P24_020

# P24_020 - q cylinder = λ L = σ (2 πRL ) = ⇒ λ = σ (2...

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20. We denote the radius of the thin cylinder as R =0 . 015 m. Using Eq. 24-12, the net electric Feld for r>R is given by E net = E wire + E cylinder = λ 2 πε 0 r + λ 0 2 πε 0 r where λ = 3 . 6nC / m is the linear charge density of the wire and λ 0 is the linear charge density of the thin cylinder. We note that the surface and linear charge densities of the thin cylinder are related by
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Unformatted text preview: q cylinder = λ L = σ (2 πRL ) = ⇒ λ = σ (2 πR ) . Now, E net outside the cylinder will equal zero, provided that 2 πRσ = λ , or σ = λ 2 πR = 3 . 6 × 10 − 9 C / m (2 π )(0 . 015 m) = 3 . 8 × 10 − 8 C / m 2 ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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