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Force between the plates of a capacitor with dielectric slab inserted
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[QUOTE="Pushoam, post: 6852870, member: 619344"] [ATTACH type="full" align="left" width="130px" alt="1675922301493.png"]321973[/ATTACH] Here is the Gaussian surface I was talking of. It includes charge of upper plate and upper part and lower part of dielectric. Electric field due to the dielectric outside the dielectric is 0. Hence, the electric field at bottom plate (which is outside the dielectric) is only due to upper plate, not the dielectric. The bound charges and the free charges are not mixed with each other. These are divided by a surface (which is invisible here as I cannot draw it here). The bottom part of my Gaussian surface is here. Now, as this surface is close to the facing surface of bottom plate, the electric field on this surface is close to the electric field acting on the charges of bottom plate. Just above the bottom part of Gaussian surface, ## \vec E = \frac{\vec E_0}K## and just below the bottom part of Gaussian surface, ## \vec E = \vec E_0 ##. The bottom plate is just below, hence here, ## \vec E = \vec E_0 ##. The bottom part of Gaussian surface is at the point of discontinuity. This approach shows that the force remains same in both cases (1) and (2). Which mistake am I making in this approach? [/QUOTE]
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Force between the plates of a capacitor with dielectric slab inserted
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