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
I answeared your question can you answear my question pleas
The force applied would be 1.05*9.8 = 10.3 N the pressure is equal to F/a area will be πr^2 = 0.002826 thus pressure will be = 10.3/0.002826= 3644.72 N/m^2
