ch22-p031 - 31. First, we need a formula for the field due...

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2 particle 0 2 arc 0 /4 2 sin( / 2)/ 4 2sin( / 2) E QR EQ R πε θ θπ ε == . With = π , we have particle arc 1.57. 2 E E π =≈ 31. First, we need a formula for the field due to the arc. We use the notation λ for the charge density, λ = Q /L . Sample Problem 22-3 illustrates the simplest approach to circular arc field problems. Following the steps leading to Eq. 22-21, we see that the general result (for arcs that subtend angle ) is [] arc 00 2s i n (/
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This note was uploaded on 06/07/2010 for the course PHYS 344 taught by Professor Bb during the Spring '10 term at The Petroleum Institute.

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