PhysicsISC Class 12 · Electrostatic Potential, Potential Energy and Capacitance

Capacitance of a Parallel-Plate Capacitor and the Effect of a Dielectric

Updated 15 October 2026 · Fact-checked

A parallel-plate capacitor with plates of area A a distance d apart in vacuum has capacitance $C = \frac{\varepsilon_0A}{d}$. Filling the gap with a dielectric of constant K multiplies the capacitance by K, because the polarised dielectric weakens the field. For a slab of thickness t, $C = \frac{\varepsilon_0A}{d - t + \frac{t}{K}}$.

Bound charges on the dielectric faces oppose the field, so E falls to E₀/K for the same plate charge, and capacitance rises K times.

Illustration drawn for QuestionBuddy.

Understand Capacitance of a Parallel-Plate Capacitor and Dielectrics

Capacitance. $C = \frac{Q}{V}$ : the charge stored per volt. It depends only on the size, shape and separation of the conductors and on the medium between them, not on Q or V.

Derivation for parallel plates. Two large plates of area A carry $+ Q$ and $- Q$ , separated by a small distance d in vacuum. The surface charge density is $σ = \frac{Q}{A}$ .
1. Between the plates the fields of the two sheets add: $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}} = \frac{Q}{ε _{0} A}$ (uniform; edge effects ignored because d is small compared with the plate size).
2. The p.d. is $V = E d = \frac{Q d}{ε _{0} A}$ .
3. So $C = \frac{Q}{V} = \frac{ε _{0} A}{d}$ .
C is directly proportional to A and inversely proportional to d: moving the plates apart decreases C, as asked in ISC 2025.

Dielectrics and polarisation. A dielectric is an insulator. In an external field its molecules become (or line up as) dipoles: the dielectric is polarised. Inside, the dipoles cancel, but bound charges appear on its faces: negative next to the positive plate, positive next to the negative plate. These induced charges set up a field opposite to the applied field, so the field inside is reduced:
$E = \frac{E _{0}}{K} .$
The free charges on the plates are unchanged by this; only the bound charges of the dielectric are induced. The induced surface charge density is $σ_{p} = σ (1 - \frac{1}{K})$ .

Dielectric constant. For a gap completely filled with dielectric,
$K = \frac{C ^{'}}{C} = \frac{ε}{ε _{0}} = ε_{r}, C^{'} = \frac{K ε _{0} A}{d} .$
K is also the factor by which the medium reduces the field or force between charges. It is 1 for vacuum, about 1 for air, and larger for glass, mica, paper or water.

Partly filled gap. A slab of thickness $t < d$ and constant K inside the gap: the field is $E_{0}$ in the air parts (total width $d - t$ ) 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 slab acts as if the gap were reduced by $t (1 - \frac{1}{K})$ . A metal slab ( $K \to \infty$ ) gives $C = \frac{ε _{0} A}{d - t}$ .

Battery disconnected (isolated capacitor): Q stays the same.
$C^{'} = K C$ , $V^{'} = \frac{V}{K}$ , $E^{'} = \frac{E}{K}$ , $U^{'} = \frac{Q ^{2}}{2 K C} = \frac{U}{K}$ (energy decreases; the slab is pulled in and the field does work on it).

Battery connected: V stays the same.
$C^{'} = K C$ , $Q^{'} = K Q$ (the battery supplies extra charge), $E^{'} = \frac{V}{d} = E$ (unchanged, because the extra free charge cancels the bound charge's effect), $U^{'} = \frac{1}{2} K C V^{2} = K U$ (energy increases).

Why dielectrics are used. They allow a larger capacitance in a small size, keep the plates apart mechanically, and raise the voltage the capacitor can withstand before breakdown.

Capacitance of a Parallel-Plate Capacitor and Dielectrics: formulas and key facts

Parallel-plate capacitor: $C = \frac{ε _{0} A}{d}$; C ∝ A, C ∝ 1/d.
Field between plates: $E = \frac{σ}{ε _{0}} = \frac{V}{d}$; Uniform field.
Dielectric constant: $K = \frac{C ^{'}}{C} = \frac{ε}{ε _{0}}$; Gap completely filled.
Filled gap: $C^{'} = \frac{K ε _{0} A}{d}$; Capacitance K times larger.
Partly filled gap: $C = \frac{ε _{0} A}{d - t + \frac{t}{K}}$; Metal slab: $\frac{ε _{0} A}{d - t}$ .
Isolated capacitor: $Q^{'} = Q, V^{'} = \frac{V}{K}, U^{'} = \frac{U}{K}$; Battery removed before inserting.
Battery connected: $V^{'} = V, Q^{'} = K Q, U^{'} = K U$; Battery supplies extra charge.

