## 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

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

$dQ = \frac{6Q}{5\pi}d\theta$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 $$+5\pi /12$$.