PhysicsCBSE Class 12 · Electrostatic Potential and Capacitance

Parallel Plate Capacitor and Dielectrics Class 12: Derivations and Effects

Updated 15 October 2026 · Fact-checked

A parallel plate capacitor has capacitance C = ε₀A/d, because the field between the plates is σ/ε₀ and the voltage is Ed. Filling the gap with a dielectric of constant K multiplies C by K; a slab of thickness t gives C = ε₀A/(d − t + t/K). A connected battery fixes V; an isolated capacitor keeps Q.

Understand Parallel Plate Capacitor and Dielectrics

A capacitor is two conductors kept apart by an insulator. Connect it to a battery and charge $+ Q$ collects on one conductor and $- Q$ on the other; the potential difference V between them is proportional to Q. The ratio $C = \frac{Q}{V}$ is the capacitance, measured in farad (1 F = 1 C/V). One farad is enormous: practical capacitors are in μF, nF or pF.

Parallel plate capacitor. Two large plates of area A, a small distance d apart, carry $+ Q$ and $- Q$ , so the surface charge density is $σ = \frac{Q}{A}$ . Treat each plate as an infinite sheet (fine when d is small compared with the plate size): between the plates the two fields $\frac{σ}{2 ε _{0}}$ add to $E = \frac{σ}{ε _{0}}$ , and outside they cancel. The field is uniform, so $V = E d = \frac{Q d}{ε _{0} A}$ , and $C = \frac{Q}{V} = \frac{ε _{0} A}{d}$ . Capacitance depends only on geometry and the medium, not on Q or V.

Dielectrics and polarisation. A dielectric is an insulator. In an external field its molecules become dipoles (non-polar molecules are stretched; polar molecules, already dipoles, are partly lined up). The dielectric is then polarised: inside it the dipoles cancel, but a layer of negative bound charge appears on the face near the positive plate and positive bound charge on the face near the negative plate. These bound charges set up a field opposite to the applied field, so the net field inside is reduced to $E = \frac{E _{0}}{K}$ , where K is the dielectric constant (relative permittivity, $K = \frac{ε}{ε _{0}} > 1$ ). The polarisation $P$ is the dipole moment per unit volume; for a slab its size equals the bound surface charge density $σ_{p}$ , with $σ_{p} = σ (1 - \frac{1}{K})$ .

Capacitor filled with a dielectric. With the same charge Q, the field and hence V fall by K, so $C = \frac{K ε _{0} A}{d} = K C_{0}$ . This gives an operational definition: $K = \frac{C}{C _{0}}$ .

Slab of thickness t (t < d). The field is $E_{0} = \frac{σ}{ε _{0}}$ in the air gap of total thickness $(d - t)$ and $\frac{E _{0}}{K}$ in the slab. So $V = E_{0} (d - t) + \frac{E _{0}}{K} t$ , and $C = \frac{ε _{0} A}{d - t + \frac{t}{K}}$ . The position of the slab between the plates does not matter. A metal slab has $E = 0$ inside (like $K \to \infty$ ): $C = \frac{ε _{0} A}{d - t}$ .

Battery connected or disconnected? This decides everything when a dielectric is put in.

Disconnected (isolated): Q stays the same. C becomes $K C_{0}$ , V and E become $\frac{1}{K}$ times, and the energy $U = \frac{Q ^{2}}{2 C}$ becomes $\frac{1}{K}$ times (the field does work pulling the slab in).
Connected to the battery: V stays the same. C becomes $K C_{0}$ , Q becomes K times (the battery supplies extra charge), E stays the same ( $E = \frac{V}{d}$ ) and $U = \frac{1}{2} C V^{2}$ becomes K times.

The energy formulas $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ are quoted without derivation in the 2026-27 syllabus.

