Solve Study Textbooks Guides

Use app

Login

>>Class 12
>>Physics
>>Electrostatic Potential and Capacitance
>>Effects of Dielectrics in Capacitors
>> $A slab of material of dielectric constan$

Question

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $3 d / 4$ , where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

Medium

Open in App

Updated on : 2022-09-05

Solution

Verified by Toppr

$C = \frac{A ϵ _{0}}{d}$

$C^{'} = \frac{A ϵ _{0}}{d - t + \frac{t}{k}}$

Put $t = \frac{3 d}{4}$ ;

$C^{'} = \frac{4 k}{3 + k} . \frac{A ϵ _{0}}{d}$

$C^{'} = \frac{4 k}{3 + k} . C$

Solve any question of Electrostatic Potential and Capacitance with:-

Patterns of problems

Was this answer helpful?

0

0

Similar questions

When a dielectric slab of thickness 6 cm is introduced between the plates of parallel plate condenser it is found that the distance between the plates has to he increased by 4 cm to restore to capacity to original value. The dielectric constant of the slab is

A parallel-plate capacitor has plates of area 0.12 m $^{2}$ and a separation of 1.2 cm. A battery charges the plates to a potential difference of 120 V and is then disconnected.A dielectric slab of thickness 4.0 mm and dielectric constant 4.8 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted?(b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab?

The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value if a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between the plates (where $d$ is the distance of separation between the plates). What is the dielectric constant of the slab?

The area of the plates of a parallel plate condenser is A and the distance between the plates is 10 mm. There are two dielectric sheets in it, one of dielectric constant 10 and thickness 6 mm and the other of dielectric constant 5 and thickness 4 mm, the capacity of the condenser is:-

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same ?

More From Chapter

Electrostatic Potential and Capacitance

View chapter >

Revise with Concepts

Basic Knowledge of Effect of Dielectric on Capacitance of Parallel Plate Capacitor

ExampleDefinitionsFormulaes

Electric Displacement

ExampleDefinitionsFormulaes

Learn with Videos

Capacitance of the parrallel plate capacitor with dielectric slab

Parallel plate capacitor with dielectric in series and parallel

Parallel plate capacitor with dielectric of different thickness

Shortcuts & Tips

Common Misconceptions

Problem solving tips

Memorization tricks

Important Diagrams

Practice more questions

Medium Questions