If we have a parallel plate capacitor whose charge is $+Q$ and the polarization charge as $Q_p$ as shown in the figure..

enter image description here

then while finding the force acting on the left plate of the capacitor for instance, shouldn't the force due to the polarized charge $-Q_p$ and $+Q_p$ together be zero and therefore the only force acting be due to the right plate of the capacitor and hence the total force acting on the left plate of the capacitor be independent of the dielectric constant of the medium?

  • $\begingroup$ The dielectrics, on being polarised, exert a force on the plates. So, you cannot neglect the presence of the dielectric. $\endgroup$
    – Aniansh
    Commented Nov 30, 2017 at 15:15
  • $\begingroup$ I never neglected the dielectric.... Isn't the force due to polarisation of dielectric zero for each plate... And the only force acting on each plate is due to the other? $\endgroup$
    – rohit_r
    Commented Nov 30, 2017 at 15:18
  • $\begingroup$ Each face of the dielectric exerts a force on both the plates of the capacitor. The resultant should be greater than if the dielectric wasn't present at all. $\endgroup$
    – Aniansh
    Commented Nov 30, 2017 at 15:22
  • $\begingroup$ Field due to polarisation is given by (sigma) /(epsilon naught).. (Sorry I'm terrible in mathjax..) which is independent of distance $\endgroup$
    – rohit_r
    Commented Nov 30, 2017 at 15:24
  • $\begingroup$ There is also an induced field density on the dielectric surface which is perfectly capable of exerting a force on the plate of the capacitor. $\endgroup$
    – Aniansh
    Commented Nov 30, 2017 at 15:36

2 Answers 2


Agreeing with @Aniansh that the two induced surface charges on both faces of dielectric are separated by a finite distance and therefore must cast a net electric field at a point beyond the dielectric boundaries because one surface will always be closer to the point than the other.
However when we are talking of capacitors it is assumed that A>>d i.e. plate area is much greater than the distance between the plates.Even when evaluating electric field inside the plates we use the formula $\frac {\sigma}{\epsilon _o}$ , this formula is valid when the electric field is being evaluated at a point just near the surface of infinite charged sheets . In such a compact setup as that of a capacitor it is safe to assume that any point inside capacitor but outside dielectric will see two equal forces being applied by the opposite faces of dielectric each being of magnitude $\frac {\sigma _d}{2\epsilon _0}$ because we are evaluating field at a point very near the surfaces of dielectric , so even if there is a separation the limiting values of forces are same because the distances from the surfaces are small enough to be considered 'very near to the surfaces'.


The nett force on the Dielectric due to the involved Electrostatic interactions is zero. There is, however, a nett force on the dielectric on the plates because the forces from both the face of the dielectric are not equal in magnitude and don’t cancel out. My solution for the problem

page 2

I have assumed the numerator of both F(-q) and F(+Q) to be equal in magnitude. As you can see, the forces are unequal because the two surfaces of the dielectric are at two different distances away from the Plate of the capacitor. Therefore, there is a nett force due to the dielectric on the capacitor plates.


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