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bolzano95

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Thread moved from the technical forums, so no Homework Template is shown

A sphere of radius

I'm trying to find the electric field distribution both inside and outside the sphere using Gauss Law.

We know that on the closed gaussian surface with spherically symmetric charge distribution Gauss Law states:

[tex]\frac{q}{ε_0}= \oint \vec{E} \cdot d\vec{A}[/tex]

1. Outside of sphere:

Logically, the charge outside of a sphere will be always on the Gaussian surface and it doesn't change, therefore the electric field outside of a sphere:

[tex]E=\frac{q}{4πε_0r^{2}}[/tex]

2. Inside of sphere:

Because the charge is symmetrically distributed on the surface and if I image a little sphere with radius r<a inside the sphere with radius r, the little sphere will have less charge on its surface.

[tex]E=\frac{q \ r}{4πε_0a^{3}}[/tex]

Is this explanation sufficient?

The problem I'm having is that in textbook is written and drawn that the electrical field inside a charged spherical shell = 0.

Isn't sphere = shell (just taking smaller shell for point 2)?

What am I missing?

*a*carries a total charge q which is uniformly distributed over the volume of the sphere.I'm trying to find the electric field distribution both inside and outside the sphere using Gauss Law.

We know that on the closed gaussian surface with spherically symmetric charge distribution Gauss Law states:

[tex]\frac{q}{ε_0}= \oint \vec{E} \cdot d\vec{A}[/tex]

1. Outside of sphere:

Logically, the charge outside of a sphere will be always on the Gaussian surface and it doesn't change, therefore the electric field outside of a sphere:

[tex]E=\frac{q}{4πε_0r^{2}}[/tex]

2. Inside of sphere:

Because the charge is symmetrically distributed on the surface and if I image a little sphere with radius r<a inside the sphere with radius r, the little sphere will have less charge on its surface.

[tex]E=\frac{q \ r}{4πε_0a^{3}}[/tex]

Is this explanation sufficient?

The problem I'm having is that in textbook is written and drawn that the electrical field inside a charged spherical shell = 0.

Isn't sphere = shell (just taking smaller shell for point 2)?

What am I missing?

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