How to solve Capacitance of a Parallel-Plate Capacitor and Dielectrics questions

First decide what is held constant: Q (isolated) or V (connected).

1Find the original capacitance $C = \frac{ε _{0} A}{d}$ with A in m² and d in m.
2Find the new capacitance: KC for a full gap, or the slab formula for a partial gap.
3If the battery was removed, keep Q fixed: $V^{'} = \frac{Q}{C ^{'}}$ , $U^{'} = \frac{Q ^{2}}{2 C ^{'}}$ .
4If the battery stays connected, keep V fixed: $Q^{'} = C^{'} V$ , $U^{'} = \frac{1}{2} C^{'} V^{2}$ .
5For the field, use $E = \frac{V}{d}$ for the air gap, and divide by K inside a dielectric.
6State in words whether each quantity increases, decreases or stays the same.

Capacitance of a 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.

Why real capacitors contain a dielectric

Suppose two capacitors are made with the same plates of area 1 cm² and a gap of 0.1 mm: one with air, one with a ceramic of dielectric constant 100.

Compare their capacitances.

Try it first, then show the solutionSolution

Air: $C = \frac{8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 4}}{1 0 ^{- 4}} = 8.85 \times 1 0^{- 12}$ F = 8.85 pF.
Ceramic: $100 \times 8.85 = 885$ pF.

Answer: 8.85 pF with air and 885 pF with the ceramic: a hundred times more charge per volt in the same size.

Memory hook: Dielectric = K times the capacitance in the same box.

Slab inserted with the battery connected

Imagine a 40 pF air capacitor kept connected to a 9 V battery while a plastic sheet of dielectric constant 2.5 is slid in to fill the gap.

How much extra charge does the battery supply?

Try it first, then show the solutionSolution

Before: $Q = 40 \times 9 = 360$ pC.
After: $C = 100$ pF, so $Q = 900$ pC.
Extra charge: 540 pC.

Answer: 540 pC flows from the battery.

Memory hook: Connected: V fixed, so Q rises with C.

Tricks to remember Capacitance of a Parallel-Plate Capacitor and Dielectrics

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

A up, d down, K up
Capacitance rises with plate area and dielectric constant, and falls with separation.
Use it when: "What happens to C if…" questions.
Isolated: Q fixed, U falls
No battery: charge is trapped, so V and U drop by K.
Use it when: Battery-disconnected problems.
Connected: V fixed, U rises
Battery on: voltage is fixed, so Q and U rise by K.
Use it when: Battery-connected problems.
Slab shrinks the gap
A slab of thickness t acts like reducing d by $t (1 - \frac{1}{K})$ .
Use it when: Partly filled capacitors.

Capacitance of a Parallel-Plate Capacitor and Dielectrics: solved examples

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

Example 1: Plates moved apart · 1 mark · MCQBoard level

The separation between the plates of a parallel-plate capacitor is doubled. Its capacitance:
(a) doubles
(b) halves
(c) is unchanged
(d) becomes four times

Try it first, then show the solutionSolution

$C = \frac{ε _{0} A}{d}$ , so $C \propto \frac{1}{d}$ .

Answer: (b) halves.

Example 2: Capacitance with and without a dielectric · 2 marks · Short answerBoard level

A parallel-plate capacitor has plates of area 0.05 m² separated by 0.5 mm of air. Find its capacitance, and its capacitance when the gap is filled with a material of K = 4. ( $ε_{0} = 8.85 \times 1 0^{- 12}$ F m⁻¹)

Try it first, then show the solutionSolution

$C = \frac{8.85 \times 1 0 ^{- 12} \times 0.05}{5 \times 1 0 ^{- 4}} = 8.85 \times 1 0^{- 10}$ F = 885 pF. (1)
With K = 4: $3.54 \times 1 0^{- 9}$ F = 3.54 nF. (1)

Answer: 885 pF; 3.54 nF.

Example 3: Derivation: parallel-plate capacitance · 3 marks · Long answerHarder board level

Derive an expression for the capacitance of a parallel-plate capacitor with vacuum between its plates.

Try it first, then show the solutionSolution

Plates of area A carry ±Q, separated by d; surface density $σ = \frac{Q}{A}$ . Between the plates, $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}}$ , uniform (d small, edge effects neglected). (1)
$V = E d = \frac{Q d}{ε _{0} A}$ . (1)
$C = \frac{Q}{V} = \frac{ε _{0} A}{d}$ . (1)

Answer: $C = \frac{ε _{0} A}{d}$ .

Example 4: Partly filled gap · 3 marks · Short answerHarder board level

A parallel-plate capacitor has plates of area 0.01 m² and a gap of 4 mm. Find its capacitance (a) with air, (b) with a 2 mm slab of K = 4 inside, (c) with a 2 mm metal slab instead.

Try it first, then show the solutionSolution

(a) $\frac{8.85 \times 1 0 ^{- 14}}{0.004} \approx 22.1$ pF. (1)
(b) $\frac{8.85 \times 1 0 ^{- 14}}{0.004 - 0.002 + 0.0005} = \frac{8.85 \times 1 0 ^{- 14}}{0.0025} = 35.4$ pF. (1)
(c) $\frac{8.85 \times 1 0 ^{- 14}}{0.002} \approx 44.3$ pF. (1)

Answer: (a) 22.1 pF; (b) 35.4 pF; (c) 44.3 pF.

Example 5: Connected or disconnected · 4 marks · Long answerHOTS / competency

A 20 pF air capacitor is charged to 100 V. A slab of K = 3 is then inserted to fill the gap. Find Q, V and the stored energy (a) if the battery was disconnected first, (b) if it stays connected.

Try it first, then show the solutionSolution

Initially $Q = 2$ nC, $U = \frac{1}{2} (20 \times 1 0^{- 12}) (10 0^{2}) = 1.0 \times 1 0^{- 7}$ J; new C = 60 pF. (1)
(a) Q = 2 nC; $V = \frac{2 \times 1 0 ^{- 9}}{60 \times 1 0 ^{- 12}} \approx 33.3$ V; $U = \frac{Q ^{2}}{2 C} \approx 3.3 \times 1 0^{- 8}$ J (decreased by 3). (1)
(b) V = 100 V; Q = 6 nC; $U = \frac{1}{2} (60 \times 1 0^{- 12}) (10 0^{2}) = 3.0 \times 1 0^{- 7}$ J (increased by 3). (1)
In (a) the field pulls the slab in, using up stored energy; in (b) the battery supplies 4 nC more charge and does work $4 \times 1 0^{- 9} \times 100 = 4 \times 1 0^{- 7}$ J, half stored and half doing work on the slab. (1)

Answer: (a) 2 nC, 33.3 V, $3.3 \times 1 0^{- 8}$ J; (b) 6 nC, 100 V, $3.0 \times 1 0^{- 7}$ J.

Example 6: A humidity sensor · 4 marks · Case-basedHOTS / competency

Suppose a humidity sensor is a parallel-plate capacitor with plates of area 1 cm² and a polymer film 10 µm thick filling the gap. The polymer's dielectric constant rises from 3.0 in dry air to 3.6 in humid air.
(i) Find the capacitance in dry air. (1 mark)
(ii) Find the capacitance in humid air and the percentage change. (2 marks)
(iii) At 5 V, how much more charge is stored in humid air? (1 mark)

Try it first, then show the solutionSolution

(i) $C_{0} = \frac{8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 4}}{1 0 ^{- 5}} = 88.5$ pF; with K = 3: 265.5 pF.
(ii) K = 3.6: 318.6 pF.
Change $\frac{53.1}{265.5} = 20%$ increase.
(iii) $Δ Q = 53.1 \times 1 0^{- 12} \times 5 \approx 2.7 \times 1 0^{- 10}$ C.

Answer: (i) about 266 pF; (ii) about 319 pF, a 20% rise; (iii) about 0.27 nC.

Where students lose marks in Capacitance of a Parallel-Plate Capacitor and Dielectrics

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

Mistake: Not knowing how C depends on d
The ISC 2025 examiners noted candidates who only said the capacitance "changes".
Do this: Quote $C = \frac{ε _{0} A}{d}$ , then conclude: d increases, C decreases.
Usually costs: 1 mark.
Mistake: Using a combination formula
In 2025 some confused this with capacitors in parallel and used a wrong formula.
Do this: A dielectric changes a single capacitor's C by K; it is not a combination.
Usually costs: 1 mark.
Mistake: Not relating C to K
Also in 2025: candidates could not relate capacitance to dielectric constant.
Wrong working
"Capacitance decreases because the dielectric is an insulator."
Right working
$C^{'} = K C$ with $K > 1$ : capacitance increases.
Do this: State the formula, then the conclusion.
Usually costs: 1 mark.
Mistake: Holding the wrong quantity fixed
Mixing up the connected and isolated cases gives every later answer wrong.
Do this: Write "Q constant" or "V constant" before any calculation.
Usually costs: 2 marks.

