Hi, I\'m having trouble with this question. I worked out the first part but if a

Question

Hi, I'm having trouble with this question. I worked out the first part but if anyone can help me with the second part it? would be greatly appreciated.

Thanks!

Consider a ring of radius 3.2 cm with a uniform linear charge density M 5.1 C/m. Calculate the E field at a distance 1 cm from the center along the axis of the ring. 2.45x10 12 V/ma You are correct. Computer's answer now shown above. Previous Tries Your receipt is 154-6793 A part of the above ring is cut off so that the remainder subtends an angle 36° at the center of the circle. Find the magnitude of the Efield at the center of this new arc, assuming that the three-fourths of the charge was lost in the cutting process. ds angle ds

Explanation / Answer

electric field at the center of the ring now. (cause the ring-part still has its symmetrically shape, so the electric field will have the direction of the middle line of the ring).
let alpha be the angle of the mid-line to particles ds.
we have
dE=kdq/r^2
dq=?'*dx=?'*r*d alpha
dE_0=dE*cos alpha = k*?'*r*d alpha *cos alpha/r^2=k*?'*d alpha* cos alpha/r.
where ?' is the new linear charge density.
?'=0.42/(7.7e-2*36*2*2pi/360)=4.34(C/m).
integrate from alpha = -36 deg to 36 deg.
we have E0=2*k*?'*sin36/r=1.062e12(N/C)

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