Effect of Dielectric on Capacitance

How inserting an insulator multiplies capacitance by the dielectric constant.

The Key Result

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.

Mechanism — Why C Increases

When charge \( Q_{\text{free}} \) is placed on the plates:

The dielectric reduces the effective electric field for the same free charge — which means a smaller potential difference — which means a larger capacitance \( C = Q/V \).

Parallel Plate Capacitor with Dielectric

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:

\[ E = \frac{\sigma}{\varepsilon} = \frac{\sigma}{K\varepsilon_0}, \qquad V = Ed = \frac{\sigma d}{K\varepsilon_0} \]

Effect on Field, Voltage, and Energy

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 \( K \)