How inserting an insulator multiplies capacitance by the dielectric constant.
When a dielectric of dielectric constant \( K \) (also called relative permittivity \( \varepsilon_r \)) completely fills the space between the plates of a capacitor:
\[ \boxed{C' = K\, C_0} \]where \( C_0 = \varepsilon_0 A / d \) is the capacitance in vacuum. The capacitance increases by factor \( K \). Since \( K > 1 \) for all dielectrics, inserting a dielectric always increases capacitance.
When charge \( Q_{\text{free}} \) is placed on the plates:
For a parallel plate capacitor completely filled with a dielectric of constant \( K \):
\[ C' = \frac{K\varepsilon_0 A}{d} = \frac{\varepsilon A}{d} \]where \( \varepsilon = K\varepsilon_0 = \varepsilon_r \varepsilon_0 \) is the absolute permittivity of the dielectric. The formulas for field and potential become:
\[ E = \frac{\sigma}{\varepsilon} = \frac{\sigma}{K\varepsilon_0}, \qquad V = Ed = \frac{\sigma d}{K\varepsilon_0} \]| Quantity | Isolated Capacitor (const. Q) | Battery Connected (const. V) |
|---|---|---|
| Capacitance \( C \) | Increases by \( K \) | Increases by \( K \) |
| Charge \( Q \) | Unchanged | Increases by \( K \) |
| Electric field \( E \) | Decreases by \( K \) | Unchanged |
| Voltage \( V \) | Decreases by \( K \) | Unchanged |
| Stored energy \( U \) | Decreases by \( K \) | Increases by \( K \) |