17_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

17_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online