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Unformatted text preview: 21. Studying Sample Problem 233, we see that the ﬁeld evaluated at the center of curvature due to a charged distribution on a circular arc is given by E= λ sin θ 4πε0 r
θ /2 along the symmetry axis
−θ/2 where λ = q/rθ with θ in radians. In this problem, each charged quartercircle produces a ﬁeld of magnitude π /4 q  q  1 √. E= = sin θ 2 r2 2 rπ/2 4πε0 r ε0 π −π/4 That produced by the positive quartercircle points at −45◦ , and that of the negative quartercircle points at +45◦ . By symmetry, we conclude that their net ﬁeld is horizontal (and rightward in the textbook ﬁgure) with magnitude Ex = 2 q  √ ε0 π 2 r 2 2 cos 45◦ = q  . ε0 π 2 r 2 ...
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 Fall '08
 SPRUNGER
 Physics, Charge

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