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apositive cgarge Q is uniformly distributed around asemicircile of radius R.find the electric field (magneitud and direction)at the center of curvature P.

  • physics -

    Charge per unit length along the semicircle is Q/(pi*R). Net field E will be perpendicular to the diameter that runs from one end to the other. Use Coulomb's Law and integrate along the semicircle.

    E = [kQ/(pi*R)]*[R*d(theta)/R^2]*
    Integral of cos theta from -pi/2 to pi/2.
    k is the Coulomb constant.

    I get E = 2 k Q/(pi*R^2)

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