INTRO:To practice Problem-Solving Strategy 25.2 for constant charge distribution problems.A directly rod of size L has a positive charge Q distributed along that is length. Find the electrical potential as result of the rod at a point located a street d from one finish of the rod along the line prolonging from the rod. (Figure 1)
 Figure 1
PROBLEM-SOLVING STRATEGY 25.2 The electrical potential of a constant distribution the chargeMODEL: model the charges together a an easy shape, such as a line or a disk. I think the fee is uniformly distributed.VISUALIZE: because that the photographic representation:Draw a photo and create a name: coordinates system.Identify the suggest P in ~ which you want to calculation the electric potential.Divide the complete charge Q into small pieces of fee ΔQ using forms for i m sorry you currently know how to recognize V. This division is often, however not always, into allude charges.Identify distances that need to be calculated.SOLVE: The mathematical depiction is V=ΣVi.Use superposition to form an algebraic expression because that the potential at P.Let the (x,y,z) works with remain as variables.Replace the tiny charge ΔQ v an tantamount expression involving a charge density and also a coordinate, such together dx, that explains the shape of fee ΔQ. This is the an important step in making the transition from a sum to one integral since you need a name: coordinates to offer as an integration variable.Express all ranges in terms of the coordinates.Let the sum end up being an integral. The integration will more than the coordinate variable that is pertained to ΔQ. The integration limits for this variable will depend on the coordinate mechanism you have chosen. Lug out the integration, and also simplify the result.ASSESS: examine that your result is continual with any type of limits because that which you recognize what the potential must be.ModelNo info is provided on the rod"s overcome section. For simplicity, assume that its diameter is much smaller 보다 the rod"s length. Friend can, then, version the rod together a line of charge.Visualize-----------------------------------------------------------------------------------------------------PART A:To most successfully solve this problem, you have to divide the rod right into pieces of fee that consists of...- thin lines of charge of length L and a very small cross section- thin "slices" that the rod reduced parallel come the axis of the rod- thin "slices" that the rod cut perpendicular come the axis of the rodSOLUTION:First, version the rod as a heat of charge. Then, divide the line into many small segments, every of size Δx. Each segment can, then, it is in modeled as a suggest charge. The potential as result of such a allude charge have the right to be established in a fairly straightforward manner.∴ option #3, thin "slices"of the rod reduced perpendicular come the axis that the rod-----------------------------------------------------------------------------------------------------PART B:Choose a coordinate system where +x is to the right and +y is upward. Place the beginning of your coordinate device at the left end of the rod, and also choose suggest P to it is in located past the right end of the rod, as presented in the snapshot to the left.What is the distance ri between point P and also a item of charge situated at place xi?Express your answer in regards to some or every one of the amounts xi, L, Q, and d.

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ri= (L+d)-xi-----------------------------------------------------------------------------------------------------PART C:Find the electric potential VPat point P.Express her answer in terms of d, L, Q, and also ϵ0.
SOLUTION:The math expression for the potential is V=ΣVi.Because that the nature the the problem, which deals with a continuous distribution that charge quite than a point-like distribution, the sum need to be treated together an integral.To collection up the summation and find what come integrate, monitor the steps detailed in the strategy ^^After splitting the full charge into tiny segments, and also modeling every segment together a allude charge, as in component A, build a mathematical expression for Vi, the electric potential as result of segment i.Treat xi, the x name: coordinates of segment i, together a change (just call it x), and also express the fee on the segment in terms of said x.Finally, let ΣVi come to be an integral and use calculus to evaluate the integral...FIND... The electric potential by means of at point P due to the charge segment located at the variable positionxThe formula because that the electric potential the a suggest charge involves the street from the fee to the suggest where the potential is to it is in computed.Recall that, in part B, you acquired an expression relating the distance between allude P and also charge segment i to the x name: coordinates of the fee segment.......The formula for electric potential is...Uelec= k q1q2/rand we all know what k is so that will certainly be temporarily ignored...The instant distance between the 2 points is ri, as figured out in part B...U = k q1q2/r ... Then simply plug in because that r asri ...Vi= k ⋅ΔQ/(L+d-xi)Next you require to know the expression because that the fee of a little segmentSo calculus is kind of annoying at first, however these topics room actually really vital to understand. Especially after that (trust me)..Because girlfriend will combine with respect to coordinate x, it is necessary to express the charge on a segment, ΔQ, in regards to its size Δx.So girlfriend gotta uncover a mathematics expression because that ΔQ, if assuming the charge is uniformly distributed along the stick (since that"s given).Recall that the complete charge ~ above the pole is Q and also the rod"s size is L...that means that the adjust in charge over the readjust in x is simply the charge split by the length, or...ΔQ/Δx = Q/Lby rearranging this to deal with for ΔQ...
plug that into your initial Viequation because that ΔQ,Vi= k ⋅Δx⋅Q/L ⋅/(L+d-xi)you now have an integrable function..Vi= ∫k⋅Q/L⋅(L+d-xi) Δx = ∫k⋅Q/L⋅(L+d-x) dxintegrate from 0 come L, since the d is currently attributed...

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k is constant, Q is constant, l is constant, so you can just pull that out...→k⋅Q/L ⋅∫1/(L+d-x) dxuse substitution...
→k⋅Q/L⋅∫1/(u) -du→k⋅Q/L⋅-ln(u) (from 0 come L)usub→k⋅Q/L⋅-ln(L+d-x) (from 0 to L)→k⋅Q/L⋅<-ln(L+d-L) - -ln(L+d-0)>→k⋅Q/L⋅<-ln(d) + ln(L+d)>using legislation of logs, subtracting them is the exact same as splitting them...→k⋅Q/L⋅ln<(L+d)/d> = k⋅Q/L⋅ln
-----------------------------------------------------------------------------------------------------PART D:Imagine that distance d is much higher than the length of the rod. Intuitively, the potential need to be around the same as the potential in ~ a street d indigenous which of the following charge distributions?- an infinitely lengthy wire with full charge Q- one infinitely lengthy wire with complete charge Qd/L- a allude charge of size Q- an electric dipole with minute QLSOLUTION:From much away, a brief rod looks an extremely much favor a an easy point charge. No surprisingly, the mathematical expression you acquired for the potential does reduce to the of a point charge if L/d≪1:In other words, since L is constant, the larger d is, the smaller sized L/d gets. When L/d is really small, the logarithm is basically simply the logarithm that 1 plus the tiny little amount. The ln(1) = 0, and oddly enough, at yes, really close proximity come 1, those worths follow the ascendancy :
if x try this with x = 0.001, because that example... Ln(0.001) = 0.0009995 ≈ 0.001therefore, Q/4Lπε0⋅ln<1+ L/d> where L/d 0⋅L/d⇒ or, ≈ Q/4dπε0Therefore, the third option is correct. Or,