Partially Filled Capacitor

When dielectric fills only part of the gap — series or parallel slab problems.

Slab Inserted Parallel to Plates (Series Combination)

Consider a parallel plate capacitor of plate area \( A \) and separation \( d \). A dielectric slab of thickness \( t \) and dielectric constant \( K \) is inserted between the plates (parallel to them), leaving gaps of total thickness \( (d - t) \) in air.

The system is equivalent to two capacitors in series:

Air gap: \( C_{\text{air}} = \dfrac{\varepsilon_0 A}{d - t} \)
Dielectric slab: \( C_{\text{diel}} = \dfrac{K\varepsilon_0 A}{t} \)

\[ \frac{1}{C} = \frac{1}{C_{\text{air}}} + \frac{1}{C_{\text{diel}}} = \frac{d - t}{\varepsilon_0 A} + \frac{t}{K\varepsilon_0 A} \] \[ \boxed{C = \frac{\varepsilon_0 A}{(d - t) + t/K} = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}} \]

Special cases:

\( K = 1 \): \( C = \varepsilon_0 A / d \) — no change (air dielectric).
\( K \to \infty \) (conducting slab): \( C = \varepsilon_0 A / (d - t) \) — same as reducing the gap by \( t \).
\( t = d \): \( C = K\varepsilon_0 A / d = KC_0 \) — fully filled.

Since \( 1 - 1/K \geq 0 \) for \( K \geq 1 \), the effective gap is reduced compared to the real gap, so \( C \geq C_0 \). Inserting any dielectric slab (even partial) always increases (or maintains) the capacitance.

Slab Inserted Perpendicular to Plates (Parallel Combination)

If the dielectric partially fills the capacitor such that it occupies area \( A_1 \) and the remaining area \( A_2 = A - A_1 \) is in air, the two regions are in parallel (same voltage):

\[ C = \frac{K\varepsilon_0 A_1}{d} + \frac{\varepsilon_0 A_2}{d} = \frac{\varepsilon_0}{d}(KA_1 + A_2) \]

If the slab fills fraction \( f = A_1/A \) of the area:

\[ C = \frac{\varepsilon_0 A}{d}(Kf + 1 - f) = C_0 [1 + f(K-1)] \]

This ranges from \( C_0 \) (when \( f = 0 \)) to \( KC_0 \) (when \( f = 1 \)), varying linearly with the fraction filled.

Two Dielectric Slabs Side by Side (Parallel)

Two dielectric slabs of constants \( K_1 \) and \( K_2 \), each of thickness \( d \) and area \( A/2 \):

\[ C = \frac{\varepsilon_0 (A/2)}{d}(K_1 + K_2) = \frac{\varepsilon_0 A (K_1 + K_2)}{2d} \]

Two Dielectric Slabs Stacked (Series)

Two dielectrics stacked on top of each other, each of thickness \( d/2 \), constants \( K_1 \) and \( K_2 \), area \( A \):

\[ \frac{1}{C} = \frac{d/2}{K_1 \varepsilon_0 A} + \frac{d/2}{K_2 \varepsilon_0 A} = \frac{d}{2\varepsilon_0 A}\left(\frac{1}{K_1} + \frac{1}{K_2}\right) \] \[ C = \frac{2\varepsilon_0 A \, K_1 K_2}{d(K_1 + K_2)} \]