Explanation Formula Used: The expression of electric potential is given by, V ( s , ϕ ) = a 0 + b 0 ln s + ∑ k = 1 α [ s p k ( a k cos k ϕ + b k sin k ϕ + s − k ( c k cos k ϕ + d k sin k ϕ ) ) ] The expression of electric potential inside the shell is given by, V i n ( s , ϕ ) = ∑ k = 1 α s k ( a k cos k ϕ + b k sin k ϕ ) The expression of electric potential outside the shell is given by, V o u t ( s , ϕ ) = ∑ k = 1 α s − k ( c k cos k ϕ + d k sin k ϕ ) The expression of normal derivative of potential with the surface charge density is given by, ( ∂ V ∂ s ) R + − ( ∂ V ∂ s ) R − = − σ ε 0 Here, σ is the surface charge density, and ε 0 is the permittivity of free space. Calculation: At one boundary condition, potential is continuous at s = R Equate the potential at inside and outside of the shell. ∑ k = 1 α R k ( a k cos k ϕ + b k sin k ϕ ) = ∑ k = 1 α R − k ( c k cos k ϕ + d k sin k ϕ ) Compare the coefficient of cos k ϕ , sin k ϕ to get a k and d k . a k R k = R − k c k c k = R 2 k a k Similarly, b k R k = R − k d k d k = R 2 k b k The derivative of potential outside the cylinder is calculated as, ( ∂ V ∂ s ) R + = ∑ ∂ ∂ s ( s − k ( c k cos k ϕ + d k sin k ϕ ) ) s = R = [ ∑ ( − k s k + 1 ) ( c k cos k ϕ + d k sin k ϕ ) ] s = R = [ ∑ ( − k R k + 1 ) ( c k cos k ϕ + d k sin k ϕ ) ] The derivative of potential inside the cylinder is calculated as, ( ∂ V ∂ s ) R − = ∑ ∂ ∂ s ( s k ( a k cos k ϕ + b k sin k ϕ ) ) s = R = [ ∑ ( c k cos k ϕ + d k sin k ϕ ) ( k s k − 1 ) ] s = R = ∑ ( c k cos k ϕ + d k sin k ϕ ) ( k R k − 1 ) The normal derivative of potential with the surface charge density is calculated as, ( ∂ V ∂ s ) R + − ( ∂ V ∂ s ) R − = − σ ε 0 ∑ ( − k R k + 1 ) ( c k cos k ϕ + d k sin k ϕ ) − ∑ ( c k cos k ϕ + d k sin k ϕ ) ( k R k − 1 ) = − σ ε 0 ∑ 2 k R k − 1 ( a k cos k ϕ + b k sin k ϕ ) = σ 0 ε 0 f o r ( 0 < ϕ < π ) ∑ 2 k R k − 1 ( a k cos k ϕ + b k sin k ϕ ) = − σ 0 ε 0 f o r ( π < ϕ < 2 π ) The value of integral of above angles at different condition is expressed as, ∫ 0 2 π sin k ϕ cos l ϕ d ϕ = 0 ∫ 0 2 π cos k ϕ cos l ϕ d ϕ = 0 f o r k ≠ 1 ∫ 0 2 π cos k ϕ cos l ϕ d ϕ = 0 f o r k = 1 Multiply with cos l ϕ on both side of the equation. 2 k R k − 1 [ ∫ 0 2 π a k cos k ϕ cos l ϕ d ϕ + ∫ 0 2 π b k sin k ϕ cos l ϕ ] = σ 0 ε 0 [ ∫ 0 π cos l ϕ d ϕ − ∫ π 2 π cos l ϕ d ϕ ] 2 k R k − 1 π a k = σ 0 ε 0 [ ∫ 0 π cos l ϕ d ϕ − ∫ π 2 π cos l ϕ d ϕ ] = [ sin l ϕ l ] 0 π a k = 0 As the value of a k is zero, multiply with sin l ϕ and integrate

4th Edition

ISBN: 9781108420419

Author: David J. Griffiths

Publisher: Cambridge University Press

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Chapter 3.4, Problem 3.45P

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The electric potential inside and outside the cylinder.

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Chapter 3.4, Problem 3.44P

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