# lecture05 - September 8 2011 Physics for Scientists&...

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Unformatted text preview: September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 1 Please register your iClicker at http://www.iclicker.com/support/ registeryourclicker/ Include the A when you enter your student ID! September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 2 Electric Field from an Electric Dipole A system of two oppositely charged point particles is called an electric dipole e vector sum of the electric eld from the two charges gives the electric eld of the dipole We have shown the electric eld lines from a dipole We will derive a general expression good anywhere along the dashed line and then get an expression for the electric eld a long distance away from the dipole September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 3 Electric Field from an Electric Dipole Start with two charges on the x-axis a distance d apart • Put - q at x = - d /2 • Put + q at x = + d /2 Calculate the electric eld at a point P a distance x from the origin September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 4 Electric Field from an Electric Dipole e electric eld at any point x is the sum of the electric elds from + q and – q Replacing r + and r- we get the electric eld everywhere on the x-axis (except for x = ± d/2) Interesting limit far away along the positive x- axis ( x >> d ) E = E + + E ! = 1 4 "# q r + 2 + 1 4 "# ! q r ! 2 E = q 4 !" 1 x # 1 2 d ( ) 2 # 1 x + 1 2 d ( ) 2 \$ % & & ' ( ) ) September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 5 Electric Field from an Electric Dipole We can rewrite our result as We can use a binomial expansion or a Taylor expansion So we can write E ≈ q 4 πε x 2 1 + d x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 1 − d x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = q 4 πε x 2 2 d x ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = qd 2 πε x 3 E = q 4 πε x 2 1 − d 2 x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 2 − 1 + d 2 x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ f ( z ) = (1 ± z ) n f ( z ) = f ( n ) ( z = 0) n ! n = ∞ ∑ ( z − 0) n α = d 2 x , 1 ± α ( ) n ≈ 1 ± n α + ... (1 −α ) − 2 = 1 − ( − 2) α ... = 1 + 2 d 2 x + ... (1 + α ) − 2 = 1 + ( − 2) α ... = 1 − 2 d 2 x + ... For small z=d/2x, stop aer n=1 September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 6 Hint for Homework Problem Exact dipole eld along x: Approximated dipole eld along x for x>>d: Relative error: E approx = qd 2 πε x 3 E ex =...
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## This note was uploaded on 02/03/2012 for the course MTH 235 taught by Professor Staff during the Fall '11 term at Michigan State University.

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lecture05 - September 8 2011 Physics for Scientists&...

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