Why does the potential difference between the plates of a capacitor decrease when a dielectric is inserted in it? How is this possible if the capacitor is connected to a DC power supply? Also, does a dielectric actually 'touch' the plates of the capacitor or is it kept in the middle?

  • $\begingroup$ Is there leakage current through the dielectric? That would certainly decrease the voltage drop, especially if your power source has a high internal resistance. $\endgroup$ – probably_someone Jun 15 '17 at 10:20

If the capacitor is isolated then the charge on the capacitor $Q$ is constant.

Adding a dielectric increases the capacitance, $\dfrac {(\epsilon \uparrow )A}{d} \Rightarrow C\uparrow$, and this in turn decreases the (electric potential) energy stored in the capacitor $\dfrac 1 2 \dfrac{Q^2}{C \uparrow} \Rightarrow \rm energy \downarrow$.

To explain where this energy goes consider the dielectric being outside the capacitor.
The charges on the plates of the capacitor induce induce charges on the dielectric and are so arranged that there is a net force of attraction between the induced charges on the dielectric and the changes on the plates of the capacitor.
If one were to release the dielectric it would accelerate as it travelled towards the capacitor, gaining kinetic energy whilst the capacitor loses energy and the dielectric has a maximum kinetic energy when it was in the middle of the capacitor.
The kinetic energy of the dielectric at this point would be the difference between the energy stored in the capacitor before the dielectric was inserted and the energy stored in the capacitor when the dielectric was inside it.
With no friction, forgetting about gravity etc, the dielectric would then move out of the capacitor but now slowing down, losing kinetic energy, with the capacitor storing more energy.
The dielectric would stop with the energy stored in the capacitor the same as at the start.
Then the dielectric would start moving towards the capacitor with the cycle of energy changes repeated.

In practice the energy of the dielectric-capacitor system would be reduced due to frictional force acting on the system (heat being produced) and/or the system doing work on something external to it (eg your hand which is holding the dielectric).

If the capacitor is connected to a battery then the potential difference across the capacitor stays constant.

Again adding a dielectric increases the capacitance but in this case this increases the (electric potential) energy stored in the capacitor $\dfrac 1 2 (C \uparrow ) V^2 \Rightarrow \rm energy \uparrow$.

Again the dielectric will be attracted into the capacitor and in an "ideal" situation the dielectric would oscillate in and out of the capacitor.
In this case the battery would supply energy, $(\text{emf} \times \text {charge moved})$, as the dielectric moved into the capacitor and that energy would be returned to the battery as the dielectric moved out of the capacitor.
So the battery supplies the extra energy to the capacitor with the dielectric inside it.


The voltage will only reduce if the capacitor is isolated. This is because the dielectric increases the capacitance. When the capacitor's terminals are not connected to anything, the charge cannot change, and hence the voltage will drop due to the capacitor equation $V=Q/C$.

On the other hand, if the capacitor is connected to a DC source, then that source will provide the extra charge needed to keep the voltage on the capacitor the same as the supply voltage, again because $V=Q/C$.

And, no the dielectric does not have to touch the plates. What happens is that the dielectric has a higher relative permittivity than air. The total permittivity of the gap between the capacitor plates is made up from that of the dielectric ans whatever air may still be present between the plates. Vacuum has a relative permittivity of 1, air is about 1.0006 and other materials have still higher permittivity. The capacitance that you measure takes all this into account.


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