Electric Field using Gaus's Law

AI Thread Summary
The discussion centers on calculating the electric field at the center of a circular hole in a uniformly charged spherical shell. The key insight is that the electric field in the hole is the superposition of the field from the original sphere and the field from a smaller sphere with a negative charge density equal to that of the hole. Participants clarify that the hole should be treated as a disk, approximating it as an infinite plane due to its small radius compared to the sphere. This approach simplifies the problem, allowing for a solvable configuration. Understanding this treatment is crucial for accurately determining the electric field in the specified region.
benndamann33
Messages
20
Reaction score
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?
 
Physics news on Phys.org
(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
 
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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top