1but sir the charges are not uniformly distributed.....in these cases how do we find out the potential at the centre without integratn?
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.
162yeah 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
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....
62u can solve it by taking flux linies in the direction of the area vector.
then some integration take an elementary ring.