- #1
pardesi
- 339
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how does coloumb's law accomadate this
the field on a say a uniformly charged shell.any point on the shell is charged so does that point contribute to the field there itself ye s the question will look stupid but does it?
further does gauss's law blow up when i take the charged sphere itself to be my gaussian surface
the field on a say a uniformly charged shell.any point on the shell is charged so does that point contribute to the field there itself ye s the question will look stupid but does it?
further does gauss's law blow up when i take the charged sphere itself to be my gaussian surface