Formula
Parallel plate capacitor
C = ε₀ A / d (vacuum). With dielectric of permittivity εᵣ: C = εᵣ ε₀ A / d. Larger area or smaller gap or higher εᵣ increases capacitance.
-- NCERT Class 12 Physics, Ch. 2, p. 69Parallel Plate Capacitor
8 MCQs1 revision card9-step worked example
Source: NCERT ElectrostaticsPYQ coverage: NEET 2020, 2024, 2025Official key: NTA-verifiedLast updated: 21 Sep 2026
Parallel Plate Capacitor, explained for NEET
The trap: forgetting the dielectric constant K, or applying it blindly. A dielectric that fills the whole gap multiplies the capacitance by K. A slab that fills only part of the gap does not, and whether V or Q changes depends on whether the battery is still connected.
Without a dielectric. Two large parallel conducting plates, each of area A, separated by a small distance d (d² << A), carry charges +Q and −Q. Between the plates the field is E = σ/ε₀ = Q/(ε₀A), directed from the positive plate to the negative plate; in the outer regions it is zero (NCERT Class 12 Physics Part I, Chapter 2, Eqs. 2.39–2.41, pages 68–69). Near the edges the field lines bend outward ("fringing"), which is ignored for d² << A. With V = Ed:
C = ε₀A/d (NCERT Chapter 2, Eq. 2.43, page 69)
C depends only on geometry: it grows with A and falls as d grows. For A = 1 m² and d = 1 mm, NCERT gets C = 8.85 × 10⁻⁹ F (Eq. 2.44, page 69), which shows that 1 F is a very large unit.
With a dielectric filling the gap. The polarised dielectric reduces the field between the plates. For the same charge Q, the potential difference becomes V = Qd/(Aε₀K) (Eq. 2.50, page 70), so
C = ε₀KA/d, and C/C₀ = K (Eqs. 2.51 and 2.54, page 70)
NCERT defines K as the factor (greater than 1) by which the capacitance increases from its vacuum value when the dielectric fills the gap completely, and notes the result holds for any type of capacitor (pages 70–71).
Bridge to NEET. Questions give C₀ or the geometry, insert a dielectric, and ask for the new C, V or Q. First ask two things: does the slab fill the gap? Is the charge fixed (battery disconnected) or the voltage fixed (battery connected)? With Q fixed, V falls to V₀/K. With V fixed, Q = CV rises to KQ₀.
Watch out: C ∝ 1/d, not 1/d², and convert mm to m and pF to F before substituting.
Can you answer these Parallel Plate Capacitor MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The capacitance of a parallel plate capacitor with vacuum between plates of area A separated by a distance d is:
Show answer and why every option is right or wrong
Answer: B. C = Q/V with V = Qd/(ε₀A) gives C = ε₀A/d (NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.43, page 69).
Why A is wrong: A is wrong because it swaps A and d; capacitance increases with plate area and decreases with separation, not the other way round.
Why C is wrong: C is wrong because it puts ε₀ in the denominator; the vacuum permittivity multiplies A/d, so a larger ε₀ gives a larger C.
Why D is wrong: D is wrong because the potential difference is V = Ed, which is proportional to d, so C falls as 1/d, not 1/d².
MCQ 2Easy RecallPractice
According to NCERT, the dielectric constant K of a substance is:
Show answer and why every option is right or wrong
Answer: A. C/C₀ = K with K > 1; NCERT calls K the factor by which capacitance increases from its vacuum value when the dielectric is inserted fully between the plates (NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.54, page 70).
Why B is wrong: B is wrong because it reverses the direction: the dielectric reduces the field and the potential difference for the same charge, so the capacitance increases.
Why C is wrong: C is wrong because it inverts the ratio; NCERT defines K = ε/ε₀ (Eq. 2.53), which is greater than 1.
Why D is wrong: D is wrong because it reverses the roles: K = 1 for vacuum, and K is greater than 1 for a dielectric.
MCQ 3Easy RecallPractice
For an ideal parallel plate capacitor with surface charge densities +σ and −σ on its plates (vacuum between them, edge effects ignored), the magnitudes of the electric field between the plates and in the regions outside the plates are:
Show answer and why every option is right or wrong
Answer: C. Between the plates the fields of the two sheets add to σ/ε₀, pointing from the positive to the negative plate; outside they cancel to zero (NCERT Class 12 Physics Part I, Chapter 2, Eqs. 2.39–2.41, pages 68–69).
Why A is wrong: A is wrong because it uses the field of one sheet only; the two plates' fields add between the plates and cancel outside.
Why B is wrong: B is wrong because it reverses the regions: the fields cancel outside the plates and add between them.
Why D is wrong: D is wrong because outside the plates the equal and opposite fields σ/2ε₀ of the two plates cancel, giving zero, not σ/2ε₀.
