Solving the Eletrodynamics Basic Homework Statement

[vivekrai]

Homework Statement

In non uniform spherical shell of mass 'm' and radius R (here E is gravitational intensity) which of the following are wrong;

(A) E = 0 (r<R)
(B) ∫ E.dS = 0 on any closed surface inside the shell
(C) E can not be zero at any point inside the shell
(D) Gravitational potential is zero at any point inside the shell.

2. The attempt at a solution
I think it should be A,C,D. A friend say C,D only. What do you say. Please help..

 

[BruceW]

I think you have got the right answer. I would interpret statement A as meaning: 'E=0 at any point inside of any non-uniform shell', which is definitely a false statement. Maybe your friend was using the rule which applies to uniform spherical shell, or maybe he/she interpreted statement A differently.

 

[vivekrai]

Please note that in the question the gravitational field has been represented by E rather the Electric field. If already noted, then please ignore.
He has this to offer as explanation :

The gravitational field(due to the shell) inside(anywhere) inside a symmetric shell=0.
This is not valid for non-symmetric objects.
Taking an example,we can consider almost total mass concentrated at a pt. and the ∂m mass forming rest of shell.Here E=0 everywhere.

This difference(from conductors) is shown because of the fact that mass unlike elec. can't move.This is also responsible for the fact that gravitation shielding doesn't exist.

Could you please elaborate what he meant to say, and how does it changes the answer.

 

[BruceW]

The gravitational field(due to the shell) inside(anywhere) inside a symmetric shell=0.
This is not valid for non-symmetric objects.

This bit is correct. So from this, it would mean statement (A) is wrong, because E is not necessarily zero inside the shell. If your friend wrote this, it seems strange that he did not choose (A).

Taking an example,we can consider almost total mass concentrated at a pt. and the ∂m mass forming rest of shell.Here E=0 everywhere.

This does not make sense. If you think about it, the situation is approximately a point mass, so does a point mass create a gravitational field?

This difference(from conductors) is shown because of the fact that mass unlike elec. can't move.This is also responsible for the fact that gravitation shielding doesn't exist.

This is right as well. But it would suggest that (A) is false, so again I don't understand why your friend didn't choose (A).

 

[vivekrai]

He said the it isn't valid for non-symmetric objects. Here the object though has mass non uniformly distributed, is symmetric. So Is it or is it not?

 

[BruceW]

Oh, the mass is symmetrically distributed? I was assuming the mass was not symmetrically distributed. OK, if the mass is symmetrically distributed, I'd say your friend is right. Although I'm not sure about the explanation he gave. I do agree with this bit though: "The gravitational field(due to the shell) inside(anywhere) inside a symmetric shell=0." (Assuming that 'symmetric shell' means the mass is symmetrically distributed). I'm guessing you've been taught this in class. So from here, would you say statement (A) is false or true?

 

[vivekrai]

What if It is a spherical shell with different mass densities along it surface. I mean shape is symmetric but the mass distribution isn't?

 

[BruceW]

It's fairly easy to think of the answer, by an example. Imagine the density on one side of the shell is much greater than on the other side. Will E be zero or non-zero?