1thanks in the given ans d option is also correct Please explain it also
an elliptical cavity is carved within a perfect conductor A positive charge q is placed at the centre of cavity The points A and B are on the cavity surface (a nearer to q) then a)electric field near A in the cavity =electric field near B in the cavity b)charge density at A = charge density at B c)potential at A = potential at B d)total electric field flux through the surface of the cavity is q/eo(here eo means epsilono) Please also explain what is charge density
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8 Answers
1since its a conductor
potential at A = potential at B
let V1=V2
q1/4pi(epsilon)r1=q2/4pi(epsilon)r2
therefore
sigma1 r1=sigma2 r2
11I M NOT SURE ABT IT.....
THE CHARGE IS NOT AT D SURFACE N IT IS IN D CENTRE..
THIS POINT SHLD BE VALID I THINK...
SRRY .. I M STILL NOT SURE ABT IT....
1OH.. Mmm....... vishal.. jus scold me as u can.... i was thinkin how q/ε0 came lik a mad.... (gauss law)
1hehe ram koi ni yaar
sometimes while thinking about big things we r not able to see small things
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no scolding but be calm while solving