# Why is a symmetric charge distribution required for application of gauss's law?

I don't want a rigorous proof of gauss's law but, I cant understand why only symmetric charge distributions are found their application in Gauss's law.

A source says;

Now imagine a sphere that is not uniformly charged. Suppose there is more charge on one side of the sphere than the other. This is a situation where Gauss' law probably won't be very useful in calculating the electric field.

Why is this so?? Reference https://www.physicsforums.com/threads/gauss-law-symmetry-of-charge-distribution.379052/

• More on Gauss's law & symmetry. – Qmechanic Mar 12 '18 at 14:22
• Gauss's law involves an integral that is usually easy to do for symmetric charge distributions. For non-symmetric charge distributions, Gauss's law still applies, but the integral may be intractable. – David White Mar 12 '18 at 14:23

Lets suppose that you have a complicated distribution of charges such that there is no symmetry in the distribution. So, basically our Electric field $\vec E$ will also be a complicated function. Now, Gauss' Law states :

$$\oint_S \vec E\cdot \vec {ds} = \frac{q_{enc}}{\epsilon_0}$$

If there is no symmetry in $\vec E$, we cannot easily integrate the left hand side, and thus Gauss' Law can't help us.

So basically, it is applied in cases where the Electric field is somehow constant so that it can be pulled out of the integration.

• Why is a surface integral used in the equation? How is it different from the usual integral in calculus . Pardon me if this is a silly question – susan J Mar 12 '18 at 14:30
• So you mean to say that if the distribution is not uniform it'll result in a non constant field which would make it difficult to chose a gaussian surface . – susan J Mar 12 '18 at 14:31
• Yes. And as for the surface integral, let me explain it a bit intuitively. Suppose you have a point charge inside of a sphere. Now, the electric field lines will obviously have to pass through the surface of the sphere. You might ask, why not the volume of the sphere? A general proof theorem from Vector calculus actually relates a volume integral to a surface integral. And it is in this surface integral that symmetrical cases can be easily handled. So the theorem is generally stated in this form. – Yuzuriha Inori Mar 12 '18 at 14:41
• I'm really sorry to be such a bother but what are volume integral and surface integral, I really want to understand how they are different from usual integrals in calculus – susan J Mar 12 '18 at 14:42
• Volume integrals are integrations that are done over the whole volume that is under consideration. It is actually a triple integration that has to be integrated over a volume. The same thing goes for a surface integral, except that it applies to surfaces or the boundaries of volumes at hand. Vector calculus is a bit more involved than normal calculus, but the methods are same. Just a small change in notation to make things look more clearer and intuitive. You can check it out here -> betterexplained.com/articles/category/math/vector-calculus – Yuzuriha Inori Mar 12 '18 at 14:49

Gauss's law is always valid. For practical use, you need a symmetric situation where the magnitude of the electric field is constant on the integration surface so that the surface integral can be evaluated by simply multiplying the constant field strength with the total area $A$ of the surface: $$\int_S \vec E d\vec a= E\cdot A=\frac {Q}{\epsilon}$$