Using gauss law to determine field Let us consider a $+q$ charge and we are trying to find out electric intensity $E$ at a distance $r$ from $+q$.The conventional way is this:
We take a gaussian sphere of radius $r$. We know the electric flux of this sphere is $\frac{q}{\epsilon_0}$. Then they use the integral definition of flux $\int E dS$ and they say that $E$ here is same due to symmetry.BUT while deriving flux of sphere using integration,we say $E$ is same because of formula $\frac{kq}{r^2}$ which is what we want to prove now. So we are using circular logic here.
I need to know how electric field is constant as mentioned in the conventional solution without using symmetry(i dont understand why electric field has to be constant for symmetric figures).
 A: Its not really circular reasoning since using gauss law to derive the field of a point charge must have preconceived knowledge of what a point charge actually means. Otherwise we can't use gauss law to find the fields. Mathematically I guess we can say that a point charge by definition has a form  $Q\delta^3(r)$ which is a point of divergence at a single point. Therefore it must be symmetric.
A: I assume that you are concerned with the field due to point charge, $Q$.
You can either use Coulomb's law
$$\mathbf E =\frac Q {4\pi \epsilon_0 r^2}\ \mathbf {\hat r}$$
or you can use Gauss's law plus symmetry. If you use this method, you don't have to use $\mathbf E =\frac Q {4\pi \epsilon_0 r^2}\ \mathbf {\hat r}$ in order to establish that the field is the same at all points on the Gaussian sphere. Instead, you simply say that, for a Gaussian sphere centred on $Q$, all points on the sphere are equidistant from $Q$, and since there is nothing in relation to $Q$ that distinguishes any point on the sphere from any other, the field must be the same in magnitude at each point. What's more, the field must be radial, as symmetry couldn't be maintained otherwise (as shown, I believe, by the hairy ball or hedgehog theorem). Hence using Gauss's law we have the same result as before:
$$\mathbf E =\frac Q {4\pi \epsilon_0 r^2}\ \mathbf {\hat r}$$
The two methods are, of course, equivalent, because for stationary charges you can derive Gauss's law from Coulomb's law. [In fact Gauss's law is more general than Coulomb's law because it applies even if the charges inside the Gaussian surface are moving.]
A: If you assume Gauss's law is valid, you can derive Coulomb's law using the spherical symmetry of the problem.
Gauss's law in integral form reads:
$$\int_S\boldsymbol{E}\cdot\hat{\boldsymbol{n}}dS=\frac{q}{\epsilon_0}\tag{1}$$
Where $\hat{\boldsymbol{n}}$ is the unit normal normal vector of the surface $S$.
The spherical symmetry suggests that the appropriate choice for $S$ would be a spherical surface with radius $R$  centered in the position of your charge, thus $\hat{\boldsymbol{n}}=\hat{\boldsymbol{r}}$. Also, by symmetry the field can only depend on the distance $r$ from $q$, and it must be radial:
$$\boldsymbol{E}(\boldsymbol{r})=E(r)\hat{\boldsymbol{r}}\implies\boldsymbol{E(\boldsymbol{r})}\cdot\hat{\boldsymbol{r}}=E(r)\tag{2}$$
Thus, combining $(2)$ and $(1)$:
$$\frac{q}{\epsilon_0}=\int_S\boldsymbol{E}\cdot\hat{\boldsymbol{n}}dS=E(r)\int_S dS=E(r)4\pi r^2\implies E(r)=\frac{q}{4\pi\epsilon_0r^2}$$
Finally, as we said the field is radial, so we field Coulomb's law:
$$\boldsymbol{E}(\boldsymbol{r})=\frac{q}{4\pi\epsilon_0r^2}\hat{\boldsymbol{r}}$$
A: The derivation of Gauss's law starts with a point charge, Coulomb's law, and all of the small solid angles which come from the charge. It shows that the flux though a closed surface which surrounds the charge is independent of the size or shape of the surface, and is proportional to the charge.  It follows from this that the flux through any closed surface surrounding any distribution of point charges will be proportional to the total charge.  Gauss's law is always true, but is generally used to calculate the field produce by a symmetrical distribution of charge.  By choosing a “Gaussian surface” which matches the symmetry of the charge, you can usually assume that the density of the flux is everywhere the same on that surface.  This lets you write a simple expression for the flux in terms of the field. With that and the total charge, you can find the field. You can do this with spheres of charge, long lines or cylinders of charge, or very wide sheets of charge.
A: The symmetry argument does not work the way you think, and the actual argument is not circular.
Remember that solving for the electric field $\mathbf E$ produced by some charge distribution $\rho$ is equivalent to solving Poisson's equation,
$$\nabla^2 V = -\frac{\rho}{\epsilon_0},$$
and taking $\mathbf E = -\nabla V$. It is a well known result (see here) from the theory of partial differential equations that, for a given charge distribution and given boundary conditions, this equation has a unique solution (it is well-posed), and this is the key thing.
