E-field & hollow non-conducting sphere

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    E-field Sphere
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The discussion centers on the electric field (E-field) inside a hollow non-conducting sphere with a uniform surface charge Q. According to Gauss's Law, the E-field inside the sphere is zero, even when considering a small hole that penetrates the material. However, within the hole, the E-field is calculated to be one half of Q divided by the square of the radius R, indicating a gradual decrease in the E-field from Q^2/R to zero as one moves through the hole. The transition in the E-field occurs over a distance proportional to the diameter of the hole. This analysis confirms that the E-field behaves consistently with Gauss's Law, demonstrating the complexity of electric fields in non-conducting materials.
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I know that the E-field around a hollow non-conducting sphere charged with Q charge comes immediately from Gauss' Law but I'm wondering what the situation is if we somehow go inside the material, we make a very small hole through the material of the sphere and go inside it. What would there be the E-field?
 
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If the charge is a uniform surface charge, then the field inside would be zero by Gauss's law.
 
Even inside the hole when being half outside, half inside the sphere?
 
Inside the hole, at radius R, the E field is one half Q/R^2.
The E field goes from Q^2/R to zero as r decreases through the hole.
The decrease in E takes place in a distance of the order of \Delta r=d, where d is the diameter of the hole.
 
So right on the surface it would be E = Q/ (4R^2 pi * epsilon_0), but when going through the hole, right inside the sphere it is already 0. Does it follow from Gauss' Law to be more obvious? E.g.: we choose the Gaussian sphere with radius R such that it goes through the charge layer. Thus, the total amount of charge inside the Gaussian sphere is Q/2 so E inside the hole is indeed 1/2 * Q/ (4R^2 pi * epsilon_0).
 
That happens anywhere at the surface, not just at the hole.
 
Interesting, thanks!
 
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