2particle02arc0/42 sin( / 2)/ 42sin( / 2)EQREQRπεθθπε==. With = π, we have particlearc1.57.2EEπ=≈31. First, we need a formula for the field due to the arc. We use the notation λfor 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 arc002sin(/
This is the end of the preview. Sign up
access the rest of the document.
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.