Parallel Plate Capacitor and Dielectrics: formulas and key facts

Capacitance: $C = \frac{Q}{V}$; SI unit farad (F); 1 pF = $1 0^{- 12}$ F.
Parallel plate capacitor (air or vacuum): $C_{0} = \frac{ε _{0} A}{d}$; Field between plates $E = \frac{σ}{ε _{0}}$ , uniform.
Completely filled with a dielectric: $C = \frac{K ε _{0} A}{d} = K C_{0}$; $K = \frac{C}{C _{0}} = \frac{ε}{ε _{0}}$ .
Dielectric slab of thickness t: $C = \frac{ε _{0} A}{d - t + \frac{t}{K}}$; Metal slab: $C = \frac{ε _{0} A}{d - t}$ .
Field inside a dielectric: $E = \frac{E _{0}}{K}, σ_{p} = σ (1 - \frac{1}{K})$; Bound charge reduces the field.
Energy stored (formula only): $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$; Use $\frac{Q ^{2}}{2 C}$ when Q is fixed and $\frac{1}{2} C V^{2}$ when V is fixed.
Dielectric inserted: summary: $Isolated: Q same, V \to \frac{V}{K}, U \to \frac{U}{K}; Battery on: V same, Q \to K Q, U \to K U$; C becomes KC in both cases.

How to solve Parallel Plate Capacitor and Dielectrics questions

Use this method for capacitance and dielectric questions.

1Convert area to m² (1 cm² = $1 0^{- 4}$ m²) and separations to m (1 mm = $1 0^{- 3}$ m).
2Find $C_{0} = \frac{ε _{0} A}{d}$ first; most answers are simple multiples of it.
3For a slab, use $C = \frac{ε _{0} A}{d - t + t / K}$ ; for a metal slab drop the $t / K$ term. For dielectrics side by side, treat them as capacitors in parallel; stacked ones as capacitors in series.
4Decide whether the battery is connected (V fixed) or disconnected (Q fixed). Write this down before changing anything.
5Find the new C, then the new Q or V from $Q = C V$ , then E from $E = \frac{V}{d}$ (or $\frac{E _{0}}{K}$ inside the dielectric).
6For energy, pick the formula that uses the quantity that stays fixed.
7Express the answer with units and, where asked, as a ratio to the original value.

Parallel Plate Capacitor and Dielectrics in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Capacitance depends on plate separation

Suppose a computer key works as a small parallel plate capacitor: a metal plate of area 1 cm² under the key sits 0.6 mm above a fixed plate, with air between. Pressing the key fully brings the gap down to 0.2 mm, and the keyboard circuit registers a press as soon as the capacitance rises by 0.5 pF.

(a) Find the capacitance with the key up and fully pressed. (b) How far must the key move before a press is registered?

Try it first, then show the solutionSolution

Key up: $C = \frac{ε _{0} A}{d} = \frac{8.854 \times 1 0 ^{- 12} \times 1 0 ^{- 4}}{0.6 \times 1 0 ^{- 3}} \approx 1.48$ pF.
Fully pressed: the gap is a third, so C is three times: about 4.43 pF.
Press registered at $C = 1.48 + 0.5 = 1.98$ pF: $d = \frac{ε _{0} A}{C} = \frac{8.854 \times 1 0 ^{- 16}}{1.98 \times 1 0 ^{- 12}} \approx 0.448$ mm.
Travel = 0.6 − 0.448 ≈ 0.15 mm.

Answer: (a) about 1.48 pF up, 4.43 pF pressed; (b) about 0.15 mm.

Memory hook: Closer plates, bigger C: a key press is a change in d.

How large a farad is

Imagine you wanted to build a 1 F capacitor from two flat metal sheets 1 mm apart with air between them.

(a) What plate area would you need? (b) Roughly how long would each side of a square plate be? (c) Suggest two changes that would shrink it.

Try it first, then show the solutionSolution

$A = \frac{C d}{ε _{0}} = \frac{1 \times 1 0 ^{- 3}}{8.854 \times 1 0 ^{- 12}} \approx 1.13 \times 1 0^{8}$ m².
Side $= 1.13 \times 1 0^{8} \approx 1.06 \times 1 0^{4}$ m, over 10 km.
Use a much smaller gap and fill it with a material of large dielectric constant; both raise C for a given area.

Answer: (a) about $1.13 \times 1 0^{8}$ m²; (b) about 10.6 km on each side; (c) reduce d and use a dielectric with large K.

Memory hook: One farad of air capacitor would cover a city: that is why we use pF and μF.

Tricks to remember Parallel Plate Capacitor and Dielectrics

Short memory aids. Say each one aloud once, then use it on a question the same day.

C likes Area, hates Distance
C = ε₀A/d: more area, more C; more distance, less C; a dielectric multiplies by K.
Use it when: Predicting changes quickly.
Cut the gap, then shrink the slab
In d − t + t/K, take the slab out of the gap (d − t) and put it back K times thinner (t/K).
Use it when: Remembering the slab formula.
Battery on: V is boss. Battery off: Q is boss
Whatever the battery fixes (V) or the isolation fixes (Q) stays; everything else changes around it.
Use it when: Dielectric and plate-separation change questions.
Metal slab = infinite K
Put K → ∞ in the slab formula and t/K vanishes.
Use it when: Conducting slab questions.

Parallel Plate Capacitor and Dielectrics: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Capacitance, charge and effect of filling · 2 marks · Very short answerBoard level

A parallel plate capacitor has plates of area 400 cm² separated by 2 mm of air. (a) Find its capacitance. (b) Find the charge on each plate when connected to 100 V. (c) Find the capacitance if the gap is filled with a dielectric of K = 5.

Try it first, then show the solutionSolution

(a) $C_{0} = \frac{8.854 \times 1 0 ^{- 12} \times 0.04}{0.002} \approx 1.77 \times 1 0^{- 10}$ F = 177 pF. (1)
(b) $Q = C_{0} V = 1.77 \times 1 0^{- 10} \times 100 \approx 1.77 \times 1 0^{- 8}$ C. (½)
(c) $C = 5 C_{0} \approx 885$ pF. (½)

Answer: (a) about 177 pF; (b) about $1.77 \times 1 0^{- 8}$ C; (c) about 885 pF.

Example 2: Derivation: capacitance of a parallel plate capacitor · 3 marks · Short answerBoard level

Derive an expression for the capacitance of a parallel plate capacitor with air between the plates. State the assumption you make.

Try it first, then show the solutionSolution

Plates of area A carry +Q and −Q, separation d; surface charge density $σ = \frac{Q}{A}$ . Assume d is small compared with the plate size, so each plate acts as an infinite sheet and edge effects are ignored. (1)
Between the plates the fields $\frac{σ}{2 ε _{0}}$ of the two plates point the same way and add: $E = \frac{σ}{ε _{0}} = \frac{Q}{ε _{0} A}$ , uniform; outside, they cancel. (1)
$V = E d = \frac{Q d}{ε _{0} A}$ , so $C = \frac{Q}{V} = \frac{ε _{0} A}{d}$ . (1)

Answer: $C = \frac{ε _{0} A}{d}$ , assuming plates large compared with their separation.

Example 3: Derivation and use: partly filled capacitor · 5 marks · Long answerHarder board level

(a) A slab of dielectric constant K and thickness t (t < d) is placed between the plates of a parallel plate capacitor of plate area A and separation d. Derive the new capacitance. (3)
(b) Evaluate it for A = 100 cm², d = 4 mm, t = 2 mm and K = 4, and compare with the air value. (2)

Try it first, then show the solutionSolution

(a) With charge Q, $E_{0} = \frac{σ}{ε _{0}} = \frac{Q}{ε _{0} A}$ in the air gaps (total thickness d − t). (1)
Inside the slab the field is reduced by polarisation to $\frac{E _{0}}{K}$ . So $V = E_{0} (d - t) + \frac{E _{0}}{K} t = \frac{Q}{ε _{0} A} (d - t + \frac{t}{K})$ . (1)
$C = \frac{Q}{V} = \frac{ε _{0} A}{d - t + \frac{t}{K}}$ . (1)
(b) $C_{0} = \frac{8.854 \times 1 0 ^{- 12} \times 0.01}{0.004} \approx 22.1$ pF. With the slab, $d - t + \frac{t}{K} = 4 - 2 + 0.5 = 2.5$ mm, so $C = \frac{8.854 \times 1 0 ^{- 14}}{2.5 \times 1 0 ^{- 3}} \approx 35.4$ pF. (1)
$\frac{C}{C _{0}} = \frac{4}{2.5} = 1.6$ . (1)

Answer: (a) $C = \frac{ε _{0} A}{d - t + t / K}$ ; (b) about 35.4 pF, 1.6 times the air value of 22.1 pF.

Example 4: Battery connected versus disconnected · 3 marks · Short answerHarder board level

A 10 pF air capacitor is charged to 50 V. A dielectric slab of K = 3 is then made to fill the gap. Find the new charge, voltage and stored energy if (a) the battery was disconnected before the slab went in, and (b) the battery stayed connected.

Try it first, then show the solutionSolution

Initially $Q_{0} = C_{0} V_{0} = 500$ pC; $U_{0} = \frac{1}{2} C_{0} V_{0}^{2} = 1.25 \times 1 0^{- 8}$ J; new C = 30 pF in both cases. (1)
(a) Q stays 500 pC; $V = \frac{Q}{C} = \frac{50}{3} \approx 16.7$ V; $U = \frac{Q ^{2}}{2 C} = \frac{U _{0}}{3} \approx 4.17 \times 1 0^{- 9}$ J. (1)
(b) V stays 50 V; $Q = C V = 1500$ pC; $U = \frac{1}{2} C V^{2} = 3 U_{0} = 3.75 \times 1 0^{- 8}$ J. (1)

Answer: (a) 500 pC, about 16.7 V, about $4.17 \times 1 0^{- 9}$ J; (b) 1500 pC, 50 V, $3.75 \times 1 0^{- 8}$ J.

Example 5: Two dielectrics in one capacitor · 3 marks · Short answerHOTS / competency

An air capacitor has capacitance 20 pF. Find its capacitance when the gap is filled with two dielectrics of K₁ = 2 and K₂ = 6 (a) side by side, each covering half the plate area across the full gap, and (b) one above the other, each filling half the gap over the full area.

Try it first, then show the solutionSolution

(a) Side by side, each part has area $\frac{A}{2}$ and gap d: two capacitors in parallel, $\frac{K _{1} C _{0}}{2} + \frac{K _{2} C _{0}}{2} = \frac{( 2 + 6 )}{2} \times 20 = 80$ pF. (1½)
(b) Stacked, each layer has area A and thickness $\frac{d}{2}$ : capacitances $2 K_{1} C_{0} = 80$ pF and $2 K_{2} C_{0} = 240$ pF in series. (1)
$C = \frac{80 \times 240}{80 + 240} = 60$ pF. (½)

Answer: (a) 80 pF; (b) 60 pF.

Example 6: Polarisation inside a dielectric · 4 marks · Case-basedHarder board level

Read the passage and answer.

The plates of a parallel plate capacitor carry a surface charge density of $8.85 \times 1 0^{- 7}$ C/m². A slab of dielectric constant 4 fills the gap.
(a) What would the field between the plates be without the slab? (1)
(b) Find the field inside the slab. (1)
(c) Find the surface density of the bound charge on the faces of the slab. (1)
(d) Explain, in terms of molecules, why the field is reduced. (1)

Try it first, then show the solutionSolution

(a) $E_{0} = \frac{σ}{ε _{0}} = \frac{8.85 \times 1 0 ^{- 7}}{8.85 \times 1 0 ^{- 12}} = 1 \times 1 0^{5}$ V/m. (1)
(b) $E = \frac{E _{0}}{K} = 2.5 \times 1 0^{4}$ V/m. (1)
(c) $σ_{p} = σ (1 - \frac{1}{K}) = 8.85 \times 1 0^{- 7} \times 0.75 \approx 6.64 \times 1 0^{- 7}$ C/m². (1)
(d) The field turns the molecules into (or lines up) dipoles; their ends leave bound charge on the slab faces, opposite in sign to the nearby plate, which sets up a field opposing the applied field. (1)

Answer: (a) $1 0^{5}$ V/m; (b) $2.5 \times 1 0^{4}$ V/m; (c) about $6.64 \times 1 0^{- 7}$ C/m²; (d) bound charges from aligned dipoles oppose the applied field.

Where students lose marks in Parallel Plate Capacitor and Dielectrics

The slips examiners see most often, and what to write instead.

Mistake: Not deciding whether the battery is connected before answering.
Students apply the 'Q fixed' result to every case.
Wrong working
Battery connected, dielectric inserted: V decreases to V/K.
Right working
Battery connected: V stays the same, Q becomes KQ.
Do this: Write 'Q constant' or 'V constant' as the first line.
Usually costs: 1 to 2 marks in a 3-mark question.
Mistake: Putting area in cm² into ε₀A/d.
Forgetting that 1 cm² = 10⁻⁴ m².
Wrong working
C = 8.854 × 10⁻¹² × 400/0.002
Right working
C = 8.854 × 10⁻¹² × 0.04/0.002
Do this: Convert area and separation to SI in the first line.
Usually costs: The answer mark.
Mistake: Writing C = Kε₀A/d for a partly filled capacitor.
Using the fully filled formula when t < d.
Do this: Use $\frac{ε _{0} A}{d - t + t / K}$ whenever the slab is thinner than the gap.
Usually costs: Most of the marks of the numerical.
Mistake: Skipping the field of each plate in the derivation.
Jumping to E = σ/ε₀ without showing where it comes from.
Do this: Write the field of each plate (σ/2ε₀), add between the plates and cancel outside.
Usually costs: 1 mark in a 3-mark derivation.
Mistake: Explaining the reduced field as 'the dielectric blocks the field'.
No mention of polarisation or bound charges.
Do this: Say: molecules become dipoles, bound surface charges appear, and their field opposes the applied field.
Usually costs: 1 mark in a reasoning question.

Exam tips

In the slab derivation, a diagram showing E₀ in the air gaps and E₀/K in the slab earns a step mark.
Quote energy formulas directly; the 2026-27 syllabus does not ask for the derivation of energy stored.
For change questions, a small table (C, Q, V, E, U before and after) is quick and easy for examiners to mark.
Board papers have set capacitor-with-dielectric questions as 3- and 5-mark items in each recent year (for example 2025 55/1/1 Q31(a) and 2026 55/1/1 Q23).
Distinguish polar and non-polar dielectrics in one line each when asked how polarisation happens.

