How Does Gauss's Law Apply to Non-Spherical Geometries?

In summary: For number 6, at R/2, you have a spherical Gaussian surface with radius R/2. The total charge inside this surface is q. So you can use Gauss's law to find the electric field at R/2.At 2R, you have a spherical Gaussian surface with radius 2R. The total charge inside this surface is 3q, since the charge is spread uniformly over the surface of the shell. Again, you can use Gauss's law to find the electric field at 2R.
  • #1
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Ok so background story, everyone failed an exam in my physics 2 class, so as a way to gain points we had to complete a packet of 35 physics2 problems, and i am having issues with only 6 problems.

1. Does a charged particle in an electric field always move along a field line? if yes, explain; if no, give an example where it is clear that the charged particle is not moving long any field line.

I'm guessing the answer is NO. But i cannot think of an example

2. Can you use Gauss's law to etermine the field of a uniformly charged cube, by drawing a cubical gaussian surface around it? if not, why not?

I'm guessing the answer is NO because gauss's law only applies to spherical objects and large flat surfaces?

3. a charge q1=+1.2C is at the origin of a reference frame, and a charge q2=+3.7C is at x=3.3m, y = 0. Compute the point or points (x,y) at which the electric field is zero.

*lost*

4. An certain flat surface is at right angles to a uniform electric field. the surface is then tilted by 60 degrees in some direction, while the electric field remains unchanged. by what percentage did the flux through the surface change as a result of the tilting?

I know you do not need to know the actual values for the electric field and surface are, because they will cancel out at the end. how ever how do you do this problem?

5. a conducting hollow sphere carries a zero net charge. in the center of the sphere there is a point charge of +5.3C. the inner and outer surfaces of the conducting sphere are concentric, and their radius's are 3.2m and 3.7m, respectively. explain why there will be uniform surface charge density on he inner surface, and also on the outer surface, and compute these two surface charge densities.

once again *lost*

6. a point charge q is at the center of a spherical shell of radius R carrying charge 2q spread uniformly over its surface. write expressions for the electric field strength at R/2 and 2R.

once again...*lost*
 
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  • #2
1. think of a free electron traveling at v, going through the plates of a parallel plate capacitor

2. the answer is no, but not how you explain it. Guass's law always applies, but is most usefull where the symmetry of the problem allows for simplifications, in the spherical case, as wevrything only depnds on r, you can assume the field only depends on r & so teh magnitude is constant for a given r. The symmetry is not there for the cubic case

3. compute the field from each cahrge & add them together, where is the 0?

4. geomtery draw a picture

5. use a few spherical gausssian surfaces, very close to the surfaces of the hollow sphere (and either side)

6. more spherical guassian surfaces
 
  • #3
I do not understand how to do 5 or 6 still. Can you be of more assistance?
 
  • #4
so due to the symmetry the fields will be radial only, so use spheres as your guassian surfaces

put one inside the hollow sphere, that contains only the point charge - use guass's law, what is the field?

now put one between the inside & outside surfaces of the hollow sphere, you know the field must be zero as it is a conductor - what charge must be on the inner surface to make it so?

and so on
 
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  • #5
For number 5, the charge induced on the surface should be equal to the charge that is inducing the charge on the surface. So if you have a charge q in the center, then the inner surface will have a charge -q induced on it. The area of the surface is [tex]4\pi r_{inner}^2[/tex]. The charge density is [tex]\frac{-q}{4\pi r_{inner}^2}[/tex] The charge density on the outer surface is [tex]\frac{q}{4\pi r_{outer}^2}[/tex]
 
  • #6
you show this with guass's law as above
 

Related to How Does Gauss's Law Apply to Non-Spherical Geometries?

What is Gauss's Law and how is it related to Electric Potential?

Gauss's Law is a fundamental law in electricity and magnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It states that the electric flux through any closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space. This law is closely related to Electric Potential, as the electric potential at a point is the amount of work needed to move a unit positive charge from infinity to that point, and it is directly proportional to the electric field.

Can Gauss's Law be applied to any charge distribution?

Yes, Gauss's Law can be applied to any charge distribution, as long as the charge is enclosed within a closed surface. This law is valid for both point charges and continuous charge distributions.

What is the difference between Gauss's Law and Coulomb's Law?

The main difference between Gauss's Law and Coulomb's Law is that Gauss's Law applies to a closed surface, while Coulomb's Law applies to a point charge. Coulomb's Law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Gauss's Law, on the other hand, relates the electric flux through a closed surface to the charge enclosed within that surface.

What is the equation for calculating Electric Potential?

The equation for calculating Electric Potential is V = kQ/r, where V is the electric potential, k is the Coulomb's constant (8.99 x 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge.

How is Electric Potential related to work and energy?

Electric Potential is directly related to work and energy. The electric potential at a point is the amount of work needed to move a unit positive charge from infinity to that point. This means that the change in electric potential between two points is equal to the work done in moving a charge between those two points. Additionally, electric potential energy is the energy a charged particle possesses due to its position in an electric field, and it is directly proportional to the electric potential at that point.

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