Consider now a problem where $\rho$ and the boundary conditions are spherically symmetric (about the origin). For the sake of contradiction, suppose we have a solution which is not spherically symmetric. Perhaps,
$$V = V_0 e^{-x^2}.$$
But due to the spherical symmetry of the problem, then certainly $V = V_0 e^{-y^2}$ is a solution as well! After all, for a spherically symmetric problem, everything would look the same if we rotated the $y$-axis into the $x$-axis. This leads to a contradiction, because we know this problem must have a unique solution.
This sort of argument works for any presumptive solution that does not have spherical symmetry, and hence we conclude that any actual solution must be spherically symmetric.
Solving for a point charge
For completeness I will show you how to use the above symmetry argument to solve for the electrostatic field from a point charge.
A point charge distribution certainly has spherical symmetry. The typical boundary conditions one imposes are that $V = 0$ at infinity, which is also spherically symmetric. Hence we know from the argument above that $\mathbf E$ must point in the radial direction and depend only on the distance $r$ from the point charge. That is, we can write
$$\mathbf E = E(r) \mathbf e_r,$$
where $\mathbf e_r$ is a unit vector in the radial direction.
Gauss' law gives us
$$\oint_S \mathbf E \cdot d\mathbf S = \frac{Q_e}{\epsilon_0},$$
where $Q_e$ is the total charge enclosed by the closed surface $S$. We choose $S$ to be a sphere of radius $r$ about the point charge $q$. Then $d\mathbf S = \mathbf e_r dS$, and we obtain
$$\frac{q}{\epsilon_0} = \oint_S \mathbf E \cdot d\mathbf S = \oint_S E(r) \mathbf e_r \cdot \mathbf e_r dS = E(r) \oint_S dS = E(r) \cdot 4 \pi r,$$
where we have used that $E(r)$ is constant on $S$ to move it out of the integral. Hence
$$E(r) = \frac{q}{4 \pi \epsilon_0 r},$$
and
$$\mathbf E = E(r) \mathbf e_r = \frac{q}{4 \pi \epsilon_0 r} \mathbf e_r.$$
A: Most of the answers to your question would tell you to accept some physical fact as the fundamental truth, the truth can be Coulomb law itself or some property of electric field or some other equation etc etc.
My argument is purely mathematical.
The law that determines $\vec{E}$ should transcend the choice of coordinate system we use to calculate $\vec{E}$.
Let's say there is a point charge at one corner of your study table and you need to calculate field at the diagonally opposite corner of the same table. Let's assume you choose a cartesian coordinate system having origin at the charge itself and x,y axes coinciding with the edges of the table. You use the law and calculte $\vec{E}=88NC^{-1}\hat{n}$. Now you mark the $\hat{n}$ direction on the table (conforming to the axes you chose) with a piece of chalk. Now if I take the axes you chose and translate them to some arbitrary distance in some arbitrary direction and while I'm at it, I also rotate the coordinate system arbitrarily in 3-d space by some Euler angles $\alpha,\beta,\gamma$ with respect to the initial  coordinate system. Now I ask you to find $\vec{E}$ at the same point as before. Well, the value of field should still be $88NC^{-1}$. As for the direction, it may now be in some different mathematical form, say $\hat{m}$. But it should be the same physically: $\hat{m}$ needs to coincide with the direction you marked on the table earlier.
Why? The coordinate system is just a mathematical tool for us to do physics, it's not the physics itself. After all, who is to decide that what kind of  coordinate system to use? Any choice of coordinate system should yield the same result, and that result should agree with the experiments.
Keeping this in mind, Value of the field should depend only on the distance between the charge$(q)$ and the point$(P)$ where it needs to be calculated. Because the position vectors of $q$ (say, $\vec{r_1}$) and $P$ (say, $\vec{r_2}$) may both change when you use different coordinate systems but $(|\vec{r_1}-\vec{r_2}|)$ stays the same. Direction of the  field may change mathematically but it should stay the same physically after translating/rotating the coordinate axes. Moving along the surface of a sphere with $q$ as the center is the same as rotating the coordinate axes in the opposite way (for example: for some constant vector in the x-y plane, rotating this vector by 30 degrees clockwise while keeping the axes fixed is same as rotating the axes by 30 degrees counter-clockwise while keeping the vector fixed). As rotation of axes should not change the direction physically, you can now intuitively tell why direction at any point on this sphere would be symmetric so to speak (this doesn't tell that the direction is radial itself, but that's besides the point). So, we've established that the value of field only depends on $|\vec{r_1}-\vec{r_2}|$ which is simply the distance between $q$ and $P$, and the direction is symmetric (the angle between the radial vector and electric field vector would be same for all points on some sphere with $q$ as center). Which means for a spherical surface having $q$ at the center,  $\vec{E}\cdot\hat{r}$ ($\hat{r}$ represents radially outwards direction with q as center) is a constant scalar $(say, C)$ for all points on the sphere.
That's why $\int\vec{E}\cdot d\vec{A}=\int\vec{E}\cdot \hat{r}dA=C\int dA$
