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=x√x2+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 becomesF≈14π0q0Qx(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.