacitance will become.

A parallel plate condenser is charged by connected it to a battery. Without disconnected the battery, the space between the plates is completely filled with a medium of dielectric constant k . Then

A parallel plate capacitor is charged by connecting is plates to the terminals of a battery. The battery remains connected to the condenser plates and a glass plate is interposed between the plates of the capacitor, then

A parallel plate capacitor is charged and then isolated. On increasing the plate separation

A parallel plate capacitor is charged completely and then disconnected from the battery. IF the separation between the plates is reduced by 50% and the space between the plates if filled with a dielectric slab of dielectric constant 10, then the potential difference between the plates

A parallel plate air condenser is charged and then disconnected from the charging battery. Now the space between the plates is filled with a dielectric then, it electric field strength between the plates

A parallel plate capacitor is first connected to a constant voltage source. It is then disconnected and ·then immersed in a liqnid dielectric, then: