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?

  • $\begingroup$ 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 ,, $\endgroup$
    – user29081
    Sep 1, 2013 at 10:42

1 Answer 1


If the plates of the capacitor are isolated then the total amount of free charge on the capacitor plates cannot change.
Put another way, if the capacitor pales are isolated where could more/less free charge come from / go to?
The situation would be different if there was a voltage source connected to the plates.
In that case the free charge on the plates would change to ensure that the potential difference across the plates is kept constant.


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