Electric Forces and Electric Fields17Thus, 3231.8710 N CxEEE+ =×Σ=and 1144 N CyEEΣ=−giving ()2231.88Rx y=Σ+Σ=×and 11tantan0.0769yxθ−−Σ4.40=−°Hence 31.8810 N C at 4.40° below the + axRx=×EJGis 15.25From the symmetry of the charge distribution, students should recognize that the resultant electric field at the center is 0R=EGR=+GGIf one does not recognize this intuitively, consider:
This is the end of the preview. Sign up
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.