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Given a parallel plate capacitor of width $w$, length $l$, with a dielectric moving along the length $l$. Let the dielectric be from $x$ onwards.

The capacitance will be $\frac{w \epsilon_0}{d} (\epsilon_r l - \chi_e x)$. Griffiths (p. 195) says that the total charge $Q$ in the $C=\frac{Q}{V}$ expression is constant as the dielectric moves. But $Q$ here refers to the free charge, and the free charge definitely increases as you move the dielectric in increasing $x$. What am I misunderstanding?

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If dielectric is placed inside the plates of a parallel plate capacitor , the capacitance of capacitor will increase ,, and as a result the voltage will be decreased ,, –  user29081 Sep 1 '13 at 10:42
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