Ben Palmer 01372065 Due Thursday, February 1, 11:59pm PHYS 2220 Spring 2024- Dr. De Grandi Homework 3 - Part 2 Instructions: To guarantee full credit show all your steps, if you are using some additional equations that are not the ones we discussed in class (not on the formula sheet) you need to be able to derive them. If you have any questions post them on the Homework 3- Questions discussion board or come to Study-hall. Problem 1: Electric field of a flat ring with considerable thickness In class, in Lecture 5, we discussed how to calculate the Electric field of a charged ring with negligible thickness (i.e. we considered it one-dimensional). In this problem instead, you’ll calculate the case of a charged ring that has a non-negligible thickness; we’ll call a the inner radius of the ring, and b the outer radius of the ring (see the figure below). The ring has an overall charge Q and is flat, therefore can be described as a two-dimensional object. a) Find an expression for the surface charge density of the charged ring. Make sure your answer is in terms of the given quantities only (a, b, Q and any given constant like k, ϵ0 , π, e, etc.). O Etat Area of Circle is b) Review how in Lecture 5 (find slides and notes under Modules: Week 3), we calculated the Electric field created by a charged disk at a point P at a distance z from the center of the disk, along an axis perpendicular to the plane of the disk. No work is needed to be shown here, just make sure you are comfortable with the calculation for the disk before moving to the next part. z JKEE.gs air d rig c) Prove that magnitude of the electric field due to the ring (see figure below) at a point P at a distance z from the center of the ring, along an axis perpendicular to the plane of the ring is given by: ! " 2Q z z √ √ EP = Edisk (z) = k 2 − b − a2 z 2 + a2 z 2 + b2 Show all the steps of your calculation. Note: you want to make sure to replace the surface charge density from part a) in terms of the given quantities (σ is NOT a given quantity in this problem), the given quantities are: a, b, Q, z and any given constant like k, ϵ0 , π, e, etc. 1 an ÉÉ o g fGtrzard EEEEII I.gg if FACEFEE zt xourh _K East d) Show that in the limit of z → ∞ the Electric field of the ring reduces to Electric field of a point charge with magnitude Q E = k 2. z Hints: you need to use the following Taylor series1 : √ 1 1 ≈1− x n r 2 1+x I where x is a small quantity (in your case x will be either az or zb which are small quantities when z → ∞). EE EEEEFiJ X HE C HEALEY IIC 1 5 13 1 More specifically called a Maclaurin series, since the expansion is around 0. i Problem 2: Who is right? Hints! Review the content of Lecture 5 that you can find under Modules: Week 3. Two parallel, infinite planes of charge have charge densities +2σ and −σ as shown in Figure 1. They are at a distance L0 from each other. Three students are asked to determine the electric field at points A and B, at a distance L above the top plane, and at a distance L below the bottom plane, respectively, as shown in the figure. Each student comes up with a different answer and a different explanation. Read carefully their statements below. The students are reminded that the electric field magnitude of an infinite plane of charge with charge density σ is E = 2ϵσ0 . Figure 1: Two equivalent views of the two infinite parallel planes • Student 1 draws the sketch in Fig. 2 [top] (on the next page) and says: “Not sure how to explain it but I think from the outside we can see it as a single infinite plane with charge density +2σ − σ = σ, so then we know the electric field must be σ/(2ϵ0 ) pointing upwards at point A and also σ/(2ϵ0 ) pointing downwards at B”. • Student 2 draws the sketch in Fig. 2 [middle] (on the next page) and says “I don’t agree with you, I think you need to consider the effect of each single plane, so at point A there will be a field upwards due to the top plane with magnitude 2σ/(2ϵ0 ) = σ/ϵ0 , and at point B there will be a field upwards due to the bottom plane with magnitude σ/(2ϵ0 )”. • Student 3 draws the sketch in Fig. 2 [bottom] (on the next page) and says “Do we just add the electric fields together? I’m going to add up the field from each plane at both points. So at A I get a field upward of magnitude σ/ϵ0 + σ/(2ϵ0 ) = 3σ/(2ϵ0 ), and at point B I get the same 3σ/(2ϵ0 ) but now pointing downward”. Evaluate each explanation and decide whether it is correct or not. If an explanation contains incorrect reasoning, explain how the argument is flawed. Figure 2: Students’ sketches. Student use I of the equations they gain to ogether .making his Student 2 correct thy fail Student S the incorrect is is field lines has drawn opposite charges acknowledge are that two right Egiolds the in can cam inaccurate because while their Egield lines are A B affected the field lines incorrect to recognize while they that are namely by both planes by failing to ragman Problem 3: Show your steps for Problem 8: Part A of Homework Part 1 After you have finished Problem 8 of Part 1 (on MasteringPhysics) The Trajectory of a Charge in an Electric Field, show here the steps of your work for Part A (not Part B): make a clear sketch of the situation including an x-y plane, list what are the given and wanted quantities, show the steps and reasoning you took to get to the final answer. É II oh o Velay V 197 kinematics equation to feel to she sort Vt E MII E É Start with apiary E MIÉELE F min qE mm a ten Eing get a solvable