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Jackson 3.7 Homework Problem Solution
Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: Three point charges ( q , -2 q , q ) are located in a straight line with separation a and with the middle charge (-2 q ) at the origin of a grounded conducting spherical shell of radius b , as indicated in the sketch. (a) Write down the potential of the three charges in the absence of the grounded sphere. Find the limiting form of the potential as a → 0, but the product qa^2 = Q remains finite. Write this latter answer in spherical coordinates. (b) The presence of the grounded sphere of radius b alters the potential for r < b. The added potential can be viewed as caused by the surface-charge density induced on the inner surface at r = b or by image charges located at r > b. Use linear superposition to satisfy the boundary conditions and find the potential everywhere inside the sphere for r < a and r > a. Show that in the limit a → 0, r , ,
Q
2 0 r (^3) ^1 −^ r 5 b (^5) P^2 cos^ SOLUTION: (a) We already know the potential due to one point charge. We just add up the potential from each point charge: r = q 4 0
∣ r − a z ∣
q 4 0
∣ r − a − z ∣
2 q 4 0
∣ r ∣ Expand the first two terms using the Legendre polynomial expansions shown below in order to get a solution in spherical coordinates: 1
∣ r − r 0 ∣
=∑ l = 0 ∞ (^) r 0 l r l 1 Pl cos^ ^ if^ r^ r^^0 x^ y z q q -2 q a a b Φ =
∣ r − r 0 ∣
=∑ l = 0 ∞ r l r 0 l 1 Pl^ cos^ ^ if^ r^ r^^0 If r a , ^ r =^ q 4 0 ∑ l = 0 ∞ a l r l 1 Pl^ cos^ ^ q 4 0 ∑ l = 0 ∞ − 1 l a l r l 1 Pl^ cos^ −^ 2 q 4 0
r r = q 4 0 [
r ∑ l = 0 ∞ 1 − 1 l a l r l 1 Pl^ cos^ ] r = 2 q 4 0 ∑ l =2,4,6... ∞ a l r l 1 Pl^ cos^ ^ (if^ r^ a^ ) If r a , ^ r =^ q 4 0 ∑ l = 0 ∞ r l a l 1 Pl^ cos^ ^ q 4 0 ∑ l = 0 ∞ − 1 l r l a l 1 Pl^ cos^ −^ 2 q 4 0
r r = 2 q 4 0 ∑ l =0,even ∞ r l a l (^1) [^1 −^ a r l , (^0) ] Pl cos (^) (if r a (^) ) Set qa^2 = Q and keep it finite. As a → 0, the only potential that matters is the r a case: r = [
2 Q
∑ l =0,2,4. .. ∞ a l − 2 r l 1 Pl^ cos] a 0 As a → 0, the higher terms in l get increasingly smaller, so that we only need to keep the first nonzero term: r =
2 Q
4 0 [^
r 3 P^2 cos^ ^ ...] r =
Q
r 3 ^3 cos 2 − 1 The l = 0 term vanishes. This makes sense because the l = 0 term is the total charge (monopole) moment of the system and in this case the total charge is zero. The l = 1 term (the dipole moment term) drops out because of the symmetry of the charges. The first non-zero term is the quadrupole moment term ( l = 2). (b) The presence of the grounded sphere of radius b alters the potential for r < b. The added potential can be viewed as caused by the surface-charge density induced on the inner surface at r = b or by image charges located at r > b. Use linear superposition to satisfy the boundary conditions and find the potential everywhere inside the sphere for r < a and r > a. Show that in the limit a → 0,
