What is the force direction on the charge placed at the centre?

[Mr confusion]

Homework Statement

a thin metallic spherical shell contains charge Q on it.A point charge q is placed at its centre and another charge q(1) is placed outside the shell collinear with the charge placed at the centre of the shell.all charges are positive. what is the force direction on the charge placed at the centre?

Homework Equations

coulomb law, gauss law

The Attempt at a Solution

the centre charge will induce some charge on the inner surface of shell. the outside charge is shielded by the metallic sphere.since the centre charge experiences equal forces from all directions ,so it will not have any net force.my problem is that i do not know whether my argument is correct.

 

[turin]

I concur: zero net force.

 

[Mr confusion]

please explain the reasoning by which u arrived at the conclusion u reached,turin. and thanks.

 

[turin]

The shell is a conductor. So, the electric field of the external charge cannot make it past the outer surface of the shell. The internal charge induces image charge on the inner surface equally in all directions, so it is attracted equally in all directions. Basically, the same as your reasoning.

 

[Bill Foster]

Use the method of images.

 

[Mr confusion]

thanks. but i am still having a problem. since the charge experiences no net force, electric field at the centre must be zero. so net E(at centre)=E(due to surface charges on the shell) +E(due to Q1)=0
since E(due to Q1) is directed towards left, E(due to surface charges on shell) must be towards right.this can only happen if there is an asymmetric distribution of charges. now, how is that happening?

 

[Bill Foster]

The charge inside the shell at the center will induce a symmetric charge on the surface.

The charge outside the shell will induce an asymmetric charge on the surface. The total charge distribution on the surface will not be uniform.

 

[turin]

The charge inside the shell at the center will induce a symmetric charge on the surface.

The charge outside the shell will induce an asymmetric charge on the surface. The total charge distribution on the surface will not be uniform.

This happens on the outer surface. Since the charge is distributed on a conductor, it is free to move anywhere it wants in the conductor. So, if there is an electric field in the conductor, it will push the charge around in such a way that the electric field reduces to zero. The conductor effectively isolates the inner surface from the outer suface.

 

[Bill Foster]

Makes sense. But it seems to me, based on section 2.2 in Jackson, and the image below, that that charge density isn't necessarily uniform.

http://img251.imageshack.us/img251/3838/fig23.gif

 

[turin]

You are correct: the charge density is certainly not uniform. However, the point that you must understand is that the charge density is arranged, due to any external charge, in just the right way to make the electric field inside the sphere vanish. So, the only thing that you need to worry about regarding the inside surface is the charge inside, and that situation is symmetrical.

 

[Mr confusion]

thanks to both of you, the problem is appearing clearer to me now. thanks again.