Jackson 2.23 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

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1 Jckson.3 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: A hollow cube hs conducting wlls defined by six plnes x =, y =, z =, nd x =, y =, z =. The wlls z = nd z = re held t constnt potentil V. The other four sides re t zero potentil. () Find the potentil Φ(x, y, z) t ny point inside the cube. (b) Evlute the potentil t the center of the cube numericlly, ccurte to three significnt figures. How mny terms in the series is it necessry to keep in order to ttin this ccurcy? Compre your numericl result with the verge vlue of the potentil on the wlls. See Problem.8. (c) Find the surfce-chrge density on the surfce z =. SOLUTION: The problem contins no chrge, so the electric potentil is described everywhere inside the cube by the Lplce eqution: = This problem hs boundry, the cube, tht best mtches rectngulr coordintes. In rectngulr coordintes, the Lplce eqution becomes: x y z = Using the method of seprtion of vribles, the most generl solution to this eqution is: x, y, z= A B x A B y A B z A B x A e i y B e i y A e z B e z, A e i x B e i x A B y A e z B e z A e i x B e i x A e i y B e i y A e z B e z Apply the boundry condition, Φ(x =, y, z) =

2 = A A B y A B z A A e i y B e i y A e z B e z, A B A B y A e z B e z A B A e i y B e i y A e z B e z This must be true for ll y nd z, so tht ech term must vnish seprtely. This forces A α = nd B α = -A α. The solution now becomes: x, y, z= B x A B ya B z B x A e i y B e i y A e z B e z, A sin x A B y A e z B e z A sin x A e i y B e i y A e z B e z Similrly, the boundry condition, Φ(x, y =, z) = leds to A β = nd B β = -A β which gives the solution: x, y, z= B x B y A B z B x A sin y A e z B e z, A sin x B y A e z B e z A sin x A sin y A e z B e z Now pply the boundry condition Φ(x =, y, z) = = B B ya B z B A sin y A e z B e z, A sin B y A e z B e z A sin A sin y A e z B e z The only wy this cn be true for ll y nd z is if B α = nd α = nπ/ where n =,,,... x, y, z= n sin n x B y A e z B e z n, sin n x A sin y A e z B e z Similrly, pply the boundry condition Φ(x, y =, z) = to get B β = nd β = mπ/ where n =,,,... x, y, z= sin n x A sin m m y A e z B e z

3 By definition we hve = so tht we now know =n m /. We cn combine severl constnts so tht we now hve: x, y, z= V = sin n x Now pply the boundry condition Φ(x, y, z = ) = V sin n x sin m y A e z n m / B e z n m / sin m y A B n, m Multiply both side by sin n ' x V sin n' x = sin m ' y sin n x sin m' y nd integrte over x nd y from to dx dy sin n' x dx sin m y sin m' y dy A B n, m n, m Due to orthogonlity, ech integrl on the right is zero, except when n = n' nd m = m' V sin n x,m B = 4V,m B = 4V sin m y dx dy= 4 A B n, m sin n x dx sin m y dy ] ] n [ n m [ m, m B n, m = 6 V n m n, m = odd Now pply the finl boundry condition Φ(x, y, z = ) = V sin n x V = Repet the process done bove to get sin m y A n, m en m / B e n m /, m e n m B e n m = 6 V n m n, m = odd Solve the system of equtions in the boxes bove to find:

4 , m = 8 V nm e n m cosh /n m B = 8 V n m The finl solution is now: x, y, z= x, y, z= e n m cosh /n m sin n x 6V n m sin n x sin m y 8V e n m e z n m / e n m e zn m / n m cosh /n m sin m y cosh / n m z/ cosh / n m (b) Evlute the potentil t the center of the cube numericlly, ccurte to three significnt figures. How mny terms in the series is it necessry to keep in order to ttin this ccurcy? Compre your numericl result with the verge vlue of the potentil on the wlls. See Problem.8. The potentil t the center of the cube is: /,/, /= 6V Φ(/,/,/)=V [ 6 n m sin n sin m π sin ( π ) sin ( π )( 3 sin sin 3 3 sin 3 sin 5 sin sin cosh /n m cosh((π/ ) )) cosh / cosh / cosh / 6...] /,/, /=V [ ] Here is solution, to three significnt figures, if you only keep the first term, or the first two terms, etc: term /,/,/=V [.348] terms /,/, /=V [.34] 3 terms /,/,/=V [.333] 4 terms /,/, /=V [.333]

5 We only hve to keep the first three terms to hve the nswer ccurte to three significnt figures. It is obvious tht the solution is converging to: /, /,/= V 3 There re six wlls on the cube nd two sides hve non-zero potentil V, so the verge vlue of the potentil on the sides of the cube is Φ ve on surf = V = V 6 3. This leds to the interesting conclusion tht: /, /,/= ve on surf (c) Find the surfce-chrge density on the surfce z =. =[ d dn ]n= We hve solved the potentil on the inside of the cube, so we cn only use tht potentil to find the surfce chrge density on the inside of the z = surfce. The norml to the inside surfce is in the negtive z direction so tht: =[ d dz ]z= =[ d dz =[ σ= 6 ϵ V π n, m odd 6V nm sin n x 6V n m sin n x n +m odd n m sin m y sin m y cosh / n m z/ cosh /n m ]z= sinh /n m z / cosh /n m /n m ]z= sin ( nπ x ) sin ( mπ y ) (tnh ((π/ ) n +m ))