- #1
Leo Liu
- 353
- 156
- Homework Statement
- Just a random question about electricity.
- Relevant Equations
- $${E} = {{kQ} \over {r^2}}$$
This page claims that "[t]he electric field outside the sphere is given by: ##{E} = {{kQ} \over {r^2}}##, just like a point charge". I would like to know the reason we should treat the sphere as a point charge, even if the charges are uniformly distributed throughout the surface of the conducting sphere. Also, is this statement valid for a non-conducting charged sphere? Thank you.
Update:
Reading this thread I realized that I can create a spherical Gaussian surface outside the conducting sphere and ##\oint{E}\mathrm{d}A = {E4\pi r^2} = {Q \over \epsilon_0} ##, which implies its electric field is the same as the electric field of a point charge, yet I still don't intuitively understand why it holds ture.
Update:
Reading this thread I realized that I can create a spherical Gaussian surface outside the conducting sphere and ##\oint{E}\mathrm{d}A = {E4\pi r^2} = {Q \over \epsilon_0} ##, which implies its electric field is the same as the electric field of a point charge, yet I still don't intuitively understand why it holds ture.
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