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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online