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anonymous

  • one year ago

A nonconducting rod with a uniformly distributed charge +Q. The rod forms a 10/24 of circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P ,by what factor is the magnitude of the electric field at P multiplied?

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  1. ParthKohli
    • one year ago
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    Kinda like this question. If the rod is collapsed at a point at distance \(R\) from the point, then it can be treated as a point charge producing an electric field of \(\frac{kQ}{R^2}\). If it is an arc, then we'll have to use calculus. 10/24th of a circle means that an angle of 150 degrees is subtended at the center.|dw:1444508331133:dw|

  2. ParthKohli
    • one year ago
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    \[dQ = \frac{6Q}{5\pi}d\theta \]We only need to calculate the vertical component of the electric field, as the horizontal would cancel out (see it for yourself!)\[dE_{y} = k \frac{dQ}{R^2}\cos \theta = \frac{6Qk}{5\pi R^2}\cos \theta d \theta \]Integrate the above from \(-5\pi/12 \) to \(+5\pi /12\).

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