# Is the field used in the Gauss Law affected by the outer charge configuration?

Following is the Gauss Law:

$$\oint E\cdot\mathrm dA = \frac {q_\text{enc}}{\epsilon_0}$$

Now consider the following scenario:

The electric field in the following case, at the sphere, is due to $$q_1+q_2$$ or $$q_1+q_2+q_3$$. Because according to Gauss law, the RHS. the term is $$q_1+q_2$$. So if we calculate the electric field by dividing flux with the area of Gaussian surface, the electric field is due to all charges or just $$q_1$$ and $$q_2$$.

• As pointed out by @RogerVadim, the contribution of $q_3$ is zero, so Gauss' Law is satisfied either way. In general Gauss' Law is not enough to determine the electric field. Example: A spherical Gaussian surface with one charge at center, and one outside of the sphere. The incorrect field $\hat{{\bf r}}q_1 /r^2$ satisfies Gauss' Law and so does the real solution. Gauss' Law is more like a constraint that the correct field must satisfy. Oct 15, 2021 at 10:45
• If you divide the flux with the area of Gaussian surface you don't find generally the electric field unless you have symmetry. For example, the flux of $\:q_3\:$ through the spherical area is zero so dividing by the sphere area you find zero electric field, untrue. So what is the question here ??? Oct 15, 2021 at 12:29

The electric field can be taken both due to all the charges or only to the charges encolosed by the Gaussian surface, whereas the charge in the right-hand-side of the Gauss law is only the enclosed charge.

In conventional statement of the law one usually implies the field due to all the charges. However, the net flux due to the charges outside of the Gaussian surface is zero: the field lines due to these charges enter the surface on one side and exit on the other.

Remark: Applying Gauss law for an arbitrary charge configuration is however tricky, since the field is not constant on the Gaussian surface. The commonly treated examples usually have symmetry (mirror, spherical, cylindrical, etc.) which allows simplifying the surface integration.

The electric field in the following case, at the sphere, is due to $$q_1+q_2$$ or $$q_1+q_2+q_3$$.

It is due to all three charges.

The electric field at a point on the sphere, or for that matter at any any location in the diagram, equals the vector sum of the electric field contributions of all three point charges. See FIG 1 below showing the electric field lines of all three charges (assumed positive) at point A of the sphere, which vectorially sum up to equal the field at that point.

So if we calculate the electric field by dividing flux with the area of Gaussian surface, the electric field is due to all charges or just $$q_1$$ and $$q_2$$.

Only if you have symmetry, such as if the enclosed charge was a point charge at the center of the sphere, and there were no field contributions from external charges, can you divide the net flux by the total area to get the magnitude of the electric field, because the field would be the same at each point on the surface. The net flux is, however, only due to the charge enclosed. $$q3$$ contributes no net flux because the total flux of the field lines into the space enclosed by the surface equals the total flux of the lines out of the space for a net flux of zero. See FIG 2.

Hope this helps.

I really think that the best way to answer this question is to think of counting electric field lines passing through a surface and that is what the integral $$\displaystyle \oint \vec E\cdot\mathrm d\vec A$$ is doing.

An electric field lines due to charges $$q_1$$ and $$q_2$$ passes through the closed surface only once and contributes to the integral however, if an electric field line from charge $$q_3$$ passes through the surface into the enclosed volume it will subsequently pass through the surface out of the enclosed volume. Thus the net flow of such an electric field line thorough the closed surface is zero and so its contribution the the integral is zero which in turn means that any charges outside the closed surface can be disregarded.

To make things more clear for you. the traditional way of solving for fields using gauss law alone relies on spherical symmetry. this is because if there is spherical symmetry about a spherical gaussian surface, then $$\mathbf E\cdot d\mathbf a$$ becomes $$|E||da|$$ as $$\mathbf E$$ is in the direction of $$d\mathbf a$$ throughout the surface (parallel vectors) so the dot product ($$\vec a\cdot \vec b = |a||b|\cos(\theta)$$). $$\theta$$ is zero for spherical symmetry so it reduces to scalar multiplication.

That is not enough to solve for fields. because it is symmetric about the origin we know that the magnitude of the E field is the same for all points on the surface as its the same distance away. therefore we can say $$\mathbf E$$ is independent of the integral so it's E* (integral da) and then we can solve for the fields.

However with a charge outside the surface or for any charge configuration at all that isn't spherical we cannot find the field via this method.

The equation states that the flux is dependant on the charge inside. this is true if any charges are outside aswell. Only int E.da is independent of charges outside not the $$\mathbf E$$ field itself.

Hope it clears that up.