Question

Two point charges q1 and q2 are separated by a distance d in vacuum the force acting decrease by 75℅then dielectric slab k and thickness d/2 placed b/w them find k

Answer 1

Answer:

s75 is the answer is the answer is the answer

Answer 2

Answer:

The dielectric constant(K) of the slab will be 4$/$3.

Explanation:

The force between two charged particles separated by distance d in a vacuum is given by,

Force (F)=Q1 × Q2$/$4π∈0 ×$d^{2}$

Q1, Q2=charge on particles 

d=distance between the charged particles

∈0=permitivity of free space=8.85 x 10^-12 m^-3Kg^-1s^4A^2

Now according to the question,

case I: Vacuum

F=q1q2$/$4π∈0 ×$d^{2}$               (1)

case II: introduction of the dielectric slab

3/4F=q1q2$/$4π∈r ×$d^{2}$            (2)

∴∈r= relative permitivity=dielectric constant(K) × ∈0

By taking the ratio of equations 1 and 2 we get,

K=4$/$3

Hence, the dielectric constant of the slab will be 4$/$3.