Department of Physics IIT Kanpur, Semester II,

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1 Department of Physics IIT Kanpur, Semester II, 7-8 PHYA: Physics II Solution # 4 Instructors: AKJ & SC Solution 4.: Force with image charges (Griffiths rd ed. Prob.6 As far as force is concerned, this problem is the same if we remove the grounded conducting plate and simply put an image charge +q at z d and an image charge q at z d. Therefore, the force on the charge +q is given by F q 4πɛ (d + q (4d + q ] (6d ẑ q 4πɛ d ] ẑ q 9 6 4πɛ d 7ẑ Solution 4.: Infinite-line image charge (Griffiths rd ed. Prob.9 This problem is an extension of the problem in which a point charge is placed above an infinite grounded conducting plane. So, in order to obtain the correct potential, we need to put an infinite line charge λ running parallel to the x-axis at a distance d from the x y plane. (see Fig.. FIG. : (a Let s first find the potential at a point (y, z on the y z plane. Assume that the distance of the point (y, z from the top infinite line charge is given by the magnitude of the vector ŝ + ; similarly the distance of the point (y, z from the bottom infinite line charge is given by the magnitude of the vector ŝ. Taking the potential V at x y plane to be the reference, the potential due to top infinite line charge can be shown to be V + λ ( s+. Similarly, the potential due to the bottom infinite line charge can be shown to be 4πɛ d V + λ ( s. Therefore the total potential at (y, z is given by 4πɛ d V (y, z λ ( s+ 4πɛ d λ ( s 4πɛ λ 4πɛ λ 4πɛ s + ( s s + + λ ( s 4πɛ d y + (z + d ] y + (z d

2 V (b The induced charge density is given by σ ɛ, where ˆn is the vector in the direction perpendicular to n the surface. In the present problem, ˆn ẑ. Therefore, V σ ɛ n ɛ V z z λ y + (z + d ]] ɛ z 4πɛ y + (z d z ] λ (z + d (z d ɛ 4πɛ y + (z + d y + (z d λ d ɛ 4πɛ y + z + d ] y + d λd π(y + d z Solution 4.: Far-field potential (Griffiths rd ed. Prob.6 In this problem, a point r on the sphere is represented in the (r, θ, φ coordinates. An observation point r is represented in the (r, θ, φ coordinates. The angle between the vectors r and r is represented by α. So, the formula for multipole expansion of potential V can be written as V (r 4πɛ n r n+ r n P n (cos αρ(rdτ Now, since we are interested in observation points along z-axis only, the angle α between the two vectors in essentially the angle θ. Therefore we can write the above expression as V (r 4πɛ n z n+ Let s calculate the above potential term by term in the limit z R. r n P n (cos θρ(rdτ. Monopole term: We have V mono (r ρ(rdτ 4πɛ z k Rr ] 4πɛ z (R r sin θ r sin θdrdθdφ 4πɛ z kr (R r sin θdrdθdφ The r-integral yields R (R rdr (Rr r R R R. Therefore, the monopole potential is zero: V mono (r 4πɛ z kr (R r sin θdrdθdφ

3 . Dipole term: We have V dip (r 4πɛ z rp (cos θρ(rdτ 4πɛ z r cos θρ(rdτ 4πɛ z r cos θ k Rr ] (R r sin θ r sin θdrdθdφ The θ-integral yields π sin θ cos θdθ sin θ π V dip (r 4πɛ z r cos θ (. Therefore, the dipole potential is also zero: k Rr (R r sin θ ] r sin θdrdθdφ. Quadrupole term: We have V quad (r 4πɛ z 4πɛ z 8πɛ z r P (cos θρ(rdτ r ( cos θ /ρ(rdτ r ( cos θ k Rr ] (R r sin θ r sin θdrdθdφ The three integrals can be calculated as follows: r integral : θ integral : φ integral : Therefore, R π π ( r r (R rdr R r4 R ( cos θ sin θdθ dφ π V quad (r π kr 8πɛ z R4 6 sin θdθ π So, the approximate potential is the potential of the quadrupole: ( π ( ( π sin 4 θdθ π 9 π ( ( R4 π (π kπ R πɛ 48z V (r V quad (r kπ R 5 4πɛ 48z Solution 4.4: Potential due to a four-charge system (Griffiths rd ed. Prob.7 The total charge of the system is zero. So, the monopole term in the potential would be zero. Now, let s calculate the dipole term see Fig. ]. We have Therefore, p (qa qaẑ + ( qa q( aŷ qaẑ V V dipole p ˆr 4πɛ r qaẑ ˆr 4πɛ r qa cos θ 4πɛ r

4 x z +q a µ a a -q a -q +q r y FIG. : Solution 4.5: Far-field Potential due to a spherical-charge distribution (Griffiths rd ed. Prob.8 (a From the symmetry in the problem it is clear that the net dipole moment will be in the z direction. Therefore p rρ(rdτ zẑρ(rdτ zρ(rdτẑ zσdaẑ (R cos θ(k cos θr sin θdθdφẑ πr k πr k π 4πR k ẑ cos θ sin θdθẑ π ( cos θ ẑ (b We find that the total charge of the system is zero. So, there will be no monopole contribution to the far-field potential. However since the dipole moment is not zero, the first contribution will be from the dipole term and the far-field potential can be written as V V dipole p ˆr 4πɛ r 4πR k ẑ ˆr 4πɛ r 4πR k cos θ 4πɛ r kr cos θ ɛ r Solution 4.6: Electric field of a pure dipole (Griffiths rd ed. Prob. The electric field of a pure dipole is given by p E dip (r ( cos θˆr + sin θˆθ 4πɛ r (p cos θˆr p cos θˆr + p sin θˆθ 4πɛ r 4πɛ r (p ˆrˆr (p ˆrˆr (p ˆθˆθ] (p ˆrˆr p] 4πɛ r Solution 4.7: Field and potential due to a three-charge system (Griffiths rd ed. Prob. 4

5 The total charge of the three-charge system is q. Therefore the monopole contribution is not zero and is given by V mono 4πɛ q r q 4πɛ r The dipole moment p of the charge system is p qaẑ + qa q( a]ŷ qaẑ. So, the dipole contribution to the potential is V dip p ˆr 4πɛ r qa cos θ 4πɛ r Therefore, the potential up to two terms in the multipole expansion is given by V V mono + V dip q ( qa cos θ + 4πɛ r r The electric field E is given by E V q 4πɛ r ˆr + ar ( ] cos θˆr + sin θˆθ x z +q a a a -q -q r y FIG. : 5