charge on hollow sphere

2009-02-18 6:08 am
explain why charges reside solely on the outer surface of the conductor but not the inner surface in a hollow sphere.
更新1:

i now know that why charge prefer to stay on outer surface but i am not sure about one thing your "charge density" is charge per line/area/volume? we often say charges are distributed evenly, but in which sense, per line/area/volume?

回答 (1)

2009-02-18 3:35 pm
✔ 最佳答案
This is because the charges would reside themselves in a position where the charge density is as lowest as possible. In this position, the electrical potential energy of the system is the lowest. In physics, the lower the energy content of a system, the more stable it is.

This is simple, since the charges are with the same sign of charges. You should know that same sign of charge repel. Therefore, the charges will repel from each other as far as possible to attain a smaller electric potential energy.

As staying at the outer surface of the conductor has a larger separation compare with that staying at the inner surface, so they reside on the outer surface.


Imagine if you hate somebody, you will try to stay away from him as far as possible. If you two are put in a room, then one of you will be at one corner in the ceiling while the other will stay at the opposite corner.

2009-02-19 09:41:29 補充:
In this case, charge density is per area.

2009-02-19 09:41:42 補充:
Because it is the surface area of the sphere.
參考: Physics king


收錄日期: 2021-04-19 13:32:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090217000051KK01955

檢視 Wayback Machine 備份