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For getting the electric field in this case we use the Gauss's law. we get the equation

$$ 2EA = pA/\epsilon_0 $$

here

  • $E$ is electric field,
  • $A$ is the cross sectional area,
  • $p$ is the uniform surface charged density,
  • $\epsilon_0$ is permittivity of the vacuum.

left hand side of the equation is understandable but in the right hand side of the equation it is $pA$, why it is not $2pA$? since infinite sheet has two side by side surfaces for which the electric field has value.

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  • $\begingroup$ More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. $\endgroup$
    – Qmechanic
    Nov 14 '18 at 13:17
  • $\begingroup$ See my revised answer. Hopefully this better answers your question. $\endgroup$
    – Bob D
    Nov 17 '18 at 15:06
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Think of an infinite plane or sheet of charge (figure at the left) as being one atom or molecule thick. So in that sense there are not two separate sides of charge. The total enclosed charge is $ρA$ on the right side of the equation.

An infinite conducting plate (figure at the right) is one having thickness that allows the charge to migrate to separate sides of the plate in response to the repulsive electrostatic forces between them. If the charge density on each side of the conducting plate of the right figure is the same as the charge density of the infinite sheet, then the total charge enclosed would be $2ρA$ on the right side of the equation.

On the other hand, if the same quantity of charge on the infinite sheet on the left were placed on the conducting plate on the right, the charge would split up making the density on each side of the plate $ρ/2$ and the total enclosed charge $ρA$, giving the same result as the infinite sheet of charge.

Hope this helps. enter image description here

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Nov 15 '18 at 5:27
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Infinte plane sheet is of only one surface. Actually it is not possible. Sheet thickness tending to zero, that is only one surface containing charge. In reality we have to consider two surfaces, 2pA must be taken.

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  • $\begingroup$ If this is so then why there is the vector addition of electric flux through two surfaces which gives 2EA in left hand side of the equation? $\endgroup$
    – ADR
    Nov 14 '18 at 8:00
  • $\begingroup$ @ADR because your Gaussian surface does have thickness $\endgroup$ Nov 14 '18 at 11:16
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The misunderstanding simply comes from mixing up what the areas are. In other words, even though both of the areas on each side of the equation have the same value, they represent different ideas.

For the left side, $2EA$, the area represents a side of the Gaussian surface parallel to the sheet of charge. This is why we have a factor of $2$, because there are two surfaces of area $A$ on our Gaussian surface through which the field has a non-zero flux.

For the right side, $\frac{\rho A}{\epsilon_0}$, the area is used to calculate the total charge enclosed by our Gaussian surface. In the case of a plane of charge, the Gaussian surface encloses a single area $A$ of the plane. Therefore, there is a factor of $1$ (not $2$).

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enter image description here Electric field due to infifinetly charged sheet

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  • $\begingroup$ Again, please do not post screenshots as answers. Please use MathJax for writing equations so that search engines can index this post. $\endgroup$ May 17 '19 at 12:14

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