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

In summary: In this case, you are assuming that the charge density of the conductor is uniform within the volume of the conductor, and so you could say that the charge density is zero at the surface of the conductor.In summary, the correct answers for this question are c and b.
  • #1
Slugger17
3
0

Homework Statement


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.

Homework Equations


none

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:
Physics news on Phys.org
  • #2
Slugger17 said:

Homework Statement


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.

Homework Equations


none

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

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?
 
  • #3
gabbagabbahey said:
Hmm... do you not think that Gauss' Law is a relevant equation for a question regarding Gauss' Law?:bugeye::wink:

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
 
  • #4
Slugger17 said:
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

Good.

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

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]

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

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.

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

It is most definitely 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.

So I'm now leaning towards b. and c. as the correct answers

Sounds good to me :approve:
 
  • #5
Yep that gave me the correct answer.
Cheers
 
  • #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.
 

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

What is Gauss's law?

Gauss's law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge.

How is Gauss's law mathematically expressed?

Gauss's law is expressed as ∮E•dA = Qenc0, where ∮E•dA represents the electric flux through a closed surface, Qenc represents the enclosed electric charge, and ε0 is the permittivity of free space.

What does Gauss's law state about the electric field?

Gauss's law states that the electric field lines emanating from a positive charge must terminate on a negative charge, and that the electric field lines are always perpendicular to the surface of a conductor.

Can Gauss's law be used to determine the electric field of a point charge?

Yes, Gauss's law can be used to determine the electric field of a point charge by considering a spherical surface surrounding the charge and using the fact that the electric flux through the surface is equal to the charge enclosed divided by the permittivity of free space.

Are there any limitations to Gauss's law?

Yes, Gauss's law has limitations in certain situations, such as when the electric field is not constant or when the charge distribution is not symmetrical. In these cases, other methods such as Coulomb's law must be used to determine the electric field.

Back
Top