1so wat is the question ?
An elliptical cavity is carved within a perfect conductor.A positive charge q is placed at the center of the cavity. the points A and B are on the surface as shown A(a,0) and B(0,b) where a and b length of the major axis and minor axis resp. the origin is the center of the cavity
Compare the electric field ,potential and charge densities at point A and B
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16 Answers
3is there any expression for charge density as a function of r from the center of the ellipse
1charge 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.
1sry I dunno proof . I just know that statement. Abhishek'll be able to answer I think.
33σ R =constant
R is radius of curvature..
But proof not striking now or may be I don't know.. let me see...
33when a conductor is charged then regions having small radius of cur will have higher charge density...
33
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 ...
33didn' read the question..
just saw the discussion and replied... :P
ok now i'll reply according to question :P
1hey 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.