Capacitance of a Parallel-Plate Capacitor with Dielectric Slab between Plates

Think of the dielectric as a "traffic buffer." When a slab is inserted between the plates, its molecules align and create an internal field that opposes part of the original field — like a crowd partially blocking wind flow through a corridor. This reduces the net electric field inside the slab region, which lowers the voltage needed for the same charge, and since C = q/V, a lower V for the same q means higher C.

Setup: Parallel plates of area A, separated by distance d, carrying charge q. A dielectric slab of thickness t (where t < d) and dielectric constant K is placed between the plates.

Step 1 - Field without dielectric (in the air gap):

Step 2 - Field inside the dielectric (reduced by K):

Step 3 - Total potential difference across the capacitor:

Step 4 - Capacitance from C = q/V:

When several dielectric slabs of thicknesses t₁, t₂, t₃,… and constants K₁, K₂, K₃,… are stacked between the plates, each acts like a capacitor in series:

Simple explanation: A 3.0 mm dielectric slab is inserted between capacitor plates, and the plate separation must increase by 2.4 mm to keep the voltage the same — this tells us the slab's dielectric constant is 5.

Why it happens: Inserting a dielectric slab reduces the field inside it, which lowers the voltage. To bring the voltage back to its original value, you must move the plates farther apart—this compensates for the slab's field-weakening effect.

Steps in the solution:

• Originally, without any slab, the voltage is simply field × distance: V₀ = E₀d.

• After inserting the slab (thickness t) and increasing the gap by d', the new voltage equation has two parts: the air gap contributes E₀(d + d′ − t), and the slab region contributes E₀t/K (weaker field inside dielectric).

• Since both expressions equal the same V_0​, they are set equal to each other and simplified.

• This simplification directly gives K = \[\frac {t}{t−d′}\].

• Plugging in the numbers: K = \[\frac {3.0}{3.0−2.4}\] = \[\frac {3.0}{0.6}\] = 5.

Answer: The dielectric constant of the slab is K = 5.