If a rod with charge density lambda is bent into a 3/4 circle what is the E field at the center?(adsbygoogle = window.adsbygoogle || []).push({});

Well if the rod is 3/4 of a circle then neither the x or y components can cancel out, thus

E= k*lambda/ R (i) + k*lambda/R (j)

right?

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# Homework Help: Uniformly charged rod in an arc

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