Question

Two charges are separated by a distance 1m. When a dielectric slab of thickness 40 cm is introduced between them, the force becomes half its previous value. The dielectric constant of medium is

Answer 1

Answer:

It has given that, Two charges are seperated by a distance 1 m. when a dielectric slab of thickness 40 cm is introduced between them the force becomes half of its previous value.

To find : The dielectric constant of medium.

solution : let dielectric constant of medium is K. and thickness of dielectric slab is t then effective thickness of dielectric slab in air is t√K.

so, the effective distance between the charges is d = (r - t) + t√K

where r is the distance between charges

so force from Coulomb's law,

this is required formula used when thickness of dielectric slab is given.

case 1 : r = 1 m  .....(1)

case 2 : t = 0.4 m , d = (1 - 0.4 + 0.4√K) = (0.6 + 0.4√K)

so,  ......(2)

a/c to question,

from equations (1) and (2) we get,

1/(1)² = 2/(0.6 + 0.4√K)²

⇒0.6 + 0.4√K = √2 = 1.414

⇒0.4√K = (1.414 - 0.6)

⇒0.4√K = 0.814

⇒K = (0.814/0.4)² = 4.14 ≈ 4

Therefore dielectric constant of medium is 4

hope it will help you...