Electric Field from a charged line?
A charged line extends from y = 2.50 cm to y = -2.50 cm. The total charge distributed uniformly along the line is -6.50 nanoCoulombs. http://session.masteringphysics.com/problemAsset/1266147/5/6754621027_copy.jpg 1. Find the magnitude of the electric field on the x-axis at x = 10.0 cm. 2. Find the direction of the electric field on the x-axis at x = 10.0 cm. 3. Is the magnitude of the electric field you calculated in part A larger or smaller than the electric field 10.0 cm from a point charge that has the same total charge as this finite line of charge? 4. At what distance does the result for the finite line of charge differ by 1.0% from that for the point charge? Thank you!
A uniform line charge extends from x = - 2.5 cm to x = + 2.5 cm and has a linear charge density of lambda = 5.?
λ= 5.8 x10^-9 C/m y = 4.5 m |x| = 2.5 cm (0.025m) dq = λ dx (dx is the infinitesimal length of the line) ------------------------------- ................| P(0;4.5) ................| ................| |dq|----------O----------|dq| The components of electric field perpendicular to the axis of the line are erased, remains only the components second y-axis: dE(P) = dE1(y) sinθ + dE2(y) sinθ by symmetry: dE(P)= 2 dE1(y) sinθ ...and definition of eletric field: dE(P)= 2k (dq/r²) sinθ Electric field in P(0;4.5) E(p)= ∫2k (dq/r²) sinθ (…extended from 0 to x…) Replace dq: E(p)= ∫2k (λ dx /r²) sinθ = 2k λ ∫ (1 /r²) sinθ dx now you need to replace the variable of integration r = y/sinθ x = y cotθ ...differential... dx = - y (1/sin²θ) dθ important note: when x → 0 then θ → (π/2) when x → |x| then θ → θ’ replace all: E(p)= - 2k λ ∫ (1/(y/sinθ)²) sinθ y (1/sin²θ) dθ (…extended from (π/2) to θ…) simplifies: E(p)= - ((2k λ)/y) ∫ sinθ dθ solve: E(p)= - ((2k λ)/y) ∫sinθ dθ = - (2k λ/y) [- cosθ] = - (2k λ/y) (cos(π/2) – cosθ’) attention to the sign: E(p)= (2k λ/y) (cosθ’ - cos(π/2)) = (2k λ/y) (cosθ’ - 0) = (2k λ/y) cosθ’ you need to find the cosine of the θ' angle (simple geometric considerations) ..................____ cosθ' = |x|/√|x|²+y² ≈ |x|/y replace and solve: E(p)= ((2k λ)/y) (|x|/y) ≈ ((2k λ)/y²) |x| ≈ 0.13 N/C