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enter image description hereI derived the equation for the force $F$ acting on a dielectric of dielectric constant $K$ being inserted into a constant voltage (connected to battery) parallel plate capacitor of capacity $C$, voltage $V$, length of plate side $l$, thickness $d$ and plate area $A$=$l^2$. $$F=\frac{-V^2C(K-1)}{2l}$$ This is constant for certain values of $V,C,K$ and $l$.

The force acting in case of constant charge (disconnected from battery after charging) capacitor is, however, variable and varies inversely with the length $x$ of dielectric inserted into the capacitor. $$F=\frac{-Q^2l(K-1)}{2C(Kx+l-x)^2}$$ where $Q$ is the charge stored on the capacitor.

I wanted to know the physical explanation behind why the force is variable in the 2nd case while it is constant in the 1st case?

P.S- I know the mathematical proof but I only want to know why this happens.

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I believe that the reason behind this is the fringe field. In case of a constant voltage capacitor, the potential remains unchanged irrespective of the the length of the dielectric inserted into the capacitor. But in case of constant charge capacitor, the potential varies depending on the the length of the dielectric inserted. Due to the variation of potential, the force is variable as well, but if potential remains constant then the force acting cannot change as well as the electric field is constant and induces the same charge on the dielectric at every point.

Please correct me if I am wrong.

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