# How is adding a charge outside a Gaussian surface consistent with Gauss' Law?

The electric field of an infinite line charge with a uniform distribution of charge could be calculated using Gauss's law. If you then add a charge outside of the constructed Gaussian surface, the amount of charge enclosed $$Q$$ remains the same. Therefore the calculation of the electric field $$\mathbf E$$ should be the same as well.

But why does the charged object not influence the electric field?

• the Q enclosed remains the same, but the configuration of the electric field cutting through your surface will not be the same Aug 6, 2021 at 3:24
• gauss law only calculates the contribution from charges inside the surface, if you have charges outside you have to add the E contributed by the external charges.
– user65081
Aug 6, 2021 at 3:25
• Does this answer your question? Electric field inside a conductor non zero
– Babu
Aug 6, 2021 at 13:27
• see
– Babu
Aug 6, 2021 at 13:27

If you then add a charge outside of the constructed Gaussian surface, the Q enclosed remains the same. Therefore the calculation and the E calculated would be the same as well.

This is incorrect. Gauss’ law only tells you about the net flux. It does not tell you about the E field. So, indeed, adding a charge outside the surface will not change the net flux, but it can and does change the field.

The only way that you can use Gauss’ law to go from net flux to field is in cases where there is a high degree of symmetry such that there can be only one possible value of the field that both satisfies Gauss’ law and the symmetry. In your example, there is such symmetry without the charge, but adding the charge removes the symmetry.

After removing that symmetry there are many configurations of field that are consistent with the net flux, but only one of those is the actual field. Which one cannot be determined by Gauss’ law alone.

• Ok, that makes sense thank you. Aug 6, 2021 at 3:43
• Wait, why can gauss' law be used only when the two sides are symmetrical? Aug 6, 2021 at 3:59
• Gauss law is only useful when there is symmetry. Aug 6, 2021 at 6:58
• @TheBluestSun Gauss’ law gives you knowledge about the net flux (which is one number). To get knowledge about the field (which is three numbers at each point) requires a very simple relationship between the net flux and the field. This is achieved when the field has some simple symmetry.
– Dale
Aug 6, 2021 at 11:10
• @TheBluestSun You can use it when there's no symmetry, just not for this type of problem. Symmetry just puts an additional constraint on what the answer could be. If I tell you, "Hey, I have 20 apples in two baskets, how many apples is in each of the baskets?", you can't really answer. But if I tell you that the number of apples is the same in each (a symmetry of sorts), or that one contains 3 times more than the other (a more general type of constraint), you can deduce the answer. But if you just care about the total number of apples, you don't need the additional info. Aug 6, 2021 at 18:40

Therefore the calculation and the E calculated would be the same as well.

No. It would not be. Even though $$Q_\text{enclosed}$$ remains the same (and so does the electric flux), the electric field will be affected.

I assume you're getting this wrong by using the "simplified" Gauss' Law. In the case of an infinite uniform conductor, you can correctly make the simplification that $$\displaystyle \oint \mathbf{\vec E}\cdot \mathrm d \mathbf{\vec A}=EA$$. This is only true because $$\bf\vec E$$ is constant (this is a neat case of symmetry). However, by introducing a charge outside the infinite conductor, the net electric field is no longer constant, in which case $$\displaystyle \oint \mathbf{\vec E}\cdot \mathrm d \mathbf{\vec A}\neq EA$$.

But why does the charged object not influence the electric field?

It does influence the field. It does not influence the net flux, but just because the net flux is unaffected doesn't mean the electric field isn't.

• Thank you. And in what cases am I safe to use the simplified "EA"? Aug 6, 2021 at 3:49
• when $\vec E \cdot d\vec A$ is constant @TheBluestSun Aug 6, 2021 at 3:50
• this isn't the only case, but usually if the $\vec E$ field is constant, the angle between $\vec E$ and $d\vec A$ are constant, and the gaussian surface is symmetrical, then $\Phi=EA$ Aug 6, 2021 at 3:52
• @TheBluestSun One can't use the simplification $EA$ if $E$ is not well defined. If $\vec E$ has more than one value, which do you plug into $EA$? One can use the "average" $\vec E$ (for the proper definition of average $\vec E$), but calculating the average $\vec E$ involves calculating the integral, so that isn't really a simplification, just an alternative formulation. Aug 7, 2021 at 1:55