{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 6 Solutions

Homework 6 Solutions - Chapter 26 1(a The charge that...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1039 Chapter 26 1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 × 10 3 C. (b) The number of electrons N is given by q = Ne , where e is the magnitude of the charge on an electron. Thus, N = q / e = (1200 C)/(1.60 × 10 –19 C) = 7.5 × 10 21 . 2. Suppose the charge on the sphere increases by Δ q in time Δ t . Then, in that time its potential increases by 0 , 4 q V r πε Δ Δ = where r is the radius of the sphere. This means 0 4 . q r V πε Δ = Δ Now, Δ q = ( i in i out ) Δ t , where i in is the current entering the sphere and i out is the current leaving. Thus, ( )( ) ( ) ( ) 0 9 in out in out 3 0.10 m 1000 V 4 8.99 10 F/m 1.0000020 A 1.0000000 A 5.6 10 s. r V q t i i i i πε Δ Δ Δ = = = × = × 3. We adapt the discussion in the text to a moving two-dimensional collection of charges. Using σ for the charge per unit area and w for the belt width, we can see that the transport of charge is expressed in the relationship i = σ vw , which leads to σ = = × × = × i vw 100 10 30 50 10 6 7 10 6 2 6 A m s m C m 2 b gc h . . 4. We express the magnitude of the current density vector in SI units by converting the diameter values in mils to inches (by dividing by 1000) and then converting to meters (by multiplying by 0.0254) and finally using 2 2 4 . i i i J A R D π π = = =
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon