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JEE Mains Physics · Electrostatics

Dielectric Slabs and Partly Filled Capacitors

A slab laid across the whole gap acts as capacitors in series, a slab placed side by side acts as capacitors in parallel; a slab of thickness t shortens the effective gap by t(1 − 1/K).

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Why this matters

Twenty-five PYQs, eighteen of them multiple choice, and five from 2026. Eighteen put a slab or a metal sheet across the gap, or stack layers of different dielectrics, so the pieces are in series. Seven place dielectrics side by side, or use plates shaped like stairs, so the pieces are in parallel.

Concept 1 of 2: Slabs and layers across the gap

A slab that covers the whole plate area but only part of the gap splits the capacitor into layers stacked one above the other. The same charge crosses every layer, so the layers are in series. Inside the slab the field is K times weaker, so a slab of thickness t acts like an air gap of only t/K. A metal sheet has no field inside at all: it simply removes its thickness from the gap.

Definition

  • Slab of thickness t over the full area: C=ε0Ad−t+t/KC = \dfrac{\varepsilon_0 A}{d - t + t/K}. Its position in the gap does not matter.
  • Metal sheet (K very large): C=ε0Ad−tC = \dfrac{\varepsilon_0 A}{d - t}.
  • Several layers: 1C=∑tiKiε0A\dfrac{1}{C} = \sum \dfrac{t_i}{K_i\varepsilon_0 A}.
  • To get the old C back after inserting a slab, move the plates apart by t(1−1K)t\left(1 - \dfrac{1}{K}\right).
  • Field in the air: E0=Vd−t+t/KE_0 = \dfrac{V}{d - t + t/K}; in the slab: E0/KE_0/K. Each layer's voltage is its field times its thickness.
  • Permittivity that changes across the gap: 1C=1A∫0ddxε(x)\dfrac{1}{C} = \dfrac{1}{A}\displaystyle\int_0^{d} \frac{dx}{\varepsilon(x)}.

Slab across the gap

C=ε0Ad−t+tK,1C=∑itiKiε0AC = \frac{\varepsilon_0 A}{d - t + \dfrac{t}{K}}, \qquad \frac{1}{C} = \sum_i \frac{t_i}{K_i\varepsilon_0 A}

Worked example

An air capacitor with plates 6 mm apart has a capacitance of 12 pF. A slab 4 mm thick with K = 4 is slid in, covering the plates. Find the new capacitance, and the capacitance if a 4 mm metal sheet is used instead.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q9Moderate

Example 1 · Electrostatics · Dielectric Slabs and Partly Filled Capacitors

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws 25%25\% more charge from the battery. The dielectric constant of mica is ____\_\_\_\_ .

Across the gap means series

A slab covering the whole plate area splits the gap into layers. Add their reciprocals; adding Kε₀A/d terms treats them as side by side.

A metal sheet removes its thickness

There is no field inside the metal, so the gap simply shrinks by t. Using t/K with some large K only approximates this.

Thickness matters, position does not

Moving a slab closer to one plate leaves C unchanged. Only its thickness and K enter the formula.
Drill 17 more on slabs and layers across the gap

Concept 2 of 2: Dielectrics side by side and stair plates

When dielectrics sit side by side, each filling the whole gap over part of the plate area, every part has the same voltage across it. The parts are capacitors in parallel, and their capacitances add, each with its own area. Plates shaped like stairs work the same way: each step is a separate strip with its own gap.

Definition

  • Parts side by side, each the full thickness: C=ε0d∑KiAiC = \dfrac{\varepsilon_0}{d}\sum K_iA_i.
  • Two dielectrics each over half the area: C=K1+K22 C0C = \dfrac{K_1 + K_2}{2}\,C_0.
  • Mixed arrangements: split the capacitor into side-by-side columns (in parallel); within each column, the layers are in series.
  • Boundary between dielectrics parallel to the plates: series. Boundary perpendicular to the plates: parallel.
  • Stair plates: each step is a strip of area AiA_i at gap did_i, and C=ε0∑AidiC = \varepsilon_0\sum \dfrac{A_i}{d_i}.
  • A full-thickness slab pushed a length x into plates of length l and width b: C=ε0bd[Kx+(l−x)]C = \dfrac{\varepsilon_0 b}{d}[Kx + (l - x)].

Dielectrics side by side

C=ε0d∑iKiAiC = \frac{\varepsilon_0}{d}\sum_i K_iA_i

Worked example

Square plates of side 10 cm are 2 mm apart. The left half of the gap is filled, full thickness, with a dielectric of K = 3; the right half is air. Find C and the energy stored at 10 V. (Answer in terms of ε0\varepsilon_0.)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q116Moderate

Example 2 · Electrostatics · Dielectric Slabs and Partly Filled Capacitors

A parallel plate capacitor with width 4 cm4\text{ }cm, length 8 cm8\text{ }cm and separation between the plates of 4 mm4\text{ }mm is connected to a battery of 20 V20\text{ }V. A dielectric slab of dielectric constant 5 having length 1 cm1\text{ }cm, width 4 cm4\text{ }cm and thickness 4 mm4\text{ }mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____\_\_\_\_ ∈0\in_{0} J. (Where ϵ0\epsilon_{0} is the permittivity of free space)

Side by side means parallel

Dielectrics that each span the whole gap share one voltage. Add their capacitances; adding reciprocals treats them as layers.

Each part uses its own area

A dielectric over half the plates contributes Kε₀(A/2)/d, not Kε₀A/d. Using the full area for every part counts the plates twice.

Read the boundary

If the surface between two dielectrics is parallel to the plates, they are in series; if it is perpendicular, they are in parallel. Look at the figure before choosing.
Drill 6 more on dielectrics side by side and stair plates

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Slabs and layers across the gap

    Slab across the gap

    C=ε0Ad−t+tK,1C=∑itiKiε0AC = \frac{\varepsilon_0 A}{d - t + \dfrac{t}{K}}, \qquad \frac{1}{C} = \sum_i \frac{t_i}{K_i\varepsilon_0 A}
  • Dielectrics side by side and stair plates

    Dielectrics side by side

    C=ε0d∑iKiAiC = \frac{\varepsilon_0}{d}\sum_i K_iA_i

Watch out for (6)

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