HW7 - Physics 3122 Fall 2010 Homework Set 7 Solutions 3.7(b...

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Unformatted text preview: Physics 3122, Fall 2010: Homework Set 7 Solutions. 3.7(b) V ( r,θ ) = Er = q 1 q 1 2 2 1/ 2 − 2 4 πε 0 [ r − 2 ra cos θ + a ] 4 πε 0 [( ra / R) − 2 ra cos θ + R 2 ]1 / 2 q ( r − a cosθ ) q ( ra 2 / R 2 − a cos θ ) − 4 πε 0 [ r 2 − 2 ra cos θ + a 2 ]3 / 2 4 πε 0 [( ra / R) 2 − 2 ra cos θ + R 2 ]3 / 2 E r r=R = q ( R − a cosθ ) q ( a 2 / R − a cosθ ) − 4 πε 0 [ R 2 − 2 Ra cosθ + a 2 ]3 / 2 4 πε 0 [ a 2 − 2 Ra cosθ + R 2 ]3 / 2 =− q (a2 − R 2 ) / R 4 πε 0 [ R 2 − 2 Ra cos θ + a 2 ]3 / 2 σ = ε 0Er = − q (a 2 − R 2 ) 4 πR [ R 2 − 2 Ra cos θ + a 2 ]3 / 2 3.9 € € V = −( λ / 2πε 0 ) ln( s) = −( λ / 4 πε 0 ) ln( s2 ) Ⱥ y 2 + ( z − ζ) 2 Ⱥ λ λ λ 2 2 2 2 V =− ln[ y + ( z − ζ) ] + ln[ y + ( z + ζ) ] = − ln Ⱥ Ⱥ 4 πε 0 4 πε 0 4 πε 0 Ⱥ y 2 + ( z + ζ) 2 Ⱥ Ⱥ y 2 + ζ 2 Ⱥ λ λ z =0 V =− ln Ⱥ 2 2 Ⱥ = − ln[1] = 0 4 πε 0 Ⱥ y + ζ Ⱥ 4 πε 0 ∂V λ 2( z − ζ) λ 2( z + ζ) λζ Ez = − = σ = ε 0 E z z=0 = − 2 2− 2 2 ∂z 4 πε 0 [ y + ( z − ζ) ] 4 πε 0 [ y + ( z + ζ) ] π ( y 2 + ζ2 ) E s 2πsL = λL /ε 0 E s = λ / 2πε 0 s 3.12 Ⱥ V0 V (0, y ) = Ⱥ Ⱥ −V0 Cn = = 0 ≤ y ≤ a /2 a /2 ≤ y ≤ a 2a 2 a /2 2a ∫ V (0, y ) sin(nπy / a)dy = a ∫ V0 sin(nπy / a)dy − a ∫ V0 sin(nπy / a)dy a0 0 a /2 a /2 a 2V0 Ⱥ − cos( nπy / a) Ⱥ 2V Ⱥ − cos( nπy / a) Ⱥ − 0 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ 0 Ⱥ a / 2 a Ⱥ nπ / a a Ⱥ nπ / a = (2V0 / nπ )[1 − cos( nπ / 2)] − (2V0 / nπ )[cos( nπ / 2) − cos( nπ )] = (2V0 / nπ )[1 − 2 cos( nπ / 2) + cos( nπ )] = (8V0 / nπ ) cos( nπ / 2)[cos( nπ / 2) − 1] / 2 Cn = 0 unless n = 4 m + 2 m integer. Then Cn = 8V0 / nπ . € 3.13 V ( x, y ) = Ex = − ȹ −nπx ȹ ȹ nπy ȹ 4V0 1 ∑ n expȹ a ȹ sinȹ a ȹ ȹ Ⱥ ȹ Ⱥ π n odd ȹ −nπx ȹ ȹ nπy ȹ ∂V 4V 1 ȹ −nπ ȹ ȹ − nπx ȹ ȹ nπy ȹ 4V0 = − 0 ∑ ȹ ȹ expȹ ȹ sinȹ ȹ = ∑ expȹ a ȹ sinȹ a ȹ ȹ Ⱥ ȹ Ⱥ ∂x π n odd n ȹ a Ⱥ ȹ a Ⱥ ȹ a Ⱥ a n odd σ = ε 0 E x x =0 = ȹ nπy ȹ 4ε 0V ∞ ȹ [2 m + 1)πy ȹ 4ε 0V0 ∑ sinȹ a ȹ = a 0 ∑ sinȹ a ȹ Ⱥ ȹ Ⱥ a n odd ȹ m =0 ∞ Note : ∞ ∑ sin([2m + 1]x ) = ∑ [e m =0 i( 2 m +1)x m =0 = limε → 0 + e ix 2i − e − i( 2 m +1)x ] / 2i ∞ ∑ (e 2ix −ε )m − limε → 0+ m =0 e − ix 2i ∞ ∑ (e −2 ix −ε m ) m =0 e ix 1 e − ix 1 = limε → 0 + 2 ix −ε − limε → 0 + 2i (1 − e ) 2i (1 − e −2 ix −ε ) 1 1 i 1 = − ix ix − ix − ix = ix − ix = 2i(e − e ) 2i(e − e ) (e − e ) 2 sin( x ) σ= € 2ε 0V0 a sin(πy / a) ...
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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 Tech.

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