# The force acting on a plate of a parallel capacitor [duplicate]

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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

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• 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 $F=EQ/2$. – G. Paily Dec 6 '14 at 17:02
• 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 $E/2$? – marmistrz Dec 7 '14 at 10:56
• Modeling the plate as an infinite sheet of charge, the field does not depend on distance. $E=\frac{\sigma}{2\epsilon_{0}}$ – G. Paily Dec 10 '14 at 16:23
• Yes, indeed, and the mentioned by you fact can be easily proved using Gauss' Law. Thanks! – marmistrz Dec 11 '14 at 20:14