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There was this question that i saw in a book and it also had an answer given.

The Question was:

If there were only one type of charge in the universe, then:

  • $\phi = \oint \boldsymbol{E}\cdot \partial\boldsymbol{A} \neq 0 $ on any surface.
  • $\phi = \oint \boldsymbol{E}\cdot \partial\boldsymbol{A} = 0$ is the charge is outside a surface.
  • $\phi = \oint \boldsymbol{E}\cdot \partial\boldsymbol{A} $ will not be defined.
  • $\phi = \oint \boldsymbol{E}\cdot \partial\boldsymbol{A} = \frac{q}{\epsilon_o}$ if the charge is inside the surface.

The answers were given as option second and the last one. (Gauss' law statements)


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Hi The-Ever-Kid and welcome to Physics Stack Exchange! Our homework tag is for more than actual homework questions, so this does qualify. To improve your question, consider this: what exactly is confusing you about why these particular options are the right ones and the others are wrong? – David Z Apr 28 '12 at 7:31
If there was only one kind of charge then there would be no nutrality in the universe i.e no bipolarity . if that was the case the where would the positive test charge go when it is repelled by the source charge . i mean i saw this video on YouTube where the teacher said that the field lines of an isolated positive charge go to infinity where the opposite charge is present so as to ensure conservation of charge. but here theres no negative charge so where would the field lines go to. – The-Ever-Kid Apr 28 '12 at 8:20
This is indeed completely ridiculous--- the "what if" has no answer, and this is a problem of "read my mind". – Ron Maimon May 1 '12 at 13:40
@RonMaimon Meaning? – The-Ever-Kid May 12 '12 at 20:00
up vote 0 down vote accepted

At the risk of being slapped for answering a homework question: it's irrelevant what the net charge is.

Consider the following model. Take a point charge and surround it with a spherical shell containing an equal and opposite charge at some large and effectively infinite distance. Now start reducing the charge of the shell, ultimately to zero, and ask what if any difference this makes to the field at a finite distance from the point charge, and whether there is any point at which Gauss' law would cease to apply.

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thanks,and yeah there isn't a point like that and neither is there a change in the field 'coz the internal field due to the shell is independent of the charge of the shell.the contributing firld of the shell is zero. – The-Ever-Kid Apr 28 '12 at 9:13

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