# HW3 - Physics 3122 Fall 2010 Homework Set 3 Solutions 2.2(b...

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Unformatted text preview: Physics 3122, Fall 2010: Homework Set 3 Solutions. 2.2 (b replaces d ) r rr 2 2 1 qiℜ i 1 qi ( r − ri ) ∑ = 4πε ∑ r r 3 4 πε 0 i =1 ℜ 2 0 i =1 r − ri r r r rr rr ˆ ˆ ˆ r = zz r1 = (b / 2) x r2 = ( −b / 2) x r − r1 = r − r2 = ( z 2 + b 2 / 4 )1 / 2 r ˆ ˆ ˆ ˆ ˆ 1 q( zz − bx / 2) 1 q( zz + bx / 2) 1 2qzz (a) E = 2 2 3/2 + 2 2 3/2 = 2 2 4πε 0 ( z + b / 4 ) 4πε 0 ( z + b / 4 ) 4 πε 0 ( z + b / 4 ) 3 / 2 r ˆ ˆ 1 (2q)z 1 Qz z >> b E = = c.f. Q = 2q at origin 2 2 4 πε 0 z 4 πε 0 z r ˆ ˆ ˆ ˆ ˆ 1 ( −q)( zz − bx / 2) 1 q( zz + bx / 2) 1 qbx (b) E = + = 2 2 3/2 2 2 3/2 2 2 4 πε 0 ( z + b / 4 ) 4 πε 0 ( z + b / 4) 4 πε 0 ( z + b / 4) 3 / 2 rr E (r ) = € 2.9 r rr ˆ ∇⋅ E = ρ /ε 0 E = kr 3 r r r 1 ∂( r 2 E r ) 1 ∂(sin θEθ ) 1 ∂E ϕ ∇⋅ E = 2 + + r ∂r r sin θ ∂θ r sin θ ∂ϕ = 1 ∂( r 2 kr 3 ) + 0 + 0 = 5 kr 2 r 2 ∂r 2π π R 0 0 0 ρ = 5ε 0 kr 2 ∫ ρdτ = ∫ ∫ ∫ 5ε kr r dr sinθdθdϕ = 4π (5ε k)[r rr ˆ ˆ Q = ε ∫ E ⋅ da = ε ∫ ∫ kR r ⋅ R sin θdθdϕr = 4 πε kR Q= 22 0 2π 0 € € 0 2.10 r r 11 ∫1 E ⋅ da = 4 6 2.14 0 r 2π π r 0 0 0 ∫ ∫∫ 0 3 2 5 0 r ∫ E ⋅ da = 24 ε r ˆ E = Err ρ = kr π 0 1q 0 r r ∫ E ⋅ da = E r 4πr 2 kr′r′2 dr′ sin θ ′dθ ′dϕ′ = 4 πk[ r′4 / 4 ]r = πkr 4 0 r ˆ E r = πkr 4 /ε 0 4 πr 2 E = kr 2 r / 4ε 0 qenc = € 5 / 5]R = 4 πε 0 kR 5 0 ...
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