State the effect of filling a dielectric in a capacitor after disconnecting the battery
Answers
If the capacitor, however, is disconnected from the circuit, say after being charged to a particular potential difference, then the charge on the plates will remain fixed, and a change in capacitance (like moving the plates together) results in a change in potential difference precisely as you point out.The voltage across the capacitor has to stay the same since it is connected to a fixed voltage supply, which means that the potential before insertion and after insertion is equal. That would mean that the electric field within the capacitor is also equal before and after (since E = -dV/dR).
However, when a dielectric is inserted, it reduces the field since the molecules of the dielectric align themselves in such a way that the moment is opposite to the external electric field, which is also supported by:
K = Eexternal/Ereduced
where K is the dielectric constant.
after disconnecting battery let us assume the charge on the plates is Q and the potential difference is V. The relation connecting capacitance C, charge Q and potential difference V is given by C = Q/V.
when dielectric material is inserted between parallel plates the capacitance increases and the new capacitance C' is given by C' = k×C, where k is dielectric constant. Charge Q remains same. Hence potential difference reduces and the new value of potential difference is V/k .
Energy stored in the capacitor before inserting dielectric = (1/2)×C×V2
after inserting dielectric, energy available in the capacitor = (1/2)×(kC)×(V/k)2 = k×(1/2)×C×V2