Practice questions on Parallel Plate Capacitor and Dielectrics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A charged parallel plate capacitor is disconnected from the battery and a dielectric slab is inserted. Which quantity stays the same?
(a) potential difference (b) charge (c) capacitance (d) stored energy
Hint
Where could charge go?
Answer
(b) charge.
Q2 · 2 marks · Very short answerBoard level
Define dielectric constant in terms of capacitance. Can it be less than 1 for any material?
Hint
Compare C with and without the material.
Answer
$K = \frac{C}{C _{0}}$ , the ratio of the capacitance with the material filling the gap to that with vacuum. No; every material reduces the field by polarisation, so K > 1 (K = 1 for vacuum).
Q3 · 2 marks · Very short answerHarder board level
A metal plate of thickness $\frac{d}{2}$ and the same area as the plates is placed in the middle of an air capacitor of plate separation d and capacitance $C_{0}$ . Find the new capacitance.
Hint
No field inside a metal.
Answer
$C = \frac{ε _{0} A}{d - d /2} = 2 C_{0}$ .
Q4 · 1 mark · Very short answerHarder board level
The plate separation of an air capacitor of capacitance $C_{0}$ is doubled and the gap is completely filled with a dielectric of constant 4. Find the new capacitance.
Hint
C ∝ K/d.
Answer
$C = \frac{4 ε _{0} A}{2 d} = 2 C_{0}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When a dielectric slab is inserted into an isolated charged capacitor, the stored energy decreases.
Reason (R): The capacitance increases while the charge stays the same, and $U = \frac{Q ^{2}}{2 C}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Isolated means Q is fixed.
Answer
(a). With Q fixed and C increased K times, U falls to $\frac{1}{K}$ of its value.
Q6 · 3 marks · Short answerHOTS / competency
A parallel plate air capacitor stays connected to a battery while its plate separation is doubled. How do (a) C, (b) Q, (c) E between the plates and (d) the stored energy change?
Hint
V is fixed.
Answer
(a) C halves ( $C \propto \frac{1}{d}$ ). (b) Q = CV halves. (c) $E = \frac{V}{d}$ halves. (d) $U = \frac{1}{2} C V^{2}$ halves.
Q7 · 3 marks · Short answerHOTS / competency
A dielectric slab of thickness half the plate separation is inserted into an air capacitor. (a) What value of K makes the capacitance 1.5 times its air value? (b) Can any dielectric slab of this thickness double the capacitance? Explain.
Hint
Use $\frac{C}{C _{0}} = \frac{d}{d - t + t / K}$ with $t = \frac{d}{2}$ .
Answer
(a) $\frac{1}{\frac{1}{2} + \frac{1}{2 K}} = 1.5$ gives $\frac{1}{2 K} = \frac{1}{6}$ , so K = 3. (b) No. For C = 2C₀ we would need $\frac{1}{2 K} = 0$ , i.e. infinite K. Only a metal slab of that thickness reaches 2C₀; any dielectric gives less.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Capacitance of a parallel plate air capacitor?Show answer
$C = \frac{ε _{0} A}{d}$ .
2.Capacitance with a dielectric slab of thickness t?Show answer
$C = \frac{ε _{0} A}{d - t + t / K}$ .
3.What happens to the field inside a dielectric of constant K?Show answer
It becomes $\frac{E _{0}}{K}$ .
4.Isolated capacitor, dielectric inserted: what happens to U?Show answer
It becomes U/K.
5.Capacitor connected to a battery, dielectric inserted: what happens to Q?Show answer
It becomes KQ.
6.Define polarisation.Show answer
Dipole moment per unit volume of the dielectric.

Parallel Plate Capacitor and Dielectrics: frequently asked questions

Why does a dielectric increase capacitance?

Polarisation creates bound charges that reduce the field between the plates. For the same charge, a smaller field means a smaller potential difference, and since C = Q/V, the capacitance rises by the factor K.

Does the position of a dielectric slab between the plates matter?

No. Only the total thicknesses of air and dielectric matter, because the voltage is the sum of field × thickness for each layer. The slab can touch a plate or sit in the middle.

What is the difference between polar and non-polar dielectrics?

Non-polar molecules (like O₂) have no dipole moment until a field stretches them. Polar molecules (like H₂O) already have dipole moments that point randomly; a field partly lines them up. Both kinds end up polarised along the field.

Do I need to derive the energy stored in a capacitor?

No. For 2026-27 CBSE lists energy stored in a capacitor as 'no derivation, formulae only'. Learn U = ½CV² = Q²/2C = ½QV and when to use each.

Why is the field outside a parallel plate capacitor zero?

Outside the plates, the fields of the positive and negative plates (each σ/2ε₀) point in opposite directions and cancel. Between the plates they point the same way and add to σ/ε₀.