...partially inserted between the plates of a parallel plate capacitor. I have a few questions about this - it is section 4.4.4 of Griffiths, 3rd edition. Imagine a parallel plate capacitor, of plate area l*d, separation d, filled with a linear dielectric of susceptibility X. Now imagine sliding the dielectric out of the plates in the direction that the plates have length l by a distance x. The idea is to find the force on the dielectric. You can do this for two cases - constant charge and constant voltage. Both give the force as F=-ε0XwV2/2d along the x direction where V is the p.d between the plates. I understand both of these derivations. However understanding the results physically is proving to be an issue. First of all if we let x=0 so the dielectric slab is perfectly between the plates, why is the force non-zero - apparently there is still a force in the negative x direction which does not make sense at all, from the symmetry of the scenario for starters. Secondly, at constant charge, F=-ε0XwV2/2d but V is not a constant and has x dependence - this makes me happy as I would expect the force to vary with x. However at constant voltage, F=-ε0XwV2/2d but now V is a constant, and so is this telling me the force is the same on the slab wherever it is in the universe (provided it remains in the correct 'slot' between the plates). This just wouldn't make sense surely because as x gets very large, the fringe field causing the force should be vanishingly small. If anyone can answer these I would be grateful! I have a few more issues about it but not understanding the above probably means I'm not in a position to worry about them just yet.