# Lecture 13 - Linking E, ϕ, and ρ

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1 Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on the inner surfce of the shell is -q. Is the surfce chrge density negtive over the entire inner surfce? Or cn it be positive on the fr side of the inner surfce if the point chrge q is close enough to the shell so tht it ttrcts enough negtive chrge to the ner side? Justify your nswer. Hint: Think bout field lines. If there were loction with positive density, then electric field lines would strt there, pointing wy from it into the sphericl cvity. But where could these field lines end? They cn t end t infinity, becuse tht s outside the shell. And they cn t end t point in empty spce, becuse tht would violte Guss s lw; there would be nonzero flu into region tht contins no chrge. They lso cn t end on the positive point chrge q, becuse the field lines point outwrd from q. And finlly they cn t end on the shell, becuse tht would imply nonzero line integrl of E (nd hence nonzero potentil difference) between two points on the shell. But we know tht ll points on the conducting shell re t the sme potentil. Therefore, such field line (pointing inwrd from the inner surfce) cn t eist. So ll of the inner surfce chrge must be negtive. Every field line inside the cvity strts t the point chrge q nd ends on the shell. A 30,000 Foot View (From Lst Time) It is importnt to relize tht lthough we hve been focusing specific chrge distributions from points, lines, sheets, nd spheres, electricity is one of the most prevlent nd influentil forces tht we encounter in our dily lives! This remrkble YouTube video highlights some mzing electricity tricks tht you cn esily test out for yourself!

2 Lecture nb Problems Complementry Section: The Missing Link By now, we re fmilir with the three fundmentl quntities in electrosttics, nmely the: Electric field (E) - The electric field E[r ] t point r equls the force per unit chrge tht point chrge would feel t r. Electric potentil (ϕ) - The potentil ϕ[r ] equls the mount of work tht must be done by n eternl gent in crrying unit of positive chrge from the reference point (usully infinity) to r without ny ccelertion. Chrge density (ρ) - The chrge distribution everywhere. These three quntities re connected by the following reltions. ρ ϕ ϕ= k dq r ϕ=- ρ ϵ 0 E= ρ ϵ 0, E=0 E=- ϕ ϕ=- E ds E= k dq r r E The only reltion tht we re not intimtely fmilir with is ϕ = - ρ ϵ 0 (1) which rises from combining E = - ϕ together with ρ = E = - ϕ - ϕ where we hve defined the ϵ 0 Lplcin ϕ = ϕ + ϕ y + ϕ z () in Crtesin coordintes. Surprisingly, Eqution (1) turns out to be one of the most useful reltions in the digrm bove becuse the Lplcin hs mny wonderful properties. For emple, s discussed in clss Eqution (1) in free spce (ρ = 0) becomes ϕ = 0 (3) whose solution ϕ[r ] equls the verge vlue of ϕ in smll sphere round r. This implies tht ϕ hs no minim or mim, nd therefore tht you cnnot construct n electrosttic field tht will hold chrged prticle in stble equilibrium. In more dvnces physics courses, this will be the primry problem solving tool in mny electrosttics problems. Complementry Section: Two Concentric Shells Tringulr E Emple Find the chrge density ρ nd potentil φ ssocited with the electric field shown in the figure below. E is independent of y nd z. Assume tht φ = 0 t = 0.

3 Lecture nb 3 Since E is independent of y nd z, we should integrte ϕ = - E[] d s long the -is. Note tht becuse the chrge distribution goes out to infinity, we don t set the zero potentil t infinity (which would not be well defined), but insted use = 0. At distnce 0 < <, ϕ = - 0 E[ ] d = - 0 E d = -E 0 - = = E0 - (4) =0 For distnces >, note tht E[] = 0 for > so tht ϕ[] = ϕ[] for this rnge. For < 0, we do similr clcultion to Eqution (4) bove, ϕ = - 0 E[ ] d = - 0 E d = -E 0 + = = -E0 + (5) =0 In summry, ϕ = E 0 < - -E < 0 E 0-0 < - E 0 The chrge density ρ = - E = - E. This is just strightforwrd derivtive which equls 0 < - ϵ 0 E 0 - < 0 ρ = - ϵ 0 E 0 0 < 0 This form of ρ shows us tht the chrge distribution is two infinity slbs with thickness nd opposite chrge densities ± ϵ 0 E 0 tht touch t the = 0 plne. With this, we hve complete picture of the setup. (6) (7) ϵ 0 E 0 ρ E 0 E ϕ E As double check, t = 0 the two infinite slbs ct effectively like sheets with chrge densities ±σ = ±ρ. They

