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How is the electric field same at any point from the planar sheet with uniform charge density which is enclosed by a gaussian surface as shown below

Here

Now I can understand the formula derivation by using equation of flux by $$E \cdot 2A = \frac{q}{\epsilon_{0}}$$ and with $\sigma = \frac{q}{A}$ one arrives at $$E=\frac{\sigma}{2\epsilon_{0}}\,, \qquad(1)$$ where $\sigma$ is the charge density of the planar sheet and $\epsilon_{0}$ is permittivity in free space.

But when I substitute the value for electric field $E=\frac{q}{4\pi\epsilon_{0} r^2} = \frac{\sigma}{2\epsilon_{0}} $ we can rearrange this to

$$\frac{1}{r^2} = \frac{2\pi\sigma}{q}\,.$$ We can find that RHS is full of constant values and $r^{2}$ is a changing value based on the distance from the planar sheet, which can't be true. Is there any explanation for this and the electric field being same for any point from the planar sheet even at infinity from the planar sheet because it doesn't seem to change with analyzing equation 1?

Edit: Here the part of the charged sheet which is enclosed by gaussian surface's electric field is only found. Not the full charged sheet. So from a far away distance that part which is enclosed by the gaussian surface is finite and can be approximated to a point charge and hence I equate both of the electric field equations.

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    $\begingroup$ You can exploit the symmetry of the problem: what does the electric field look like if you shift the plate by any amount in any direction? $\endgroup$ Apr 14, 2023 at 7:00
  • $\begingroup$ then the area wont cancel out in both equation we would get components and many more i think @FlatterMann $\endgroup$
    – Naveen V
    Apr 14, 2023 at 7:02
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    $\begingroup$ The charge density looks exactly the same if you shift the plate, hence the field has to look the same. $\endgroup$ Apr 14, 2023 at 7:09
  • $\begingroup$ For your edit: The trouble with that is that outside the gaussian surface the charge inside the gaussian surface is not the only source of the electric field. So you can't use the point charge idea. $\endgroup$
    – kricheli
    Apr 16, 2023 at 12:34
  • $\begingroup$ what if I move in a way that im inside the gaussian surface that is sideways so that charge enclosed by the rectangular box is the charge enclosed by the gaussian surface and hence the only charge. Then even If I dont move with the reasoning the charges from outside will affect the charges inside right but that isnt the case , so if i move sideways in such a way that im inside the gaussian surface the other charges mustn't affect it is what my thinking is. Please correct me if I am wrong @kricheli $\endgroup$
    – Naveen V
    Apr 16, 2023 at 12:40

3 Answers 3

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$E = \frac1{4\pi\epsilon_0} \frac q{r^2}$ is the expression for a point charge. Here you have a constant surface charge density. You cannot use the formula that assumes one thing, in the place of another formula assuming another, incompatible, thing.

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  • $\begingroup$ we cant assume the charge enclosed by gaussian surface to be q for a far away distance so that its a point charge ? and then equate both of the equations (because at a long away distance the field doesnt change with distance right ) $\endgroup$
    – Naveen V
    Apr 14, 2023 at 6:38
  • $\begingroup$ No, because in order for the Gaussian pillbox argument to be strictly true in this case, the plate must be infinitely large. There is no "infinitely far away" because the E field strength is not decreasing no matter how far you move away from the plate. $\endgroup$ Apr 14, 2023 at 6:45
  • $\begingroup$ but the plate is infinitely long as told in the title right so there is "infinitely far away" in this case ? Im confused $\endgroup$
    – Naveen V
    Apr 14, 2023 at 6:49
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    $\begingroup$ Look, either E is constant no matter where you are away from the plate, or E is decaying as 1/rr. It cannot be both at the same time. $\endgroup$ Apr 14, 2023 at 6:54
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Consider a plane $P$ that is normal to the charged planar sheet - for simplicity start with the $x$-$z$-plane. The charge density is mirror symmetric with respect to this plane and, in consequence, so is the electric field. You can do this with any plane normal to the charged plane, and as a result the field vector is normal to the charged planar sheet, i.e. the electric field only has an $x$-component. Further use either those mirror symmetries with respect to arbitrary normal planes, or simply the invariance of the charge density under translations in the $z$ or $y$-direction to find that the electric field cannot depend on $z$ or $y$.

