A thin glass rod is bent into a semicircle of radius R. A charge +Q is uniformly

Question

A thin glass rod is bent into a semicircle of radius R. A charge +Q is uniformly distributed along the upper half, and a charge -Q is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field E (in component form) at point P, the center of the semicircle. Two uniformly charged insulating rods are bent in a semicircular shape with radius r = 10.0 cm. If they are positioned so that they form a circle but do not touch and if they have opposite charges of +1.00 mu C and -1.00 mu C, find the magnitude and the direction of the electric field at the center of the composite circular charge configuration. A uniformly charged rod of length L with total charge Q lies along the y-axis, from y = 0 to y = L. Find an expression for the electric field at the point (d,0) (that is, the point at x = d on the x-axis).

Explanation / Answer

22-36)

field due to semicircle = 2K* lambda / R

her lambda = charge density = Q / ( pi*R)

here we have two semicircles, one with charge density lambda . other with charge density - lambda.

so, net field = (2K* lambda / R ) - (2K* lambda / R ) =0

net electric filed at centre = 0 N/C

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