1@swaraj:i`ve solved it by solid angle....looking for another method....forgot to mention
@nishant sir:yes,sir i`ve undrstood now.....actually the pattern in which the qstn has been put confused me a bit....
1.A point charge Q is located on the axis of a disc of radius R at a distance a from the plane of the disc.If 1/4th of the flux from the charge passes through the disc,then find the relation between a and R.....seems pretty easy but cant seem to come up with a solution
2. A non-conducting ring of radius 0.5m carries a total charge of 1.11 x10^-10 distributed non-uniformly on its circumference producing an electric field E everywhere in space.The value of the line integral:
\int_{l=\propto }^{l=0}{-Edl} in volts is? [l=0 being the centre of the ring]
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6 Answers
62The second part is simple? no?
E can be found by symmetry without any integration
also you can use simple energy definition. Just the potential energy difference between the center and infinity.
potential at center is kq/r
at infinity is zero.
1but sir the charges are not uniformly distributed.....in these cases how do we find out the potential at the centre without integratn?
62yeah but still potential for each elementary charge will be
kdq/r
and it will noly be a scalar which can be added.
so the integral will be k/r integral of dq
which will be kq/r
21for the 1st one
we know solid angle subtended by a cone of max angle 2θ = 2Π(1-cosθ)
here flux is 1/4th so solid angle is 4Î /4=Î
therefore Π=2Π(1-cosθ)
θ=60
now tanθ=R/a
162u can solve it by taking flux linies in the direction of the area vector.
then some integration take an elementary ring.