We have thus found quite some info already just by exploiting symmetries of the system, and only now do we start into the calculation you present above, i.e. we use Gauß' law (generally valid, a statement that the electric flux through a surface is equal to the enclosed charge) for the surface $S$ you show above, i.e. $$ \int_S \vec{D} \cdot \vec{n}\, \text{d}A = q\,. $$ After using $\vec{D} = \varepsilon_0 \vec{E}$ and $q = A \sigma$, we turn our attention to the integral. Since, as we have found above, the electric field is normal to charged sheet, only on the faces labelled 1 and 2 is the scalar product $\vec{E} \cdot \vec{n}$ of electric field and surface normal non-zero. Even better: it is constant on that part of the surface because $\vec{E}$ does not depend on $y$ or $z$. Which means we can evaluate $$ \int_S \vec{E} \cdot \vec{n}\, \text{d}A = 2 A \left|\vec{E}\right|\,. $$

This is, in a bit more detail, what is going on in the derivation you are referencing. You see how important it is to exploit the symmetry, right?

Now, for the point charge issue. The electric potential - and thus the electric field as well - is calculated from the charge distribution in the whole of $\mathbb{R}^3$, not just a part in some Gaussian surface. As in (the solution to Poisson's equation) $$ \varphi(\vec{x}) = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{x}')}{\left|\vec{x}-\vec{x}'\right|} \text{d}^3\vec{x}'\,, $$ $$ \vec{E}(\vec{x}) = \frac{1}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\vec{x}') (\vec{x}-\vec{x}')}{\left|\vec{x}-\vec{x}'\right|^3} \text{d}^3\vec{x}'\,. $$ You can see this as an infinite superposition of the fields of point charges all over the charged plane.

Now, what we could have done instead of the "trick shot" involving Gauß' law is just to use the charge density of the infinite plane $$ \rho(\vec{x}) = \sigma \delta(x) $$ ($x$ without arrow meaning the $x$-component of $\vec{x}$), plug this into the integral and also find the constant field from $$ \vec{E}(\vec{x}) = \frac{\sigma}{4 \pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\delta(x) (\vec{x}-\vec{x}')}{\left|\vec{x}-\vec{x}'\right|^3} \text{d}^3\vec{x}'\,. $$ (Which of course is a bit more tricky, an exercise in evaluating integrals... ;) )

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  • $\begingroup$ I got It , thank you for making me understand for trying to help me from start about where I went wrong ! $\endgroup$
    – Naveen V
    Apr 16, 2023 at 16:49
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When using Gauss' law, we consider the electric field generated by all the charges in the system - not just the charges enclosed in the Gaussian surface. After all, we want to find the electric field from the entire plane, not just a patch! Gauss' law is extremely useful in systems like this with high degrees of symmetry. If we only consider the charges contained in the Gaussian surface, the translational symmetry of the system is destroyed and the electric field is much more complicated than $\sigma/2\epsilon_0$. Far away we cannot say the field looks like a point charge, we need to consider all the charges in the infinite plane, which still looks like an infinite plane from far away.

As you mentioned in a comment above, yes there are infinite charges in the plane. However as there is a uniform finite surface charge distribution, each charge only contributes an infinitesimal amount to the electric field. The whole triumph of calculus is that you can add an infinite number of these infinitesimal contributions and get a finite number.

The field was derived above with Gauss' law, but it can also be calculated with Coulomb's law if you really wish to use that formula. Consider the infinite plane with coordinates as shown below.

We can calculate the electric field from a small patch on the plane, then sum up (integrate) many patches to find the field due to the entire plane. The symmetry allows us to only consider contributions to the field in the $z$-direction (there is always a point opposite to the one we consider that cancels the horizontal contribution to the field).

\begin{align} E_z &= \int \cos\theta\cdot dE = \int\frac{1}{4\pi\epsilon_0}\frac{dq}{r^2}\cos\theta = \frac{\sigma}{4\pi\epsilon_0}\int_{-\infty}^\infty\int_{-\infty}^{\infty}\frac{dxdy}{r^2}\cos\theta \\ &= \frac{\sigma z}{4\pi\epsilon_0}\int_{-\infty}^\infty\int_{-\infty}^{\infty}\frac{dxdy}{(x^2+y^2+z^2)^{3/2}} \\ &= \frac{\sigma}{2\epsilon_0} \end{align}

and you recover the electric field found using Gauss' law.

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  • $\begingroup$ Thank you for the answer $\endgroup$
    – Naveen V
    Apr 16, 2023 at 16:49

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