This question already has an answer here:

How can you prove (using high-school-knowledge) that the force which acts on one plate of a parallel capacitor is equal

$F = \frac{E \cdot Q}{2}$

Where $E = \frac{U}{d}$ is the magnitude of the electric field beside the plates and $Q$ is the charge on each of the plates?

I can calculate basic integrals (although they're not in high school).


marked as duplicate by user36790, Michael Seifert, ACuriousMind Dec 15 '16 at 23:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ If $E$ is the total field between the plates, then each plate contributes $E/2$. But we argue that a charged plate can't feel a force from its own electric field. So it only feels a force from the other plate, giving $F=EQ/2$. $\endgroup$ – G. Paily Dec 6 '14 at 17:02
  • $\begingroup$ This doesn't convince me. Suppose we have a test charge (+) near the (+) plate. Then the most of the field is being contributed by the (+) plate (as the distance from it is much smaller that from the (-) plate and the electrostatic force is inversely proportional to distance, according to Coulomb's Law). Analogically, if the test charge is put very near to the (-) plate, most of the force acting on the test charge comes from the (-) plate. So why then should each plate contribute $E/2$? $\endgroup$ – marmistrz Dec 7 '14 at 10:56
  • 1
    $\begingroup$ Modeling the plate as an infinite sheet of charge, the field does not depend on distance. $E=\frac{\sigma}{2\epsilon_{0}}$ $\endgroup$ – G. Paily Dec 10 '14 at 16:23
  • $\begingroup$ Yes, indeed, and the mentioned by you fact can be easily proved using Gauss' Law. Thanks! $\endgroup$ – marmistrz Dec 11 '14 at 20:14