Gauss's law:E vs r graph (two cases of shell)

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gracy
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Homework Statement


Ok.Then there was a question
If r is distance measured from center of a charged shell and R is it's radius then the graph which may correctly represent variations of electric field is

Two E versus r graphs labelled (A) and (B). In (A), E is zero for r below R, jumps up at R, then decreases with r. In (B), E rises linearly from zero to a constant value up to R, then decreases with r. Axes are E and r, with R marked on each.

Two E versus r graphs labelled (C) and (D), each with origin O and radius R marked. In (C), E rises linearly from O to a peak at R, then falls off gradually. In (D), E rises smoothly from O to a peak at R, then falls linearly.


Homework Equations


E vs R graph

The Attempt at a Solution


The answer is given to be A but I want to ask why not C ?There are two cases
1-shell with all charges on surface in this case E for r<R is zero so Option A fits in
2-Shell with charges throughout the volume(volumetric distribution)
In which E for r<R has formula
E=ρr/3ε0 (ρ=CHARGE PER UNIT VOLUME)
it is clear E ∝ r
For this C seems correct.And the question does not clarify which type of shell(type 1 or type 2) is to be referred
So why answer is A?Why it can't be C?
 
"Shell" is usually intended to mean an infinitesimal (limit-zero thickness) surface.
 
blue_leaf77 said:
(limit-zero thickness) surface.
so no volume ?nearly 2d?
 
Yeah, no volume. The shell alone has no volume, but of course it encompasses certain volume inside it.
 
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