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Consider two infinite parallel plans of opposite charge density Let's say $+\sigma$ for the left plane and $-\sigma$ for the right plane. Why is the electric field calculated this way? $$ E=\frac{\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0} $$ I understand that between the planes the vector $\vec{E}_+$ will point to the right toward the negatively charged plane. The same goes for the vector $\vec{E}_-$ that goes toward the negatively charged plane.

What I don't understand is why do we not consider the $-\sigma$ value in the equation?

Qmechanic
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Harshit Khullar
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1 Answers1

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I assume that you mean "... $-\sigma$ for the right plane".

The method used is to sum the fields that the left and right hand sheets would generate, if each sheet were the only one.

Using Gauss's law and symmetry we have, with arrows to show field directions,...

Field to the right of left sheet due to left sheet = $\frac 1 {2\epsilon_0} \sigma\rightarrow$.

Field to the left of right sheet due to right sheet = $\frac 1 {2\epsilon_0} \sigma \rightarrow$.

So total field between sheets = $\frac 1 {2\epsilon_0} \sigma \rightarrow + \frac 1 {2\epsilon_0} \sigma \rightarrow=\frac 1 {\epsilon_0} \sigma \rightarrow$.

You should check that the total field to the left of the left hand sheet and to the right of the right hand sheet is zero.

Philip Wood
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  • Why did you plus it instead of subtracting??? – Harshit Khullar Jul 24 '23 at 13:37
  • Because, between the sheets, the field due to the left hand sheet points to the right and the field due to the right hand sheet also points to the right. I've changed the notation in my answer, to try and make this clearer. – Philip Wood Jul 24 '23 at 17:32