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Let's assume we have a capacitor of capacitance $C$ and potential difference $U$. After charging it we disconnect it. Then we put a dielectric between the plates. I know that capacitance will increase by $C * k$, however what happens with the charge and potential difference on it.

Let's say $k = 2$. Will $q$ double or will $ U $decrease to a half ?

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If the plates are disconnected, the charge has nowhere to go. Rather U will have to change. What happens is the charged capacitor does work on the dielectric (pulling it in), resulting in a change in the energy stored in the capacitor.

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  • $\begingroup$ Many thanks. Can you please explain further what you mean with "pulling it in" ? $\endgroup$ Mar 20, 2014 at 14:41
  • $\begingroup$ Sure. The charge of the plates causes the dielectric to become polarized: positive charges in the dielectric get pulled a little bit towards the negative plate, and negative charges in the dielectric get pulled a little bit towards the positive plate. These induced dipoles in the dielectric are attracted by the electric field of the plates. $\endgroup$ Mar 20, 2014 at 15:12
  • $\begingroup$ I'm actually a bit confused now. According to page 5-22 of web.mit.edu/viz/EM/visualizations/coursenotes/modules/…, the voltage is observed to decrease by a factor of $k$, but the capacitance increases by a factor of $k$. This means that energy $\frac{1}{2}CV^2$ decreases by a factor of $k$. Somehow that seems strange, is that correct? According to question 3 of ilt.seas.harvard.edu/students/lectures/printablelecture/…, this is the case... $\endgroup$ Mar 20, 2014 at 15:22
  • $\begingroup$ E = 1/2 * q * U so energy decreases by factor of k. The equation you stated is obtained by q = C * U, U and C being the initial potential difference and capacitance. I apologize if I'm incorrect. $\endgroup$ Mar 20, 2014 at 15:27
  • $\begingroup$ @DumpsterDoofus That sounds right to me. Why does it seem strange? $\endgroup$ Mar 20, 2014 at 15:29

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