Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P23 021

Transcript

21. Studying Sample Problem 23-3, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by
= E
θ/2 λ sin θ 4πε0 r −θ/2 along the symmetry axis where λ = q/rθ with θ in radians. In this problem, each charged quarter-circle produces a field of magnitude
π/4   |q| |q| 1 
 √ . = sin θ E  = 2 rπ/2 4πε0 r ε0 π r 2 2 −π/4 That produced by the positive quarter-circle points at −45◦ , and that of the negative quarter-circle points at +45◦ . By symmetry, we conclude that their net field is horizontal (and rightward in the textbook figure) with magnitude   |q| |q| √ . Ex = 2 cos 45◦ = ε0 π 2 r 2 ε0 π 2 r 2 2