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 Potential V inside nonconducting sphere with cavity
Reply to thread
Message
[QUOTE="haruspex, post: 5025231, member: 334404"] It isn't. The potential at r is that at r[SUB]2[/SUB] minus the integral from r to r[SUB]2[/SUB]. The potential at infinity is the same as it would be at r if the body were uncharged, or if r[SUB]1[/SUB] = r = r[SUB]2[/SUB]. It was evaluated as (something like) ##\frac{4\pi}3\left(r_2^2-\frac{r_1^3}{r_2}\right)\frac{\rho}{4\pi\epsilon_0}## and combined with the result of the integral. Write out the steps and you'll see the result is correct. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Electric Potential V inside nonconducting sphere with cavity
Back
Top