{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW5 - Physics 3122 Fall 2010 Homework Set 5 Solutions 2.39...

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

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

Unformatted text preview: Physics 3122, Fall 2010: Homework Set 5 Solutions. 2.39 E 2πsL = λL /ε 0 C = λL /V = 2πε 0 L / ln(b / a) € L W = 1 ε 0 ∫ E 2 dτ = ( λ2 / 8π 2ε 0 ) ∫ 0 2 2π b −2 0 a ∫∫ b s sdsdϕdz = ( λ2 / 8π 2ε 0 ) L2π [ln( s)] a C = Q2 / 2W = λ2 L2 / 2W = 2πε 0 L / ln(b / a) 2.40 P = 1 ε0 E 2 2 F = PA = 1 ε 0 E 2 A 2 U = 1 ε 0 E 2 As 2 € C / L = 2πε 0 / ln(b / a) 2.39 alt E = λ / 2πε 0 s as before W = λ2 L ln(b / a) / 4 πε 0 € a V = − ∫ b ( λ / 2πε 0 s) ds = ( λ / 2πε 0 ) ln(b / a) E = λ / 2πε 0 s dW = Fd = 1 ε 0 E 2 Aε 2 dU = 1 ε 0 E 2 Ads = − 1 ε 0 E 2 Aε 2 2 3.1 ˆ q at zz, z < R q 4 πR V = ∫ Vda = 4 πε 0 sphere 2 qR = 2ε 0 € 00 2 2 1/ 2 π V= Ⱥ ( R − 2 Rz cosθ + z ) Ⱥ Rz Ⱥ q 4 πε 0 R π ∫ (R 0 2 sin θdθ − 2 Rz cos θ + z 2 )1 / 2 Ⱥ qR Ⱥ | R + z | − | R − z | Ⱥ qR 2 2 z qR Ⱥ = Ⱥ Ⱥ = 2ε Rz = ε Ⱥ Rz Ⱥ 0 2ε 0 Ⱥ 0 0 2 qj qi +∑ outside 4 πε 0 ri inside 4 πε 0 R ∑ 3.3 V = V ( r) V = V ( s) € ∫∫ 1 qR 2 2 R sin θdθdϕ = ˆ r − zz 2ε 0 All charges : 2 V= 2π π 1 d ȹ 2 dV ȹ dV C r2 = −C1 V = 1 + C2 ȹ r ȹ = 0 2 r dr ȹ dr Ⱥ dr r 1 d ȹ dV ȹ dV ∇ 2V = s = C1 V = C1 ln( s) + C2 ȹ s ȹ = 0 s ds ȹ ds Ⱥ ds ∇ 2V = ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online