P25_007 - 7 The potential difference between the wire and...

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: 7. The potential difference between the wire and cylinder is given, not the linear charge density on the wire. We use Gauss’ law to find an expression for the electric field a distance r from the center of the wire, between the wire and the cylinder, in terms of the linear charge density. Then integrate with respect to r to find an expression for the potential difference between the wire and cylinder in terms of the linear charge density. We use this result to obtain an expression for the linear charge density in terms of the potential difference and substitute the result into the equation for the electric field. This will give the electric field in terms of the potential difference and will allow you to compute numerical values for the field at the wire and at the cylinder. For the Gaussian surface use a cylinder of radius r and length , concentric with the wire and cylinder. The electric field is normal to the rounded portion of the cylinder’s surface and its magnitude is uniform over that surface. This means the electric flux through the Gaussian surface is given by 2πr E , where E is the magnitude of the electric field. The charge enclosed by the Gaussian surface is q = λ , where λ is the linear charge density on the wire. Gauss’ law yields 2πε0 r E = λ . Thus, λ E= . 2πε0 r Since the field is radial, the difference in the potential Vc of the cylinder and the potential Vw of the wire is r r w c rc λ λ ∆ V = Vw − V c = − E dr = ln , dr = 2πε0 r 2πε0 rw r r c w where rw is the radius of the wire and rc is the radius of the cylinder. This means that λ= and E= 2πε0 ∆V ln(rc /rw ) ∆V λ = . 2πε0 r r ln(rc /rw ) (a) We substitute rc for r to obtain the field at the surface of the wire: = ∆V 850 V = −6 m) ln [(1.0 × 10−2 m)/(0.65 × 10−6 m)] rw ln(rc /rw ) (0.65 × 10 = E 1.36 × 108 V/m . (b) We substitute rc for r to find the field at the surface of the cylinder: E = ∆V 850 V = rc ln(rc /rw ) (1.0 × 10−2 m) ln [(1.0 × 10−2 m)/(0.65 × 10−6 m)] = 8.82 × 103 V/m . ...
View Full Document

Ask a homework question - tutors are online