Home
Class 12
PHYSICS
A parallel plate capacitor is charged by...

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is now introduced between the two plates to occupy the space completely. State the effect on the following:
(i) the capacitance of the capacitor.
(ii) potential difference between the plates.
(iii) the energy stored in the capacitor.

Text Solution

AI Generated Solution

To solve the problem, we will analyze the effects of introducing a dielectric slab into a parallel plate capacitor that has been charged and then disconnected from the battery. We will examine the changes in capacitance, potential difference, and energy stored in the capacitor. ### Step-by-Step Solution: 1. **Initial Setup**: - Let the initial capacitance of the capacitor (without the dielectric) be \( C_0 \). - The potential difference across the capacitor when connected to the battery is \( V_0 \). - The charge stored on the capacitor is \( Q_0 = C_0 V_0 \). ...
Promotional Banner

Similar Questions

Explore conceptually related problems

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected ? Justify your answer in each case.

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab having dielectric constant (relative permittivity) K, is now introduced between its two plates in order to occupy the space completely. State, in terms of K, its effect on the following: The capacitance of the capacitor

A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,

A parallel plate condenser is charged by connected it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then

To reduce the capacitance of parallel plate capacitor, the space between the plate is

A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected ? Justify your answer.

A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then:

A parallel plate capacitor is charged and then the battery is disconnected. When the plates of the capacitor are brought closer, then

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