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In a sphere of uniformly distributed charge, we can use Gauss Law, but to do that, $E$ must be the same magnitude over the surface.

I understand why $E$'s magnitude will be the same on the surface, but to get flux, we also need cosa. My question is why $cosa$ is 1 for every $E$ and normal to the surface? I understood why it is 1 when charge was at the center as its $E$ would be in the exact same direction as radius (hence perpendicular), but in sphere of charges, 99% of the charges are not in the center so their $E$ must make different angle with the normal of the surface. Why are we not including this in the calculation? I get that for every charge(not in center), some of its $E$ lines cancel out, but couldn't imagine which lines actually not get cancelled. Visual representation would be appreciated.

Qmechanic
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Giorgi
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2 Answers2

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If the charge distribution has spherical symmetry, so must the electric field. You are right that the individual contributions to the E-field from the different parts of the sphere will point in all kinds of directions, but because of the spherical symmetry we know that all the "odd" directions must cancel.

Imagine if this were not the case, i.e. the E-field were not radial. Let us say that it had an azimuthal component. Now imagine that we rotate the entire charged sphere 180 degrees, so that the old north pole of our coordinate system now is the south pole as illustrated in the figure. Since the whole physical configuration has rotated, so must the field, and we notice that the azimuthal component of the field now points in the opposite direction as before. But due to the spherical symmetry of the charge distribution, nothing changed when we rotated the sphere, and we should therefore have the same electric field before and after applying the rotation. This shows that the electric field cannot have an azimuthal component, and the only remaining option is that it must be purely radial. enter image description here

Jakob KS
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Why are we not including this in the calculation?

You can but one does not necessary have to go back to first principles.

Consider the electric field produced by a spherical shell of charges.
Using symmetry arguments and Gauss you can ascertain that there is no electric field due to those charges within the shell and the electric field outside the shell is radial and produced as though all the shell of charges were concentrated at the centre of the shell as shown here.

If you are not satisfied with this approach then you can find the electric field from "first principles" as described in the Wikipedia article Shell Theorem were is done for another inverse square law - gravitation.

Then using the principle of superposition you can find the electric field due to a uniformly charges sphere of charge.

Farcher
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  • well, funny thing with sphericall shell charges is, inside , not sure why E is 0. I mean horizontal components of E for charges definitely cancel out, (wherever you stand), but vertical components not quite easily. If I stand at the very most left side, the vertical components of near charges will be bigger then vertical components of far away charges at this exact point. You could say that since there're more far away charges, their fields are more abundant(even though they're weaker) and overally, they cancel, but still why exactly cancel out ? – Giorgi Jun 15 '23 at 08:34
  • Gravitational field intensity inside a hollow sphere and the same for an electric field as the force follows and inverse square law. – Farcher Jun 15 '23 at 08:53