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Two gaussian surfaces for a charge outside a charged spherical shell

I feel like I am missing something simple here, but how exactly are gaussian surface determined? Looking at the case of a charge outside a sphere, why don't we pick a gaussian surface not including the sphere itself. Would then the flux inside this sphere be zero? Would that not mean the electric field is zero, since the flux vanishes (like in the case inside a spherical shell)? Also if we take an infintesmally small surfaces, doesnt that imply that the electric flux is always zero except inside a distribution of charge (assuming continuous distribution of elctric charge)?

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2 Answers 2

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When applying Gaussian surfaces to problems of this type, generally one tries to find surfaces that exploit the symmetry of the situation to simplify the calculation of the E-field. If the surface doesn't include the charges (say outside the sphere above), it means that the net flux through the surface is zero, not necessarily that the E-field is zero everywhere. It would be more difficult to determine the actual E-field since the symmetry of the sphere is not being exploited.

You can find some more discussion and figures here: Gauss's Law and Flux through Sphere vs. Shell

I hope this helps.

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  • $\begingroup$ For a simplified calculation, you would like the E field to be either constant or zero on all parts of the Gaussian surface. $\endgroup$
    – R.W. Bird
    Commented Oct 20, 2020 at 13:06
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If you want to know the field at $\vec r_p$, you need to pick a Gaussian surface that goes through $\vec r_p$ AND is such that the $\vec E$-field is constant or $0$ everywhere on the surface so that $$ \oint \vec E\cdot d\vec S = \vert \vec E\vert \times S\, . \tag{1} $$ Usually, the choice of the surface is dictated by the symmetries of the charge distribution since from the symmetries of the charge distribution one can usually infer the direction of the field.

Thus, spherically symmetric charge distributions will result in spherically symmetric fields $\vec E$ depending only on the radial distance from the center of the distribution. Choosing a spherical Gaussian surface then makes $d\vec S$ always parallel (or antiparallel) to $\vec E$ at every point on the sphere and (1) holds. There then remains to evaluate the charge enclosed by that Gaussian sphere.

The same logic applied to an infinitely long cylindrically symmetric distribution shows that the field will be perpendicular to the cylinder and can only depend on the distance to that cylinder, so that a cylindrical Gaussian surface would be warranted in that case.

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