Question

Force between two identical charges placed at a distance of r in vacumm is F.Now a slab of dielectric constant 4 is ijnserted between these two charges . If the thickness of the slab is r//2, then the force between the charges will becomes

Answer 1

Answer:

I am sorry I didn't know that answer

Answer 2

answer : option (c) 4F/9

explanation :Let charge on each particle is q.

given, force between particles placed at a

distance of r in vaccum is F.

so, F = kq²/r² ......(1)

now, a slab of dielectric constant , k = 4 is

inserted between these two charges. and

also given thickness of slab , t = r/2

so, now force will be F' = kq²/(r + t√k - t)²

= kq²/(r + r/2 × √4 - r/2)²

= kq²/(r + r - r/2)²

= kq²/(3r/2)²

= 4kq²/9r²......(2)

from equation (1),

F' = 4F/9

hence, force between the charges will

become 4F/9.