HW4 - Physics 3122 Fall 2010 Homework Set 4 Solutions 2.20(assume k 0 r rr(a E = k xyx 2 yzy 3 xzz E = k(0 2 y x k(0 3z y k(0 x z 0 No V r rr(b E =

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Physics 3122, Fall 2010: Homework Set 4 Solutions. 2.20 (assume k 0) ( a ) r E = k [ xy ˆ x + 2 yz ˆ y + 3 xz ˆ z ] r × r E = k (0 2 y x + k (0 3 z y + k (0 x z 0 No V ( b r E = k [ y 2 ˆ x + (2 xy + z 2 y + 2 yz ˆ z ] r × r E = k (2 z 2 z x + k (0 0)ˆ y + k (2 y 2 y z = 0 E x dx 0 x y = 0 z = 0 + E y dy 0 y x = x z = 0 + E z dz 0 z x = x y = y = k [0 + xy 2 + yz 2 ] = V V = k ( xy 2 + yz 2 ) r E = r V = k [ y 2 ˆ x + (2 xy + z 2 y + 2 yz ˆ z ] 2.21 r < R 4 π r 2 E = 4 3 r 3 ρ / ε 0 = qr 3 / 0 R 3 E = qr /4 πε 0 R 3 V = V 0 qr 2 /8 0 R 3 r > R 4 r 2 E = q / 0 E = q 0 r 2 V = V 0 q 0 R ( q 0 r 2 ) d r R r V = V 0 q 0 R − − q 0 r [ ] R r = V 0 3 q 0 R + q 0 r V ( ) = V 0 = 3 q 0 R r < R V = ( q 0 R )(3 r 2 / R 2 r V = ˆ r V / r = qr ˆ r 0 R 3 = r E r > R V = q 0 r r V = q ˆ r 0 r 2 = r E 2.21 alt. Centrally symmetric: V ( r , θ , ϕ ) = V ( r r E = E ( r ) ˆ r 2 V = 1 r 2 r r 2 V r + 1 r 2 sin ∂θ sin V + 1 r 2 sin 2 2 V ∂ϕ 2 = 1 r 2 d dr r 2 dV dr r R 2 V = 1 r 2 d dr r 2 dV dr = 0 =
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This note was uploaded on 01/22/2012 for the course PHYS 3122 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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HW4 - Physics 3122 Fall 2010 Homework Set 4 Solutions 2.20(assume k 0 r rr(a E = k xyx 2 yzy 3 xzz E = k(0 2 y x k(0 3z y k(0 x z 0 No V r rr(b E =

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