Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.
dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m
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In this exercise we have to use the knowledge of electric field to calculate the charge in a quarter of the arc of the circumfernica, so:
[tex]E=2.19*10^{14[/tex]
So we can say that as it is an arc of a circle we have symmetry in both axes, so:
[tex]dEy=kp/r^2*sin(a) \\E=\int\limits^\pi_0 {2kq/(pi*r^3)} \, \\E=\int\limits^\pi_0 {2(9*10^9)(9.8)/(\pi*(0.093)^3)}\\E=2.19*10^{14}[/tex]
See more about eletric field at brainly.com/question/14438235
