Recent content by Kosta1234

First of all, I've calculated the partition function:Z=1h3∫e−βH(q,p)d3pd3q=1h3∫e−β(p22m−12mrω2)d3prdrdθdz=2πL(2mπh2β)3/2e12βmω2R2−1ω2mβThe probability of being of one particle in radius $r_0$ is: p(r=r0)=1Z∫e−βHd3pd3q=∫1Z2πL(2mπh2β)3/2eβmrω22rdr So I've thought that because, by definition, the...

Thanks.

okay, thanks. And the reason that the capacitor in the attached file is fully charged immediately after closing the switch is because there is no resistence (R = 0) so ## e^{-\frac{t}{RC}} ## goes to 0? And the circuit on this case is considered open because that the voltage on the...

Why when a capacitor is fully charged the circuit is acting like an open circuit? And what is the meaning of "fully charged", if the charge as a function of time equation is: $$ Q = CV_b [1 - e^{\frac{t}{RC}}] $$ so by this equation the charge on the capacitor will never reach exactly, Q = CV...

Thank you!.

Because the net flux is zero? Using Gauss law is not only when I got symmetry in the problem?

why is that?

Hello. I will be glad if someone can help me with this: I've a grounded conductive shell with outer radius of ## R_2## and inner radius ##R_1##. a charge ## Q ## is located inside of the shell, in distance ## r<R_1 ## , and a charge ## q ## is located in distance ## a > R_2 ## outside of the...

Thank you I hope I solved it right. On one of the next questions I've been asked to figure out what is the area charge density on the edges of the grounded shells if the shells were with width ## \Delta a ## and ## \Delta b ##. How can I figure out this? I know that the charge density with...

Why are those inverse? Is thia because the amount of charge that coming to the inner grounded conductive shell?

Problem Statement: conductive and grounded shells Relevant Equations: ## E\cdot dS = \frac {q}{\epsilon_0} ## Hi. I'll be glad if you can help me with this question.I've two conductive and grounded shells with radius 'a' and radius 'b' with their center on the same point. And another...

thank you.

Well I ment to write: $$ \int d \psi = \int_0^R ( \frac {1}{r^2} \int_0^R - \frac {\rho_0 r^3}{R \varepsilon _0} dr) dr $$ If that what you ment.also, here is what I'm getting: $$ \psi = - \int \frac {1}{r^2} ( \cdot \frac {\rho_0 r^4}{4R \varepsilon _0} + C ) dr $$ ** Do I've to work with...

Actually I tried that, but I didn't knew what I'm doing: $$ \frac {\partial}{\partial r } (r^2 \cdot \frac {\partial \psi}{\partial r})=- \frac {\rho_0 r^3}{R \varepsilon _0} $$ $$ \int \partial \psi = \int_0^R ( \frac {1}{r^2} \int_0^R - \frac {\rho_0 r^3}{R \varepsilon _0} dr) dr $$ I'm...

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