Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
How Do You Calculate Current Density Between Coaxial Cylinders?
Reply to thread
Message
[QUOTE="RyanP, post: 5598165, member: 606089"] [h2]Homework Statement[/h2] [/B] Two long, coaxial metal cylinders are separated by a material of conductivity sigma and dielectric constant epsilon. The radius of the inner cylinder is a, the radius of outer cylinder is b, and the length of both is L. Suppose that the inner conductor is held at a potential Vo with respect to the outer one. What is the current density J(r) between the two? [h2]Homework Equations[/h2] J(r) = I/A = sigma*E C=Q/V R=s/(A*sigma) [h2]The Attempt at a Solution[/h2] First attempt: I found resistance of the dielectric to be ln(b/a)/(2pi*L*sigma). As a function of radius r, R(r) = ln(r/a)/(2pi*L*sigma). Current = V/R = Vo / (ln(b/a)/(2piLsigma)) = Vo*2pi*L*sigma / ln(r/a) as a function of r. Then divide by Area=2pi*L*r to get current density J(r) = Vo*sigma / (r*ln(r/a)). Second attempt: Vo = Q/C. I found capacitance of the configuration to be 2pi*epsilon_o*L*epsilon / ln(b/a) using the expression for resistance. Then Q = Vo*2pi*L*epsilon*epsilon_o / ln(b/a). EA = Q/epsilon_o, so E = Vo*epsilon / (r*ln(b/a)). Using J(r) = E*sigma, I get J(r) = sigma*Vo*epsilon / rln(b/a). The difference in the two is the presence of epsilon (dielectric constant) in the numerator, and b rather than r in the ln term of the denominator. Are either of these approaches correct? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
How Do You Calculate Current Density Between Coaxial Cylinders?
Back
Top