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Which of the following statements about Gauss's law are correct?

  1. Aug 3, 2012 #1
    1. The problem statement, all variables and given/known data
    There may be more than one correct choice
    a. Only charge enclosed within a Gaussian surface can produce an electric field at points on that surface.
    b. If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.
    c. The electric flux passing through a Gaussian surface depends only on the amount of charge inside that surface, not on its size or shape.
    d. If there is no charge inside of a Gaussian surface, the electric field must be zero at points of that surface.
    e. Gauss's law is valid only for symmetric charge distributions, such as spheres and cylinders.

    2. Relevant equations
    none

    3. The attempt at a solution
    My answer is B but it is apparently incorrect. So I'm not sure if it's wrong or there is another that is also correct
     
    Last edited: Aug 3, 2012
  2. jcsd
  3. Aug 3, 2012 #2

    gabbagabbahey

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    Hi Slugger17, welcome to PF!:smile:

    Hmm... do you not think that Gauss' Law is a relevant equation for a question regarding Gauss' Law?:bugeye::wink:

    How about you give your reasoning behind ruling out a,c,d, & e; so we can see where you are going wrong?
     
  4. Aug 3, 2012 #3
    Yeah good point.

    a. if a Gaussian surface is placed in an electric field, the field passes through it so there must be a field at points on the surface. There's a lecture slide that illustrates this so I'm confident that a. is false

    c. I'm starting to think this is true since [itex]\phi[/itex]=[itex]\oint[/itex][itex]\vec{E}[/itex].d[itex]\vec{A}[/itex]=[itex]\frac{q}{\epsilon_{0}}[/itex] is the formula for flux

    d. this seems to me to be a restatement of a. so by the same reasoning I believe this to be false

    e. in our lectures we used superposition to apply gauss's law to asymmetric shapes so this must be false.

    So I'm now leaning towards b. and c. as the correct answers
     
  5. Aug 3, 2012 #4

    gabbagabbahey

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    Good.

    I agree. I think the wording on this is a little tricky, since the calculation of [itex] \oint \mathbf{E} \cdot d \mathbf{A}[/itex] certainly depends on the size and shape of the surface you are integrating over (in general), but the net flux can always be determined entirely from the charge enclosed since [itex] \Phi = \frac{ Q_{ \text{enc} }}{ \epsilon_0 }[/itex]

    Right again. Only the total flux over the surface must be zero, the electric field can be non-zero at any given point the surface.

    It is most definately false. Gauss' law is always true (it is one of Maxwell's equations). It is only useful for determining E when applied to charge distributions with certain types of symmetry (when you used the superpostition principle in class, I'm sure each of the component shapes had such symmetries), but it is always true.

    Sounds good to me :approve:
     
  6. Aug 4, 2012 #5
    Yep that gave me the correct answer.
    Cheers
     
  7. Aug 4, 2012 #6
    Sentence b sound ambiguous to me.

    "b. If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."

    Someone could interpret the words "completely inside" to mean that this sentence allows you to have a little environment in a hole that's inside a conductor. In that case, if there is an unknown charge distribution inside that hole, you could have a nonzero electric field on the Gaussian surface.

    If they mean that the surface is completely immersed in the conductive material, then b is true.
     
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