Timeline for Electrostatic Potential Energy
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2012 at 21:25 | comment | added | Ron Maimon | @The-Ever-Kid: The integral of the force depends on the sign of the test charge ("the polarity" as you call it), but the potential energy is not the potential, the potential energy is the test charge times the potential, so there is no paradox. | |
| Apr 30, 2012 at 17:00 | comment | added | The-Ever-Kid | the derivation states that the sign of U(r) will depend on the polarity i agree as physically this is the case | |
| Apr 30, 2012 at 16:58 | comment | added | The-Ever-Kid | no im concerned that the previous results dont hold true when we add the cos theta term to it | |
| Apr 30, 2012 at 16:09 | comment | added | John Rennie | Are you concerned that there is an ambiguity in the sign of $U(r)$ i.e. that it's positive if the charges are the same but negative if they're different? | |
| Apr 30, 2012 at 15:54 | comment | added | The-Ever-Kid | okay tell me if both the charges are the same then what will cos theta be? | |
| Apr 30, 2012 at 15:53 | comment | added | Vijay Murthy | I am confused. Why does the result change? | |
| Apr 30, 2012 at 15:38 | comment | added | The-Ever-Kid | But you see when i do that the equation becomes $(kCos\theta q q_o)/(r)$ now when the charges are the same then the force and the displacement are opposite to each other so $Cos\pi = -1$ this changes the result itself BTW the angle theta is the angle between the force on the charge and the direction of incremental displacement right? BTW you're Indian too? | |
| Apr 30, 2012 at 15:32 | history | answered | Vijay Murthy | CC BY-SA 3.0 |