A thin glass rod is bent into a semicircle of radius R. A charge 0 is uniformly

Question

A thin glass rod is bent into a semicircle of radius R. A charge 0 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, in terms of k, Q, R, and . (Assume that point P is located at the origin of a coordinate system where the positive x A thin glass rod is bent into a semicircle of radius R. A charge +Q is uniformly Assume that point P is located at the origin of a coordinate system where the positive r -direction is to the right and the positive y- direction is upwards.) + O

Explanation / Answer

here

dq = lambda = ds

dEr = dE * cos(theta) = ( 1 / 4 * pie * e0) * lambda * cos(theta) ds / r^2

ds = r * d(theta)

Er = integration from +45deg to -45deg [ (1/4 * pie * e0) * lambda * cos(theta) * r * d(theta) / r^2)

= ( lambda / 4 * pie * e0 * r) * ( integration from +45deg to -45deg [sin(theta) ])

= (sqrt(2) * ( 4 * q / 2*pie*r) / (4*pie*e0*r) ) = q / (sqrt2 * pie^2 * e0 * r^2)

E = (E+) + (E-) = +sqrt(2 * Er) i

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