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01Series Approximations: Keeping One or Two Terms02Integrals as Sums: Along a Line and a Curved Path03Line Integrals: Work as the Integral of F · ds04Closed Line Integrals: Work Around a Loop05Double Integrals: Area, Charge and Volume06Closed Surface Integrals: The Sphere and Its Charge07Electric Flux Through Surfaces and Volume IntegralsProblem set0/10Problem set 20/10MIT problem set0/4
01Field Theory Versus Action at a Distance02Scalar Fields: Contour, Color and Relief Maps03Vector Fields, Sources, Sinks and Divergence04Curl: Measuring the Circulation of a Field05Gravitational, Electric and Magnetic Fields06Drawing Vector Fields: Arrows, Lines and Grass SeedsProblem set0/10Problem set 20/10MIT problem set0/1
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01Charge Density and the Field on a Rod's Axis02The Field on a Charged Rod's Perpendicular Bisector03The Fields of a Charged Ring and a Charged Disk04How Field Lines Carry Force, Energy and Momentum05Electric Pressure and Tension: The Stress ε₀E²/2Problem set0/10Problem set 20/10MIT problem set0/6

The Field on a Charged Rod's Perpendicular Bisector

Why does symmetry cancel the sideways field beside a charged rod, and how does a tan substitution finish the integral?


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    1. 1

      Symmetry cancels ExE_xEx​, leaving dEy=dEcos⁡θ′dE_y = dE\cos\theta'dEy​=dEcosθ′

      Write dq=λ dx′dq = \lambda\, dx'dq=λdx′ and dEdEdE at distance x′2+y2\sqrt{x'^2 + y^2}x′2+y2​, sketch paired elements at ±x′\pm x'±x′ with their contributions at PPP, and reduce dEydE_ydEy​ to 14πε0λy dx′(x′2+y2)3/2\frac{1}{4\pi\varepsilon_0}\frac{\lambda y\, dx'}{(x'^2 + y^2)^{3/2}}4πε0​1​(x′2+y2)3/2λydx′​.

    2. 2

      The substitution x′=ytan⁡θ′x' = y\tan\theta'x′=ytanθ′ and the result

      Integrate dEydE_ydEy​ from −ℓ/2-\ell/2−ℓ/2 to ℓ/2\ell/2ℓ/2, carry the substitution and new limits ±θ\pm\theta±θ through to 2sin⁡θy2\frac{2\sin\theta}{y^2}y22sinθ​, and write Ey=14πε02λsin⁡θyE_y = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta}{y}Ey​=4πε0​1​y2λsinθ​ in terms of ℓ\ellℓ and yyy.

    3. 3

      The limits y≫ℓy \gg \elly≫ℓ and ℓ≫y\ell \gg yℓ≫y

      Take each limit of the result, showing y≫ℓy \gg \elly≫ℓ gives the point-charge field Q4πε0y2\frac{Q}{4\pi\varepsilon_0 y^2}4πε0​y2Q​ and ℓ≫y\ell \gg yℓ≫y gives the infinite-line field 2λ4πε0y\frac{2\lambda}{4\pi\varepsilon_0 y}4πε0​y2λ​, with units checked.

    Attempt 1 of 2