electrostatics doubts

Is the answer for first one (A) ?

In second one, I'm a bit confused. Why would the empty region between the plates contain the charge. Charge is residing on the shells na?

1st one is (C) because the dielectric will cancel some amount of electric field due to Q.

2nd one A,B,C,D all match with III uncharged!
If the spheres are continuous!

1. answer given is none of these.
2. i think the question is somewhat unclear

For One: How could this be... I was pretty much sure of it.. Anyways let me digg some more or wait for an expert comment.

for first one...i think..it shud be D

cant be A coz wen theres a region of dielectric...the electric field reduces a bit...option A wud hav been correct wen there was no region of dielectric

cant be B coz at a distance "r"...relative permittivity bears no significance coz its a free space region

cant be C coz of the same reason as B

hence D shud be the ryt one....

I consider that to be a Lame reasoning.

" the electric field reduces a bit "

How does it reduce it and by how much.

" relative permittivity bears no significance "

If it reduces the Electric field then why it bears no Significance.

then what should be the correct expression for the field ?

well vivek...yes its a bit lame reasoning for A bt not really for B n C

for B n C...luking at the question....the region with the dielectric ends at R...bt in options...its given upto r...

even i was a bit speculative abt A...

bt for B n C its pretty sure...coz at point "P"...relative permitivity really bears no significance at all...its a region of free space ryt??

please someone provide me with the right expression for Q1.

let us consider that we have a dielectric medium of relative permitivity k

let two charges be there of magnitude Q₁ and Q₂ , seperated by a distance r.

the force between the two charges inside the dielectric is given by:

F =Q₁Q₂4Ï€kε₀r²

If we remove the dielectric , then we can consider the 2 charges to be placed in vacuum and seperated by a distance D

If the force acting between them remains the same,

=Q₁Q₂4πε₀D² =Q₁Q₂4Ï€kε₀r²

=> D= r√k
Thus we can keep two charges in vacuum at a distance r√k, which is equivalent to placing them a distance r apart in a dielectric.

Using this in the given question,
D=R√k + r-R

Thus the electric force between the two two charges is

F = Q₁Q₂4πε₀(R√k-R +r)²

Divide by Q₂ to find the field and the answer comes out to be (D)

Vikrant! don't know why I'm not satisfied with your one!