a.) Consider a ring of radius 7.7 cm with a uniform linear charge density = 3.5

Question

a.) Consider a ring of radius 7.7 cm with a uniform linear charge density = 3.5 Cm . Calculate the E field at a distance 7 cm from the center along the axis of the ring.

b.) A part of the above ring is cut off so that the remainder subtends an angle 59° at the center of the circle. Find the magnitude of the E field at the center of this new arc, assuming that the three-fourths of the charge was lost in the cutting process.

I figured out part a by integrating kdq/(x^2+R^2) cos from 0 to 2 getting

E= KQx/(x^2+R^2)^3/2

plugging in my numbers I got 9.46E11 N/C which is correct but I can not figure out part b. The wording is throwing me off.

Thanks for the help

Explanation / Answer

total charge of the ring. Q=?*2pi*R where ? be the charges density. only Q/4 left. so Q0=Q/4=?*2pi*R/4=0.42(C). ----------------- 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*59*2*2pi/360)=5.3(C/m). integrate from alpha = -59 deg to 59 deg. we have E0=2*k*?'*sin59/r=1.062e12(N/C)

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