Exam tips

Convert mm to m and cm² to m² before substituting; most numerical slips are units.
Give capacitances in pF or µF with the power of ten shown.
In explanation questions, mention polarisation and induced (bound) charges.
Remember the syllabus: energy formula only, no derivation.

Practice questions on Capacitance of a 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 dielectric slab is inserted between the plates of a capacitor that remains connected to a battery. The charge on the plates:
(a) decreases
(b) stays the same
(c) increases K times
(d) becomes zero
Hint
V is fixed.
Answer
(c).
Q2 · 1 mark · Very short answerBoard level
A parallel-plate capacitor of 5 pF in air is completely filled with a dielectric of K = 4, and its plate separation is then doubled. What is the new capacitance? (A variation on ISC 2025.)
Hint
$C = \frac{K ε _{0} A}{d}$ .
Answer
$C = 5 \times \frac{4}{2} = 10$ pF.
Q3 · 2 marks · Short answerHarder board level
An air capacitor has capacitance 5 µF. A slab of K = 2 with thickness half the gap is inserted. Find the new capacitance.
Hint
$d - \frac{d}{2} + \frac{d}{4}$ .
Answer
$C = \frac{ε _{0} A}{0.75 d} = \frac{4}{3} \times 5 \approx 6.67$ µF.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Inserting a dielectric slab into an isolated charged capacitor reduces its stored energy.
Reason (R): $U = \frac{Q ^{2}}{2 C}$ , Q stays constant and C increases.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Isolated means Q is fixed.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
The field between the plates of an isolated air capacitor is $3 \times 1 0^{4}$ V m⁻¹. A slab of K = 6 fills the gap. Find the field in the slab and the induced surface charge density as a fraction of the free surface charge density.
Hint
$σ_{p} = σ (1 - \frac{1}{K})$ .
Answer
$5 \times 1 0^{3}$ V m⁻¹; $\frac{5}{6}$ of σ.
Q6 · 3 marks · Short answerHOTS / competency
Two identical 2 µF capacitors are connected in parallel to a 12 V battery, which is then removed. A dielectric of K = 3 is inserted into one of them. Find the new common p.d., the charge on each and the change in total energy.
Hint
Total charge 48 µC is shared by 2 µF and 6 µF.
Answer
$V = \frac{48}{8} = 6$ V; 12 µC and 36 µC; energy falls from 288 µJ to 144 µJ.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a variable capacitor has plates of area $2 \times 1 0^{- 3}$ m², 1 mm apart in air. It is charged to 50 V, disconnected, and its plates are then pulled apart to 2 mm.
(i) Find the initial capacitance and charge. (1 mark)
(ii) Find the new capacitance and p.d. (1 mark)
(iii) Find the initial and final stored energy. (1 mark)
(iv) Where does the extra energy come from? (1 mark)
Hint
Q stays fixed.
Answer
(i) 17.7 pF, 885 pC. (ii) 8.85 pF, 100 V. (iii) $2.2 \times 1 0^{- 8}$ J and $4.4 \times 1 0^{- 8}$ J. (iv) From the work done against the attraction of the plates while pulling them apart.

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 capacitor in vacuum?Show answer
$\frac{ε _{0} A}{d}$ .
2.Define dielectric constant in terms of capacitance.Show answer
$K = \frac{C ^{'}}{C}$ : capacitance with the dielectric filling the gap ÷ capacitance with vacuum.
3.Isolated capacitor, dielectric K inserted: what happens to V?Show answer
It becomes $\frac{V}{K}$ .
4.Plates 0.1 m² and 1 mm apart in air: C?Show answer
885 pF.
5.Capacitance with a metal slab of thickness t?Show answer
$\frac{ε _{0} A}{d - t}$ .

Capacitance of a Parallel-Plate Capacitor and Dielectrics: frequently asked questions

Why does a dielectric increase capacitance?

The polarised dielectric develops bound charges that oppose the applied field. For the same free charge, the field and p.d. are smaller, so C = \frac{Q}{V} is larger.

Does a dielectric change the charge on the plates?

Only if the capacitor is connected to a battery. If isolated, the free charge stays the same and only V, E and U fall.

What happens if a metal slab is inserted instead?

A metal has no field inside, so it acts as if the gap is reduced by its thickness: C = \frac{\varepsilon_0A}{d - t}.

Is the partly filled capacitor formula in the ISC syllabus?

Yes. The 2027 syllabus gives C' = \frac{\varepsilon_0A}{d - t + t/\varepsilon_r} for a capacitor partially filled with dielectric.

How was this tested recently?

In 2025 a 2-mark question asked how capacitance changes when the plate separation is increased and when a dielectric slab is inserted.