I had a doubt about why inside a capacitor having equal and opposite charges on both plates have a electric field $Q/A€$ and not $2Q/A€$ (where $€$ stands for permittivity).

But a question at stack exchange answer this by saying that superposition will not apply for when electric field terminate at the negative charge


i want to know why not the same concept is applied in this case of infinitely long line charges having opposite charges

why we are using superposition to find electric field at a point between the two line charges

enter image description here

  • $\begingroup$ Because before superposition the field is $Q/(2A\epsilon_0)$ and after superposition it is $Q/(A\epsilon_0)$. There is no conflict with the superposition principle. $\endgroup$
    – verdelite
    Commented Mar 18, 2021 at 22:39

1 Answer 1


As @verdelite suggest, the electric field from the positive (or negative) charges is $Q/(2A\epsilon_0)$ (i.e. the electric field of infinity uniform charged plate) and using superposition principle (note that this is just a fancy name for electric field additivity) the total electric field is $Q/(2A\epsilon_0)$.

  • $\begingroup$ I have two questions (1) are you not considering any thickness of sheet (2) I am not asking question about an isolated infinite sheet I have a question about the field in a capacitor because if the sheet is thick and all the charges will be on inside surface to make the field zero in the mass of metal in the case of a capacitor so field inside the capacitor due to any one plate will be Q/AE0 and directed towards the negative plate $\endgroup$ Commented Mar 19, 2021 at 9:49
  • $\begingroup$ (1) yes, I'm not considering any thickness of the sheet because the charges inside conductor arranged in the surface. (2) you must consider the capacitor plate as infinity plate if you want to get simple uniform electric filed (we can use this assumption because the plate in capacitor are very close compare to the distance between them). $\endgroup$
    – ziv
    Commented Mar 19, 2021 at 11:18
  • $\begingroup$ Note that this assumptions are usually done when we describe the model of capacitor. $\endgroup$
    – ziv
    Commented Mar 19, 2021 at 11:35
  • 1
    $\begingroup$ ok got your point $\endgroup$ Commented Mar 19, 2021 at 11:39
  • $\begingroup$ good, welcome to physics stack exchange $\endgroup$
    – ziv
    Commented Mar 19, 2021 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.