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

AI Thread Summary
The discussion revolves around applying Gauss's Law to various physics problems, particularly in non-spherical geometries. It clarifies that charged particles do not always follow electric field lines, with an example of a free electron in a parallel plate capacitor. Gauss's Law can be applied to any shape, but its utility is maximized in symmetrical situations, like spheres, where the electric field depends solely on distance from the center. The conversation emphasizes the importance of using Gaussian surfaces to analyze electric fields and charge distributions, especially in complex scenarios involving hollow conductors. Overall, understanding symmetry and charge distribution is crucial for solving these physics problems effectively.
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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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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
 
I do not understand how to do 5 or 6 still. Can you be of more assistance?
 
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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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 4\pi r_{inner}^2. The charge density is \frac{-q}{4\pi r_{inner}^2} The charge density on the outer surface is \frac{q}{4\pi r_{outer}^2}
 
you show this with guass's law as above
 
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