ch22-p024 - 24. Studying Sample Problem 22-3, we see that...

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net, 22 00 92 2 1 2 122 | | 14 | | 2 cos45 44 (8.99 10 N m C )4(4.50 10 C) 20.6 N/C. (5.00 10 m) x qq E rr πε π ⎛⎞ = ⎜⎟ ⎝⎠ ×⋅ × == × (b) By symmetry, the net field points vertically downward in the ˆ j direction, or 90 ° counterclockwise from the + x axis. 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 0 sin 4 E r θ λ ε = π G along the symmetry axis, with
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