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Electricity and Magnetism
Mathematical Tools and Fields
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

Line Integrals: Work as the Integral of F · ds

How does the dot product F · ds pick out just the part of a force that does work along a path?


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Your summary note

    1. 1

      The dot product F⋅ds=∣F∣ ∣ds∣cos⁡θ\mathbf{F} \cdot d\mathbf{s} = |\mathbf{F}|\,|d\mathbf{s}|\cos\thetaF⋅ds=∣F∣∣ds∣cosθ

      Write the definition as the work element dW=F∥ ∣ds∣dW = F_{\parallel}\,|d\mathbf{s}|dW=F∥​∣ds∣, sketch the force, displacement and angle θ\thetaθ, and record the Cartesian form Fx dx+Fy dy+Fz dzF_x\,dx + F_y\,dy + F_z\,dzFx​dx+Fy​dy+Fz​dz.

    2. 2

      The polar form Fr dr+Fθ r dθF_r\,dr + F_\theta\,r\,d\thetaFr​dr+Fθ​rdθ

      Sketch the radial and angular components of both F\mathbf{F}F and dsd\mathbf{s}ds, derive the arc length r dθr\,d\thetardθ for the tangential step, and state that a central force has Fθ=0F_\theta = 0Fθ​=0.

    3. 3

      Reducing U(b)−U(a)=−∫abF⋅dsU(b) - U(a) = -\int_a^b \mathbf{F} \cdot d\mathbf{s}U(b)−U(a)=−∫ab​F⋅ds to one ordinary integral

      Write the general definition, then reduce it to −∫abFx dx-\int_a^b F_x\,dx−∫ab​Fx​dx when dy=dz=0dy = dz = 0dy=dz=0 and to −∫abFr dr-\int_a^b F_r\,dr−∫ab​Fr​dr when Fθ=0F_\theta = 0Fθ​=0, noting which point sits in each limit.

    4. 4

      Worked example: repulsive inverse-square force, U(r)=k/rU(r) = k/rU(r)=k/r

      Integrate −∫∞rk r−2 dr-\int_\infty^r k\,r^{-2}\,dr−∫∞r​kr−2dr with U(∞)=0U(\infty) = 0U(∞)=0, check that the positive sign matches pushing in against repulsion, and write the analogue V(b)−V(a)=−∫abEr drV(b) - V(a) = -\int_a^b E_r\,drV(b)−V(a)=−∫ab​Er​dr in volts.

    Attempt 1 of 2