Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.