net,220092212122||14||2cos4544(8.99 10 N m C )4(4.50 10C)20.6 N/C.(5.00 10 m)xqqErrπεπ−−⎛⎞=°=⎜⎟⎝⎠×⋅×==×(b) By symmetry, the net field points vertically downward in the ˆj−direction, or 90−°counterclockwise from the +xaxis. 24. Studying Sample Problem 22-3, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by 0sin4Erθλε−=πGalong the symmetry axis, with
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