# Capacitor with dielectric between the plates

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 ?

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.
• 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... Mar 20, 2014 at 15:22