17_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

17_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

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Electric Forces and Electric Fields 17 Thus, 3 23 1.87 10 N C x EEE + = × Σ= and 1 144 N C y EE Σ =− giving () 2 2 3 1.88 Rx y + Σ = × and 11 tan tan 0.0769 yx θ −− Σ 4.40 = ° Hence 3 1.88 10 N C at 4.40° below the + ax R x E JG is 15.25 From the symmetry of the charge distribution, students should recognize that the resultant electric field at the center is 0 R = E G R =+ GG If one does not recognize this intuitively, consider:
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This document was uploaded on 10/20/2011.

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