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26.71. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet. Visualize: Solve: (a) Divide the sheet into many long strips parallel to the y-axis and of width ∆x. Each strip has a linear charge density λ = η∆x and acts like a long, charged wire. At the point of interest, strip i contributes a small field r Ei of strength Ei = 2λ 2η∆x = 4πε 0 ri 4πε 0 ri By symmetry, the x-components of all the strips will add to zero and the net field will point straight away from the sheet along the z-axis. The net field’s z-component will be Ez = ∑ ( Ei ) z = ∑ Ei cosθ i i ( The distance ri = x + z 2 i ) 2 1/ 2 i ( and cosθ i = z ri = z x + z Ez = ∑ i 2η∆x 4πε 0 ( x + z 2 i 2 i ) 2 1/ 2 . Thus, z ) (x 2 1/ 2 2 i +z ) 2 1/ 2 = 2ηz 4πε 0 ∑x i 2 i ∆x + z2 If we now let ∆x → dx, the sum becomes an integral ranging from x = ∆L/2 to x = L/2. This gives Ez = 2ηz 4πε 0 L/2 dx ηz 1 −1 x η = tan = 2 2 ∫ x +z z − L / 2 2πε 0 2πε 0 z − L/2 L/2 −1 L −1 − L tan 2 z − tan 2 z From trigonometry, tan −1 ( −φ ) = − tan −1 (φ ) . So finally, Ez = (b) As z → 0 m, r L η η L tan −1 ⇒ E = tan −1 kˆ πε 0 2z πε 0 2z r η π ˆ η ˆ π L ⇒E→ k= k tan −1 → 2z πε 0 2 2ε 0 2 This is the electric field due to a plane as you can see from Equation 26.26. We obtain this result because in the limit as z → 0 m, the dimension L becomes extremely large. As z → ∞, r η L ˆ 1 2λ ˆ L L ⇒E→ k= k tan −1 → 2z πε 0 2 z 4πε 0 z 2z where we have used ηL = λ as the charge per unit length of the sheet. This is the electric field due to a long, charged wire. We obtain this result because for z >> L, the infinitely long sheet “looks” like an infinite line charge. (c) The following table shows the field strength Ez in units of η/ε0 for selected values of z in units of L. A graph of Ez is shown in the figure above. η ε0 z L Ez 0 0.25 0.50 1.0 2.0 3.0 4.0 0.5 0.35 0.25 0.15 0.08 0.05 0.04