a thin half ring of radius 20 centimetre is uniformly charged with a total charge q =0.7nc . the magnitude of the electric field strength at the centre of this half ring is? answer is

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Vikas TU · Answer

Already answered this qstn. as U have asked it twice,

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.

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a thin half ring of radius 20 centimetre is uniformly charged with a total charge q =0.7nc . the magnitude of the electric field strength at the centre of this half ring is? answer is

a thin half ring of radius 20 centimetre is uniformly charged with a total charge q =0.7nc . the magnitude of the electric field strength at the centre of this half ring is? answer is

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