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Suppose there is a uniformly charged solid sphere of radius $R$ and we choose a gaussian surface of radius $r$ centered about the center of the solid uniformly charged sphere where $R>r$.

Then is the value of $E$ (electric field) due to the part of solid charged sphere outside the gaussian surface zero at a 'particular point' on the gaussian surface?

OR: Is the net value of $E$ on the gaussian surface as a whole has zero value, while the value of $E$ on a particular point on the gaussian surface has a finite value? And they cancel out each other when we consider the gaussian surface as a whole?

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  • $\begingroup$ Some formatting may make this clearer... $\endgroup$ – user207455 Jul 7 at 7:59
  • $\begingroup$ please suggest some $\endgroup$ – p0803 Jul 7 at 8:00
  • $\begingroup$ Si Thomas has showed you... much clearer and easy to read. $\endgroup$ – user207455 Jul 7 at 9:10
  • $\begingroup$ Which do you mean, a sphere with a inform surface or volume charge? $\endgroup$ – my2cts Jul 7 at 10:48
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If I'm understanding the question correctly, this depends on how you choose the Gaussian surface. If this surface is a sphere centered about the center of the charged sphere, then the E-field due to the region of the charged sphere lying outside the Gaussian surface is zero everywhere on the surface. This follows by Gauss' Law (obviously none of the charge lying outside the Gaussian surface is enclosed by this surface) and symmetry arguments (the E-field must have the same value everywhere on the surface since no direction is special).

If the Gaussian surface is not centered about the center of the charged sphere, Gauss' Law still holds so the integral of the E-field on the Gaussian surface is still zero, but we can no longer make use of symmetry. The E-field due to the region outside the surface is no longer necessarily zero everywhere on the surface.

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  • $\begingroup$ you mean that at any point on the gaussian surface centered about the center of the charged sphere the field due to the portion of the charged sphere outside the gaussian surface is 0. $\endgroup$ – p0803 Jul 7 at 9:20
  • $\begingroup$ And would be a net electric field on a point on the gaussian surface if the volume charge density of the charged sphere was not constant ? $\endgroup$ – p0803 Jul 7 at 9:24
  • $\begingroup$ The answer to your first question is yes. As for your second question, as long as the spherical symmetry is retained, the E-field would still be zero. For instance, if the charge density was a non-uniform function, but a function of only the distance from the center of the sphere (and not the direction), the charge distribution would still be spherically symmetric, and the E-field on the Gaussian sphere due to the charge outside it would be zero. $\endgroup$ – Puk Jul 7 at 9:29
  • $\begingroup$ And if say, volume charge density was the function of say, the x co ordinate, the value of E would not be zero ' at a point ' on the gaussian surface due to the outside portion , as the value of E would not be uniform on the gaussian surface as a whole and the symmetry would not be retained , am I right? $\endgroup$ – p0803 Jul 7 at 9:33
  • $\begingroup$ That is correct. $\endgroup$ – Puk Jul 7 at 9:36

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