4 4 Lecture nb effectively chrge They crete field pointing to the right with mgnitude σ, so the totl field is ρ = ρ. Since we found tht ϵ 0 ϵ 0 ϵ 0 ρ = ϵ 0 E 0, the field equls E 0, in greement with the given vlue. Stisfying Lplce Emple Does the function f[, y, z] = + y stisfy Lplce s eqution? Does the function g[, y, z] = - y? Suppose g[, y, z] represented n electric potentil nd tht g[, y, z] stisfies Lplce's eqution. Wht cn you sy bout the grdient g? Here is plot of f[, y, z] nd g[, y, z]. In Crtesin coordintes, the Lplcin =, does not stisfy Lplce s eqution while y, z. Therefore, f [, y, z] = + 0 (8) g[, y, z] = - = 0 (9) does stisfy Lplce s eqution. If g[, y, z] were n electrosttic potentil, it would correspond to the electric field which is shown below for constnt z, - g[, y, z] =, - y, 0 (10)

5 Lecture nb 5 6,-y,0 4 y Becuse the Lplcin equls zero (or equivlently, the divergence of the grdient of g[, y, z] equls zero), there is zero net flu out of ny closed volume. You should convince yourself tht this is true of the figure bove. Complementry Section: Grounding Shell Emple A conducting sphericl shell hs chrge Q nd rdius R 1. A lrger concentric conducting sphericl shell hs chrge -Q nd rdius R. Wht is the potentil t ll points in spce? If the outer shell is grounded, eplin why nothing hppens to the chrge on it. If insted the inner shell is grounded, find its finl chrge. Before either sphere is grounded, the electric field outside both sphericl shells is zero, nd hence the outer sphere hs the sme potentil s infinity. This implies tht when we ground the outer sphere, nothing will hppen. If some negtive chrge did flow off, then there would be net positive chrge on the two shells, nd hence n outwrdpointing field for r > R, which would drg the negtive chrge bck onto the outer shell. Similrly, if some positive chrge flowed off, then there would be n inwrd-pointing field for r > R which would drg the positive chrge bck onto the shell. If the inner shell is grounded, it must end up with the mount of chrge tht mkes its potentil equl to the potentil t infinity. Whtever chrge distribution ultimtely results must stisfy E = 0 inside of both sphericl conductors. Hence we know tht there must be no chrge on the inner surfce of the smller sphere, nd ll of the chrge Q 1 on the smller sphere must reside on its outer surfce. The outer sphere must therefore hve chrge -Q 1 on its inner surfce (to stisfy E = 0 inside), nd hence the remining chrge Q 1 - Q must be eqully distributed on the outer surfce of this lrger sphere. We cn now clculte the potentil everywhere. First, the potentil ϕ[r ] t the surfce of the outer sphere must equl ϕ[r ] = k(q 1-Q) R (11) since this is identicl to the potentil from point chrge Q 1 - Q t the origin (using superposition). Net, the

6 6 Lecture nb potentil point chrge origin (using superposition). potentil difference ϕ[r 1 ] - ϕ[r ] between the inner sphere nd the outer sphere equls ϕ[r 1 ] - ϕ[r ] = k Q 1 R 1 - k Q 1 R (1) which is identicl to the potentil difference from point chrge Q 1 t the origin (by superposition nd the fct tht the chrge Q 1 - Q on the outer surfce of the outer sphere genertes no electric field for r < R ). Therefore the potentil ϕ[r 1 ] on the surfce of the inner sphere, which must be zero t equilibrium, is given by 0 = ϕ[r 1 ] = k(q 1-Q) R + k Q 1 R 1 - k Q 1 R (13) which we cn solve to obtin Q 1 = R 1 R Q (14) Intuitively, if none of the chrge leves (so Q 1 = Q), then the inner shell is t higher potentil thn the outer shell, which in turn is t the sme potentil s infinity in this cse. On the other hnd, if ll of the chrge leves (so Q 1 = 0), then the inner shell is t the sme potentil s the outer shell, which in turn is t lower potentil thn infinity in this cse. So, by continuity, there must be vlue of Q 1 tht mkes the potentil of the inner shell equl to the potentil t infinity. Recommended Problems This is list of ecellent problems (with solutions) in Dvid Morin s book Why leve? Advnced Section: Imge Chrges Mthemtic Initiliztion

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