Dielectric in a capacitor

Problem

Qualitative: dielectric in a capacitor. Explain in five lines why, inserting a dielectric between the plates of a charged, isolated capacitor (not connected to the battery), the potential difference decreases. What happens instead if the capacitor is connected to a battery?

Solution

Isolated capacitor (constant charge). Disconnected from the battery, the charge $Q$ on the plates cannot change: it is “trapped”. Inserting the dielectric, the material’s molecules polarise and generate a polarisation field opposite to that of the plates. The net internal field is reduced by a factor $\kappa$ (dielectric constant): $\ev{E = \frac{E_0}{\kappa}}.$ Since the distance $d$ is fixed and $V = E\,d$, the potential difference also drops by the same factor: $\ev{V = \frac{V_0}{\kappa}}.$ The capacitance $C = Q/V$ increases as a result: $C = \kappa C_0$.

Capacitor connected to the battery (constant voltage). Now it is $V$ that is imposed by the battery and cannot change. The capacitance still increases to $C = \kappa C_0$, so to maintain the same $V$ the battery pushes more charge onto the plates: $\ev{Q = \kappa\,Q_0}.$ The field $E = V/d$ stays unchanged, but the stored charge (and energy) increase. In summary: at constant charge, $V$ drops; at constant voltage, $Q$ rises. In both cases the dielectric “helps” the capacitor store more.

Connections

Topics: Electric field and potential Concepts: Dielectrics and polarisation · Capacitance and capacitor Skills: Superposition principle Methods: Gradient method Objects: Parallel-plate capacitor