cavity

so wat is the question ?

oops forgot to type the question

is there any expression for charge density as a function of r from the center of the ellipse

charge density is inversely propotional to the radius of curvature at that point . so , charge density at B is more. and so is the potential and electric field.

can u prove that sri

but in the soln for the pbm the charge density remains same

sry I dunno proof . I just know that statement. Abhishek'll be able to answer I think.

σ R =constant

R is radius of curvature..
But proof not striking now or may be I don't know.. let me see...

in which cases σ R=constant

when a conductor is charged then regions having small radius of cur will have higher charge density...

Ok i'm trying to explain..

Consider only for a wire type of structure (for simplicity)
in the two fig (lines)

charge at end is greater for equilibrium..

If its not clear i'll try to xplain in a bit detail ...

but in this question charges are produced by induction

didn' read the question..

just saw the discussion and replied... :P

ok now i'll reply according to question :P

Pot at A and B will be same....

hey ya Abhi's right . otherwise . there will be current.
pot A=pot B.
but EB > EA . I think this is bcoz E = -dV/dr r is less then , E is more.