Electromagnetism Answers to Problem Set 3 Spring Jackson Prob. 2.1: Charge above a grounded plane. (z d)

Transcription

1 Electromagnetism 76 Answers to Problem Set 3 Spring 6. Jackson Prob..: Charge above a groune plane (a) Surface charge ensity E z (ρ, z) = φ z = q [ (z ) [ρ + (z ) (z + ) 3/ [ρ + (z + ) 3/ Evaluate E z at z = an multiply by ɛ to fin σ. σ(ρ) = q π [ρ + 3/ Plot of σ with = an q = π (Actually we show -σ) (b) The irection of the force on the plane is along z axis. Its magnitue is F z = πρρ qσ(ρ) ρ + ρ + = q ρρ (ρ + ) 3 = q 4. This is precisely the force on the image charge preicte by Coulomb s law. (c) The force obtaine by integrating σ /(ɛ ) is F z = ɛ q (π) πρρ (ρ + ) 3 = q () Work one to move charge to infinity. q (e) Potential energy of charge an image. z 4z = q 4 V = q. ρρ (ρ + ) 3 = q 4.

2 This is twice the value naively expecte. The work-energy relation for a pair of charges iscusse in Sec.. assume that one charge was fixe! In this case the both charges move as work is one on the system. (f) W for an electron at = Å from surface.. Jackson Prob.: q 4 ( ) = 4 π = J = 3.6eV (a) Assuming that the charge is on the z axis at istance from the origin, the potential at points insie the sphere is Φ(r, θ) = { q r r cos θ + q }, r r cos θ + where q = qa/ an = a /. (b) Inuce charge ensity: First, we etermine the raial electric fiel E r = Φ r = { q(r cos θ) [r r cos θ + q (r } cos θ) 3/ [r r cos θ + 3/ The surface charge ensity σ is ɛ E r, evaluate at r = a. (In this case, the normal points inwar!). After simplification, this becomes σ(θ) = q /a 4πa [ (/a) cos θ + /a 3/ (c) Magnitue an irection of force on q: F z = q πa = q a(a ) = q a (a ). σ(µ) (aµ ) µ [a aµ + 3/ (aµ ) µ [a aµ + 3 The force on q is in the +z irection. exerte on q by the image charge q. This is precisely the force

3 () Changes in the solution: i. Sphere at a fixe potential V. In this case, Φ(r, θ) Φ(r, θ) + V, av r, r < a r a There is an aitional uniformly istribute charge Q = q + V on the sphere. Thus σ σ + q + V 4πa. The uniformly istribute charge exerts no aitional force of q since (aµ ) µ [a aµ + = 3/ ii. Sphere has a fixe charge Q. In this case, an aitional charge Q + q is again uniformly istribute over the surface. Therefore 3. Jackson Prob.5: Φ(r, θ) Φ(r, θ) + Q + q, a r < a Q + q, r r a The aitional uniformly istribute charge is σ σ + Q + q 4πa, an, as above, there is no ae force on q. (a) Quasistatic force neee to balance charge q above a groune sphere. F y = q ay y a Work one to remove charge to infinity q F y y = 8πɛ a a = qq 8πɛ This is / of the (negative of) the potential energy of the charge an its image. Here again the image is not fixe as the charge moves out, so the work-energy theorem, in its usual form, is not vali. 3

4 (b) Quasistatic force neee to balance charge q above an isolate sphere carrying charge Q. F y = q [ Q y qa3 (y a ) y 3 (y a ) Work neee charge to remove charge to infinity = F y y = [ q(q + q ) + [ Qq q a + q a a qq ( ) The first term is the negative of the potential energy of the ae charge Q + q an the charge q. This term has the correct sign as the uniformly istribute ae charge charge is effectively at the origin. The secon term is the negative of the charge-image potential with the factor of / associate with the fact that the image moves along with the original charge. Aenum on the work-energy theorem: In the plane an spherical image problems worke out above, we foun that the work neee to bring the charge q in from infinity was / the potential energy of the charge an its image. To explain this factor /, let us examine the general expression for energy of a charge istribution 8πɛ ρ(r )ρ(r ) r r τ τ = Φ(r)ρ(r) τ For two fixe charges, q an q i, we have Φ = Φ q + Φ i an ρ = qδ(r r q ) + q i δ(r r i ). Here, Φ q (r) = q r r q, with a similar expression for Φ i. One fins q Φ q(r q ) + q Φ i(r q ) + q i Φ q (r i ) + q i Φ i (r i ) The first an fourth terms are (infinite) self-energy terms an must be exclue from the sum. The secon an thir terms have ientical values an lea to the well-known expression for the interaction energy between two charges qq i r i r q, For a charge q an a surface istribution σ such as we have in the present case, the energy expression becomes q Φ i(r q ) + q (r)σ(r)a + SΦ Φ i (r)σ(r)a, S 4

5 where we have omitte the self-energy of q. Since the two contributions to the potential Φ q an Φ i precisely cancel on the surface, the secon an thir terms above cancel an we are left with q Φ i(r q ) = qq i 8πɛ r i r q. This is, as expecte, just / of the charge-image interaction energy. 5