x.Hme=\"margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px

Question

x.Hme="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; padding-top: 0px; padding-right: 0px; padding-bottom: 0px; padding-left: 0px; list-style-type: none; list-style-position: initial; list-style-image: initial; "> A charged insulating rod is bent in a circular arc of radius "r" as show. Assume that it is thin enough that one value of "r" describes the radius. The rod has a linear charge density which varies with the angle ( measured counterclockwise from the +x-axis) according to the equation sin , where is constant. a) find the magnitude of the x-component of the electric field " " at the origin in terms of "", "r" and "K". Hint: For a circle, s = r =>ds =rd. Then = dQ/ds. b) Find the magnitude of the y-component of the electric field "

Explanation / Answer

Let dEx is the x-component electric field that is caused by dQ thatis on the radius. dEx=kdQ/r^2 *cos dQ=*ds=*r*d=0rsindd. -From this part onward is the integra calculating dEx=k0r/r^2 *cossind.---- cos*sin=(sin2)/2 and d2=2d socossind=sin2*d2 /4       =k0/4r *sin2d2 integrate , we have ------ the integral of sinxdx is -cosx so ihave the equation follow (from 0 to 2/3) Ex=k0/4r *(cos(0)-cos*(2/3))=3k0/8r. Similar to dEx dEy=k0/r *sin^2d=k0/r*(1-cos2)/2*d----this part i use the equationsin^2=(1-cos2)/2 so Ey=k0/r *(/6-sin(2/3)/4)----(1-cos2)/2*d=d/2-cos2d2/4, intergralof cosxdx is sinx and of dx is x. Ey=k0/r *0,31 c)so E=k0/r *sqrt((3/8)^2+0,31^2)=0,49*k0/r=132,3(V/m)

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