Timeline for Energy of a charge inside spherical conducting shell
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2014 at 14:07 | comment | added | Drake Marquis | The inner shell will not contribute energy because the electric field inside the shell is zero. The shell is an equipotential object, so potentian gradient vanishes. You only have to consider the energy stored in the electric field outside the shell and you should get the correct energy. | |
| Nov 20, 2014 at 8:29 | comment | added | levitt | Hi, I've added some correction. Is that right? | |
| Nov 20, 2014 at 7:08 | comment | added | Jold | The potential energy while it's inside the sphere is $qV_0$. At infinity the potential vanishes. By conservation of energy the work done in bringing the charge to infinity must be $W=qV_0$. (Otherwise where did the energy go?) | |
| Nov 20, 2014 at 6:07 | comment | added | levitt | I did not understand this "energy of the charge is kq2/R". If I assume I am bringing the charge q from infinity to the center of spherical shell, total work done is qV (V at the center of conducting shell due to the charges already present there.) How do you know V when it is uncharged. Charge gets induced on both the surfaces of the shell in the process which we need to take into account, right? | |
| Nov 20, 2014 at 5:23 | history | answered | Drake Marquis | CC BY-SA 3.0 |