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
Electric field in two concentric conducting cylinders
Reply to thread
Message
[QUOTE="Ryaners, post: 5470581, member: 577398"] [h2]Homework Statement [/h2] Two concentric cylindrical conducting shells of length L are separated by a vacuum. The inner shell has surface charge density +σ and radius r[SUB]a[/SUB]. The outer shell has radius r[SUB]b[/SUB]. Using Gauss’ Law, as a function of radius r find: The direction and magnitude of electric field inside and outside the shells. Be sure to clearly state the Gaussian surfaces that you are using! Find an expression for the voltage between the shells. [h2]Homework Equations[/h2] Electric flux = ∫ EdA = q[SUB]encl[/SUB] / ε[SUB]0[/SUB] σ = q/A [h2]The Attempt at a Solution[/h2] I'm using a cylindrical Gaussian surface. I understand that when r < r[SUB]a[/SUB] the field E = 0. Here's what I've gotten so far for when r is between r[SUB]a[/SUB] and r[SUB]b[/SUB]: E = q[SUB]encl[/SUB] / ε[SUB]0[/SUB]⋅A[SUB]gauss[/SUB] = q[SUB]encl[/SUB] / ε[SUB]0[/SUB]⋅2πrl ⇒ E = σ2πr[SUB]a[/SUB]l / ε[SUB]0[/SUB]⋅2πrl ⇒ E = σr[SUB]a[/SUB] / rε[SUB]0[/SUB] That doesn't seem very... right. I don't think there should be a dependence on r[SUB]a[/SUB] for starters. Would the expression for the flux through a Gaussian surface with r > r[SUB]b[/SUB] be the same also, as there isn't any non-induced charge on it? Also, for the last part of the question I don't understand why there would be a voltage between the shells when the charge on the outer shell is just induced by the inner cylinder, ie the net charge on the outer cylinder is 0. Isn't it? Sorry if I'm not making much sense, I'm almost through my exams and my brain is extra melty. :frown: [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Electric field in two concentric conducting cylinders
Back
Top