Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P23 025

Transcript

25. Consider an infinitesimal section of the rod of length dx, a distance x from the left end, as shown in the diagram below. It contains charge dq = λ dx and is a distance r from P . The magnitude of the field it produces at P is given by 1 λ dx . 4πε0 r2 1 λ dx sin θ The x component is dEx = − 4πε0 r2 1 λ dx cos θ . and the y component is dEy = − 4πε0 r2 dE = y dq x R .. ... ... ... . . .. ... ... ... ... . . .. ... ... ... ... . . .. ... ... ... ... . . .. ... ... ... ... . . .. ... ... ... ... . . . ... ... ... ... ... x r θ . ...• P ... . . .. ... ........... d E  We use θ as the variable of integration and substitute r = R/ cos θ, x = R tan θ and dx = (R/ cos2 θ) dθ. The limits of integration are 0 and π/2 rad. Thus,
λ Ex = − 4πε0 R and Ey = − λ 4πε0 R π/2 sin θ dθ = 0
π/2 λ λ cos θ0 = − 4πε0 R 4πε0 R π/2 cos θ dθ = − 0 π/2 λ λ sin θ0 = − . 4πε0 R 4πε0 R  makes an angle of 45◦ with the rod We notice that Ex = Ey no matter what the value of R. Thus, E for all values of R.