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problemset3

problemset3 - 6.641 Electromagnetic Fields Forces and...

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6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 Problem Set 3 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 3.1 A r - r r + x (x,y,z) z +q d 2 d 2 -q Figure 1: Addition of potential contributions from 2 point charges that form an electric dipole. (Image by MIT OpenCourseWare.) We can simply add the potential contributions of each point charge: q q Φ = 4 πε 0 r + 4 πε 0 r d r + = x 2 + y 2 + ( z ) 2 2 d r = x 2 + y 2 + ( z + ) 2 2 q 1 1 Φ = ± ± 4 π 0 x 2 + y 2 + ( z d 2 ) 2 x 2 + y 2 + ( z + d 2 ) 2 B p = qd . must make some approximations. As r → ∞ , ± r + , ± r , and ± r become nearly parallel. Thus, d r + r a = r cos θ 2 d r + r (1 cos θ ) 2 r 1
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± C Problem Set 3 6.641, Spring 2009 r - r r + x z θ a= cos θ d 2 Figure 2: Differences in lengths between r + , r , and r (Image by MIT OpenCourseWare.) Similarly, d r r (1 + 2 r cos θ ) ± By part (a): Φ = q 4 πε 0 1 r + 1 r . If | x | 1, then 1 1+ x 1 x d | 2 r cos θ | 1 so ² ³ 1 1 1 1 d r + r 1 d 2 r cos θ r 1 + 2 r cos θ ² ³ 1 1 1 1 d r r 1 + d 2 r cos θ r 1 2 r cos θ 1 1 1 d d r + r r r cos θ = r 2 cos θ qd cos θ p cos θ Φ 4 πε 0 r 2 = 4 πε 0 r 2 , p = qd dipole moment E Φ i r 1 Φ i θ 1 Φ i φ = −± Φ = ˆ ˆ ˆ ∂r r ∂θ r sin θ ∂φ Φ p cos θ Φ p sin θ = 2 πε 0 r 3 ; ∂θ = 4 πε 0 r 2 Φ = 0 E = p cos θ ˆ i r + 1 p sin θ ˆ i θ 2 πε 0 r 3 r 4 πε 0 r 2 −→ E = p 2cos θ ˆ i r + sin θ ˆ i θ 4 πε 0 r 3 2
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1 Problem Set 3 6.641, Spring 2009 D dr E r 2cos θ = = = 2cot θ rdθ E θ sin θ dr = 2 cot θdθ r 1 dr = 2 cot θdθ r ln r = 2 ln(sin θ ) + k r = C sin 2 θ ; when θ = π ,r = C = r 0 2 Thus, C = r 0 , r = sin 2 θ r 0 Figure 3: The equi-potential (dashed) and feld lines (solid) ±or a point electric dipole calibrated ±or 4 πε 0 /p = 100. The equi-potential lines and the electric feld lines are perpendicular to each other. 3
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Problem Set 3 6.641, Spring 2009 Plots of Equipotential and Field Lines -2 -1 1 2 0.75 0.5 0.25 -0.25 -0.5 -0.75 r 0 = 1 r 0 = .5 r 0 = 0.25 r 0 = 2 θ E Field Lines Figure 4: Polar plot of dipole electric ±eld lines r 0 sin 2 θ for 0 θ π and for r 0 = 0.25, 0.5, 1, and 2 meters with 4 πε 0 = 100 volt 1 -m 2 (Image by MIT OpenCourseWare.) p -1 1 Φ = .01 Φ = 0.04 Φ = 0 Φ = .0025 Φ = .16 -0.5 0.5 1 2 -2 -1 Equipotential Lines Figure 5: Polar plot of equipotential lines Φ = p cos θ for 0 θ π, Φ = 0 , ± 0 . 0025 , ± 0 . 01 , ± 0 . 04 , ± 0 . 16 , 4 πε 0 r 2 and ± 0 . 64 volts
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problemset3 - 6.641 Electromagnetic Fields Forces and...

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