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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 $q/\epsilon_0$. Then they use the integral definition of flux $\int E \mathrm{d}S$ 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 $kq/r^2$ 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 don't understand why electric field has to be constant for symmetric figures).

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  • $\begingroup$ Can you explain why you think $kq/r^2$ is the only way to show spherical symmetry? $\endgroup$ Commented Dec 27, 2021 at 20:06
  • $\begingroup$ Could you please how to derive it the other way. $\endgroup$
    – madness
    Commented Dec 28, 2021 at 2:44
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    $\begingroup$ A good question @madness and it has a good answer. Spherical symmetry of the object (could be a point charge, or a ball of uniform charge density) means that, for any two points in space, I can find an axis through the center of the ball where rotation moves one point into the other. Because the ball of charge is unchanged by this rotation, the electric field it generates must also be unchanged by this rotation. If the field at the two chosen points had different E magnitudes, the E field would be changed by their swap. Therefore E must be same on any spherical surface about the center. $\endgroup$
    – Ben H
    Commented Mar 16 at 10:14
  • $\begingroup$ I meant any two points at the same distance from the center of the object. I've added an answer. $\endgroup$
    – Ben H
    Commented Mar 16 at 10:22

8 Answers 8

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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: $$\boldsymbol{E}(\boldsymbol{r})=\frac{q}{4\pi\epsilon_0r^2}\hat{\boldsymbol{r}}$$

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  • $\begingroup$ That's exactly my concern,why can we exploit symmetry in this case?Why does symmetry imply $S$ is constant?Also the fact yoh stated $E$ here only depends on $r$ seems an ok but non rigorous proof to me,what would be the complete rigorous proof? $\endgroup$
    – madness
    Commented Dec 28, 2021 at 2:47
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    $\begingroup$ The system we are considering is a point charge. If set the origin in the position of the charge, any rotation of the frame of reference would put you in the exact same situation (indistinguishable). On the other hand if the intensity of the electric field in a point were to depend also on the direction (i.e. polar angles) the previous rotation would require it to change under axis rotation despite the fact the new system is indistinguishable? Similarly we deduce it must be radial. Does that clarify your doubts? If not, what do you mean by "complete rigorous proof"? $\endgroup$ Commented Dec 28, 2021 at 14:46
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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.

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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.

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  • $\begingroup$ Actually you are missing my point. I am using Gauss law to prove elementary things,yes but why does a symmetric surface imply that $E$ is constant? $\endgroup$
    – madness
    Commented Dec 28, 2021 at 2:48
  • $\begingroup$ Consider what happens to the field as you move from point to point on a symmetric surface. Why would it change? $\endgroup$
    – R.W. Bird
    Commented Dec 28, 2021 at 14:50
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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

$$\mathbf E = E(r) \mathbf e_r = \frac{q}{4 \pi \epsilon_0 r} \mathbf e_r.$$

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  • $\begingroup$ Could you please prove that $E$ is constant for symmetric surfaces without using the heavy machineries?I am in grade $12$ correctly and we haven't been taught poissons equation yet but i am aware of the fact that $E=-\frac{dV}{dr}$ but this was also probably derived using $E=\frac{kq}{r^2}$ which is what we want to prove. $\endgroup$
    – madness
    Commented Dec 28, 2021 at 2:50
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    $\begingroup$ @madness We need some machinery if we want to be rigorous. But the main idea is actually pretty simple: 1. We know (from mathematical theory) that a solution exists and is unique. 2. If we somehow had one nonsymmetric solution, we could make more by rotating it, yielding a contradiction. 3. Hence the solution must be symmetric. $\endgroup$
    – ummg
    Commented Dec 28, 2021 at 13:02
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    $\begingroup$ @madness I understand that it is frustrating to have to rely on a mathematical result that you haven't learned yet, but maybe you can accept it for now, with the knowledge that you can learn it later? $\endgroup$
    – ummg
    Commented Dec 28, 2021 at 13:03
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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.]

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  • $\begingroup$ That's actually my question,how do we assume $E$ is constant?I mean the symmetry argument is not kind of a rigorous proof,i want to prove it rigorously. $\endgroup$
    – madness
    Commented Dec 28, 2021 at 2:45
  • $\begingroup$ @madness Do you consider all arguments based on symmetry to be non-rigorous? $\endgroup$ Commented Dec 28, 2021 at 11:13
  • $\begingroup$ I wonder if Coulomb, when he discovered the inverse square law of forces between small charged spheres, checked whether or not the spheres' individual orientations affected the force between them. [Just an aside.] $\endgroup$ Commented Dec 28, 2021 at 13:10
  • $\begingroup$ @PhilipWood That is a good question! However, surely it must have been done by the time of Faraday? $\endgroup$ Commented Mar 16 at 19:22
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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 $\int\vec{E}\cdot d\vec{A}=\int\vec{E}\cdot \hat{r}dA=C\int dA$

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The spherical symmetry of the object (it could be a point charge, or a ball of uniform charge density) means that, for any two points in space ($A$ and $B$) that are the same distance $d$ from the center of the object, I can find an axis through the center of the ball that, under a $\pi$-rotation, will rotate point $A$ into point $B$ (and vice versa). The ball of charge is unchanged by this rotational transformation. Therefore, the electric field it generates must also be unchanged by this rotation. Now, if it happened that $\left|\vec{E}_A\right| \ne \left|\vec{E}_A\right|$, then the rotation would have swapped the magnitudes at these two points, and the electric field would be changed by such a transformation. But we already said that the electric field cannot be changed by the transformation (that is the essence of the symmetry argument: if the object generating the field is unchanged by a transformation, then the field it produces cannot be changed by the transformation). Therefore the electric field must have $\left|\vec{E}_A\right| = \left|\vec{E}_A\right|$. Because the two points I chose were arbitrary, it is true that any two points a distance $d$ from the center of the object must have the same magnitude of the electric field, so: $\left|\vec{E}\right|$ is uniform on every concentric sphere.

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As a purely theoretical endeavor, it is possible to envision a universe where symmetric charge distributions simply do not give rise to symmetric fields. There is no fundamental reason that the symmetries of the source must be the symmetries of the field. So if you do not know the law obeyed by the field, how do you know a priori that symmetric sources give rise to symmetric fields? To be frank, you do not know.

However, the theory of the electromagnetic field was not developed as a purely theoretical endeavor, rather, decades of experimental researches preceded the work of Maxwell. Before Gauss' law, people knew that the symmetries of the electric field were also the symmetries of the charge distribution or source. Thus, from the advantage point of empirical knowledge of the nature of the electric field, it is perfectly rigorous to argue from the symmetry of the source to the symmetry of the field. Gauss' law and Coulomb's law are two equivalent statements of the same thing; they are empirically established natural laws that have been codified into mathematical form.

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