Physics 1B Spring 2015 Discussion Solutions - Physics 1B...

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Physics 1B: Oscillations, Waves, Electric and Magnetic FieldsSpring 2015Discussion: Week 4Agnieszka WergielukGraduate TA[email protected]Theoretical IntroductionThe force between two charges,q1andq2, separated by a distancer, is given by the Coulomb law:F=14π0q1q2r2ˆr.(1)The principle of superposition tells us that if we have many chargesq1,q2,. . .,qnthe force on thei-th of them isgiven byFi=qi4π0nXk=1,k6=iqkr2kiˆrki,(2)whererki=ri-rk. For continuous distributions of charge the above equation becomesF=q4π0Zdr0ρ(r0)(r-r0)2r-r0|r-r0|.(3)Example: Force on a point chargeq0located at pointPon the axisof a uniformly charged ring of total chargeQWe start by writing down the linear charge density:λ=Q2πa.(4)Let us consider the forcedFat pointP(which is some distancexfrom the center of the ring) due to an element ofchargedq:
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Agnieszka Wergieluk, Physics 1B Discussion: Week 4 - Spring 2015The distancerbetween the pointPand the element of chargedqis given byr=px2+a2.(5)The element of charge is nothing else thandq=λdl .(6)Therefore the force exerted by by the element of chargedqis given bydF=q04π0dqr2=q04π0λdlx2+a2(7)At this point we consider the symmetry of the problem and realize that they-component of the force produced bydqwill be canceled by they-component of the force produced by corresponding elementdqat the opposite side of thering. Therefore we only need to add up (i.e, integrate!) thex-components of all the force elements produced by allelements of charge. Now, thex-component of the force produced bydqis justdFx=dFcosθ=q04π0λdlx2+a2cosθ .(8)A quick look at the picture convinces us thatxandθare not independent variables (the biggerxis, the smallerθbecomes, and vice versa). It is customary (and in many cases alsonecessary) to express the integrand by means ofone these variables only. Easily,cosθ=xx2+a2.(9)Then we havedFx=q04π0λx dl(x2+a2)3/2.(10)We are ready to integrate! This time we’re lucky: no quantity on the right hand side depends ondland thereforethe integral over this variable yields just 2πa. ThusF=Fx=q04π0λx2πa(x2+a2)3/2.(11)Finally, we use the fact thatλ=Q/2πato finally obtainF=14π0q0Qx(x2+a2)3/2.(12)Note that in the limitxathis becomesF14π0q0Qx(x2)3/2=14π0q0Qx2.(13)What does it mean? It means that from a great distance a hoop uniformly charged with a chargeQlooks (andacts)almost like a point charge of magnitudeQ.
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Agnieszka Wergieluk, Physics 1B Discussion: Week 4 - Spring 2015Test questions

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