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Ex. Electric Potential due to a Finite Charged Rod Find the electric potential some distance y above a uniformly charged finite rod r =x +y 2 P λ= dy 2 Q 2L l = 2L 2 ⇒ Total Linear Charge Density ⇒ Total Length of the Rod r y dq x Since this is a uniform charged rod Æ λ= dq represents the amount of charge on the rod piece of length dx. dx dq dx dq = λ dx Î At point P: V = ke dq ∫r L V = ke λ dx ∫ x +y 2 −L 2 Since x2 is an even function, we can change the limits of integration from –L to L to 2 times the integral from 0 to L. L V = 2λ k e dx ∫ x +y 2 0 2 From integration tables: dx ∫ Î x +a 2 ( V = 2λ ke ln L + V= k eQ L 2 ( = ln x + x 2 + a 2 L +y 2 2 L + L2 + y 2 y ln ) ) − ln ( y ) Using λ = a & ln = ln a − ln b 2L b Q Alternate Integration: We can also determine the electric potential by using the electric field for a finite charged rod. Recall: E= 2ke λ L y y +L 2 y 2 yˆ for a finite charge rod K ∫ ∞ Using V = − E ⋅ dl y V=− ∫ Remember: Our V = 0 point is at r = ∞ (or y = ∞ in this case). 2ke λ L ∞ y' K y' + L 2 y V = −2keλ L ∫ ∞ y' 2 yˆ ⋅ dy ' K dy ' Note: yˆ ⋅ dy = yˆ dy ' cosθ = dy ' , since θ = 0 . y' + L 2 2 From integration tables: ∫ Î a + u2 + a2 = − ln a u u u2 + a2 du 1 1 L + L2 + y 2 V = −2keλ L − ln + ln(1) y L L + L2 + y 2 V = 2keλ ln y Note: ln(1) = 0 L + L2 + y 2 V= ln L y Using λ = keQ Which is the same result we obtained before. Q 2L