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# A thin half ring of radius 20m is uniformly charged with a total charge q=0.7nC.The magnitude of the electrica field at the centre of the half ring is

Vikas TU
14149 Points
4 years ago

Reposting:
We would need to calculate the electric field in both direction that is x and y respeectively.
Therefore for x direcn.,
dE = k’*k*dl*cos(thetha)/r^2
where k’ be the charge densite over the semi arc.
Now integerate both sides from theta 0 to pi.
O integerating we get,
Ex = (k*k’/r)(sinpi – sin0) = 0
Similarly in y direcn. we get,
dEy= k’*k*dl*cos(thetha)/r^2
Integerating both sides,
Ey = 2*K*K’/r
Net electric field => E = Ey => 2KK’/r
where K = 1/4pie
and K’ = 0.7 x 10^-9/pi*0.2.