electrostatics doubts

16 Answers

71
Vivek @ Born this Way ·

Is the answer for first one (A) ?

71
Vivek @ Born this Way ·

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?

1
Athenes Analyst ·

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

1
Athenes Analyst ·

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

262
Aditya Bhutra ·

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

71
Vivek @ Born this Way ·

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.

1
buddyboyyash ·

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....

71
Vivek @ Born this Way ·

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.

21
Shubhodip ·

I think vivek is correct

apply alternate form of gauss law

∫kE.ds = Qfree/ε0

262
Aditya Bhutra ·

then what should be the correct expression for the field ?

1
buddyboyyash ·

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??

262
Aditya Bhutra ·

please someone provide me with the right expression for Q1.

1
vikrant001 ·

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 =Q1Q24πkε0r2

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 =Q1Q24πkε0r2

=> 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)

1
Athenes Analyst ·

VIKRANT THANKS!

i dint take hardship to try this one :(

71
Vivek @ Born this Way ·

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

1
vikrant001 ·

@Vivek , why , in this answer i have effectively removed the dielectric , for making it easier to solve mathematicaly, though i agree that i am not able to solve this question by using gauss' law, but still i believe my answer is correct , i would appreciate if someone solved it using gauss' law!!

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