Electric Field using Gaus's Law

In summary, the conversation discusses a uniformly charged spherical shell with a circular hole in its surface and the electric field at the center of the hole. It is mentioned that the field within the hole is a superposition of the field due to the original uncut sphere and a sphere with a negative charge density. The radius of the hole is smaller compared to the radius of the sphere, and the trick to solving the problem is treating the cutout as an infinite plane rather than a sphere. The conversation ends with a thank you for help.
  • #1
benndamann33
22
0
A uniformly charged spherical shell with surface charge density omega contains a circular hole in its surface. The radis of the hole is smalle compared with the radius of the sphere. What is the electric field at the center of the whole(Hint: the field within the whole is the suprposition of the field due to the original uncut sphere, plus the field due to a sphere the size of the hole with a uniform negative charge density -omega)

I don't undestand this because the electric field, if the sphere were solid, doesn't depend on the radius, it's just q/(epsilon_0). Any idea how this works out?
 
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  • #2
(Hint: the field within the whole is the superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of the hole with a uniform negative charge density -omega)

The original charge I guess is q, and one superimposes a field of similar charge density (, but negative, as in -omega) on the smaller sphere of the diameter of the hole.

For a sphere the electric field outside the sphere is the same as a point charge of the same magnitude.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html#c2
 
  • #3
Yeah, the trick to the problem was actually that you can't treat the cut out as a sphere, you treat it as a disk and inherently an infinite plane. The hint about radius cutout being very small was that you treat it as one dimensional. The radius cancels out if you treated them both as a sphere and without a specific radius for each I don't believe the problem would be solvable if you went about it as that superposition. But if you treat it as an infinite plane then it was solvable. Thanks for your help
 

1. How do you calculate electric field using Gauss's Law?

Gauss's Law states that the electric field through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. This can be written mathematically as E = Q/ε0, where E is the electric field, Q is the charge enclosed, and ε0 is the permittivity of free space.

2. What is the significance of using a Gaussian surface in Gauss's Law?

A Gaussian surface is an imaginary surface that is used to simplify calculations in Gauss's Law. It allows us to take advantage of symmetry in a system and make use of the fact that the electric field at every point on the Gaussian surface is the same. This makes the calculations much easier and helps us to find the electric field at a specific point without having to consider the entire charge distribution.

3. Can Gauss's Law be used for any charge distribution?

Yes, Gauss's Law can be used for any charge distribution as long as it is enclosed by the Gaussian surface. This includes point charges, line charges, and surface charges. However, the charge distribution must have some form of symmetry in order for the Gaussian surface to be useful in simplifying the calculations.

4. How does the electric field change with distance from a point charge using Gauss's Law?

According to Gauss's Law, the electric field at a point due to a point charge decreases inversely with the square of the distance from the charge. This means that as the distance increases, the electric field decreases. This relationship is the same as the inverse square law, which can also be derived using Coulomb's Law.

5. Can Gauss's Law be used in situations where there is no symmetry?

No, Gauss's Law can only be used in situations where there is some form of symmetry present. Without symmetry, it is difficult to determine the value of the electric field at a specific point using a Gaussian surface. In these cases, Coulomb's Law or other methods must be used to calculate the electric field.

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