# Force on plate of parallel plate capacitor with dielectric

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

then while finding the force acting on the left plate of the capacitor for instance, shouldn't the force due to the polarized charge -Qp and +Qp 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?

• The dielectrics, on being polarised, exert a force on the plates. So, you cannot neglect the presence of the dielectric. – Aniansh Nov 30 '17 at 15:15
• 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? – physics123 Nov 30 '17 at 15:18
• 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. – Aniansh Nov 30 '17 at 15:22
• Field due to polarisation is given by (sigma) /(epsilon naught).. (Sorry I'm terrible in mathjax..) which is independent of distance – physics123 Nov 30 '17 at 15:24
• 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. – Aniansh Nov 30 '17 at 15:36

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'.