Let the linear density of charge, = q/0.5*pi*R
The charge on the element of arc length, dq = .dS
Electric field, dE at the center due to the charge, dq is given by,
dE = k.dq/R2 directed radially outward.
Resolving this along x and y axis.
dEy = dE cos = k.dq cos/R2
= k..dScos /R2
But dS = R.d
Hence dEy= k. .R.d. cos/R2
= (k./R).cos. d
Integrating from = 0 to = pi/2
Ey = (k./R).[sin ] for = 0 to = pi/2
= k./R
Using, = 2q/pi*R
Ey+= 2kq/pi*R2
Similarly for the lower quadrant.
But the x components cancel out. The y components add up.
So, Total El field, E = 2.E+ = 4k.q/pi*R2
where, k = 1/4pi*o
Evaluating,
E = 4* 9*109 *1.38*10-12/3.14*(1.67*10-2)2
E = 56.73 N/C
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