2. A thin plastic rod is bent into a semi-circular arc of radius R in the xy-pla

Question

2. A thin plastic rod is bent into a semi-circular arc of radius R in the xy-plane and with the center at the origin. The rod carries a charge that is distributed non-uniformly along the rod. The charge distribution is a function of the angle 0 and is given by where M is the charge density amplitude (in Coulombs) a- M cose, do a. What is the charge distribution along the rod, ie. au b. What is the direction of the electric field vector at the origin? c. What is the magnitude of the electric field at the origin?

Explanation / Answer

2. given a thin plastic rod

bent into semicircular arc

radius R

dq/d(phi) = Mcos(phi) ( M is charge density amplitude in coloumbs)

a. dl = rd(phi)

dq/d(phi) * dl/dl = rdq/dl

dq/dl = Mcos(phi)/R

b. electric field vector at the origin is directed at angle thetas to the x axis

now for some angle phi

dEx = k*cos(phi)dq/R^2 = k*cos^2(phi)M*d(phi)/R^2

dEy = -k*sin(phi)*dq/R^2 = -k*sin(2*phi)*d(phi)/R^2

integrating from theta = 0 to theta = pi

Ex = kM*pi/2R^2

Ey = 0

hence net electric field at origin is towards +x axis

c. |E| = Ex = kM*pi/2R^2

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