Why Isn't Induced Charge Shown in Electrostatics Problems?

[vijayramakrishnan]

Homework Statement

Please see this illustration

http://www.physicsgalaxy.com/lectures/1/57/1006/Solved-Example-1#6 (see question only)

Homework Equations

potential due to a hollow sphere at it's centre = kq/r

The Attempt at a Solution

here won't there be an induced positive charge on the outer surface of inner shell and an induced negative charge on the inner surface of outer shell also to maintain the electrical neutrality of spherical shell?

Why isn't it shown here? And the net potential is also affected by the positive positive charge on the outer surface of inner shell and an induced negative charge on the inner surface of outer shell so won't the answer be different

 

[TSny]

Hello. Welcome to PF.

I believe there is only one shell, not two.

R₁ is the radius of the inner surface of the shell and R₂ is the radius of the outer surface of the shell.

 

[Simon Bridge]

The question in the link is:

Respected sir, can you explain me why the distribution of positive induced charge is symmetrical at outer surface? I can understand that due to eccentric position of charge distribution of negative induced charge will not be uniform. Is the relation charge density is inversely proportional to the radius of curvature valid for induced charges also? please explain in detail? I will be highly thankful to you.

... there's also a video.
The video problem has a thick conducting shell (inner and outer radii labelled on the diagram) with a small charged placed off-center inside it.

The thing to remember about a conductor is that charges in them can move around. They always move so as to cancel out any electric field that may otherwise have been inside. Thus the solid interior of the conductor must have zero static electric field.

The space between the inner and outer surfaces has no charges - all charges have moved to the surfaces. Negative charges have moved to the inner surface, attracted by the positive charge. The positive charges left behind all repel each other so they go to the outer surface.

 

[vijayramakrishnan]

Hello. Welcome to PF.

I believe there is only one shell, not two.

R₁ is the radius of the inner surface of the shell and R₂ is the radius of the outer surface of the shell.

iam extremely sorry sir,for not reading the question properly.my sincere apologies.

 

[vijayramakrishnan]

The question in the link is:
... there's also a video.
The video problem has a thick conducting shell (inner and outer radii labelled on the diagram) with a small charged placed off-center inside it.

The thing to remember about a conductor is that charges in them can move around. They always move so as to cancel out any electric field that may otherwise have been inside. Thus the solid interior of the conductor must have zero static electric field.

The space between the inner and outer surfaces has no charges - all charges have moved to the surfaces. Negative charges have moved to the inner surface, attracted by the positive charge. The positive charges left behind all repel each other so they go to the outer surface.

thank you very much for replying sir.
iam extremely sorry sir,for not reading the question properly.my sincere apologies.

 

[TSny]

iam extremely sorry sir,for not reading the question properly.my sincere apologies.

That's nothing to worry about. We all do that from time to time.