MCQ 4Direct ApplicationPractice
A parallel plate capacitor has plates of area 2.0 × 10⁻² m² separated by 2.0 mm of air (treat as vacuum). Taking ε₀ = 8.85 × 10⁻¹² F m⁻¹, its capacitance is about:
Show answer and why every option is right or wrong
Answer: D. C = ε₀A/d = (8.85 × 10⁻¹²)(2.0 × 10⁻²)/(2.0 × 10⁻³) = 8.85 × 10⁻¹¹ F ≈ 8.9 × 10⁻¹¹ F (NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.43, page 69).
Why A is wrong: A is wrong because 4.4 × 10⁻⁸ F comes from dividing by d² = 4.0 × 10⁻⁶ m² instead of d; C is proportional to 1/d.
Why B is wrong: B is wrong because 8.9 × 10⁻¹⁴ F comes from substituting d = 2.0 (millimetres) without converting to 2.0 × 10⁻³ m: (8.85 × 10⁻¹²)(2.0 × 10⁻²)/2.0.
Why C is wrong: C is wrong because 8.9 × 10⁻¹³ F comes from ε₀d/A = (8.85 × 10⁻¹²)(2.0 × 10⁻³)/(2.0 × 10⁻²), swapping area and separation.
MCQ 5Direct ApplicationPractice
An air-filled parallel plate capacitor has capacitance 5.0 pF. A dielectric of dielectric constant 4.0 is inserted so that it completely fills the space between the plates. The new capacitance is:
Show answer and why every option is right or wrong
Answer: D. A dielectric that fills the gap multiplies the capacitance by K: C = KC₀ = 4.0 × 5.0 pF = 20 pF (NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.54, page 70).
Why A is wrong: A is wrong because 1.3 pF comes from C₀/K = 5.0/4.0 = 1.25 pF; the dielectric increases capacitance, it does not divide it.
Why B is wrong: B is wrong because 5.0 pF ignores the dielectric completely, using C = ε₀A/d without the factor K.
Why C is wrong: C is wrong because 9.0 pF comes from adding K to C₀ (5.0 + 4.0); K is a multiplying factor and is dimensionless, so it cannot be added to a capacitance.
MCQ 6Direct ApplicationPractice
A parallel plate capacitor with vacuum between its plates has capacitance 12 pF. The separation between the plates is doubled, with the plate area unchanged. The new capacitance is:
Show answer and why every option is right or wrong
Answer: A. C = ε₀A/d, so C is inversely proportional to d; doubling d halves C: 12/2 = 6.0 pF (NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.43, page 69).
Why B is wrong: B is wrong because 24 pF treats C as proportional to d; a larger separation gives a smaller capacitance.
Why C is wrong: C is wrong because 3.0 pF = 12/4 treats C as proportional to 1/d²; C falls as 1/d, so doubling d only halves it.
Why D is wrong: D is wrong because it assumes capacitance does not depend on the separation; NCERT states C depends on the geometry of the system, and d is part of that geometry.
MCQ 7CalculationPractice
An air-filled parallel plate capacitor is charged to a potential difference of 120 V and then disconnected from the battery. A dielectric of dielectric constant 3.0 is then inserted to fill the gap completely. The new potential difference between the plates is:
Show answer and why every option is right or wrong
Answer: B. Disconnected, the charge Q stays fixed. The capacitance becomes 3.0C₀, so V = Q/(3.0C₀) = 120/3.0 = 40 V; equivalently V = Qd/(Aε₀K) falls by the factor K (NCERT Class 12 Physics Part I, Chapter 2, Eqs. 2.50 and 2.54, page 70).
Why A is wrong: A is wrong because 360 V comes from multiplying V by K (120 × 3.0); at fixed charge, a larger capacitance means a smaller potential difference.
Why C is wrong: C is wrong because 120 V assumes the potential difference stays fixed; that happens only if the battery remains connected, and here it was disconnected, so Q is fixed instead.
Why D is wrong: D is wrong because 13 V comes from dividing by K twice (120/9.0 ≈ 13 V); the potential difference falls by the factor K once.
MCQ 8CalculationPYQ Pattern
An air-filled parallel plate capacitor of capacitance 10 pF stays connected to a 50 V battery while a dielectric of dielectric constant 5.0 is inserted to fill the gap completely. The charge on the positive plate after insertion is:
Show answer and why every option is right or wrong
Answer: C. With the battery connected, V stays 50 V. The new capacitance is C = KC₀ = 5.0 × 10 pF = 50 pF = 5.0 × 10⁻¹¹ F, so Q = CV = (5.0 × 10⁻¹¹)(50) = 2.5 × 10⁻⁹ C (NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.54, page 70, with C = Q/V, Eq. 2.38, page 67).
Why A is wrong: A is wrong because 5.0 × 10⁻¹⁰ C = C₀V = (1.0 × 10⁻¹¹)(50) ignores the dielectric; the capacitance has become K times larger, so at fixed V the charge is K times larger.
Why B is wrong: B is wrong because 1.0 × 10⁻¹⁰ C = C₀V/K = (5.0 × 10⁻¹⁰)/5.0 divides by K; the dielectric increases the capacitance, so the charge drawn from the battery increases.
Why D is wrong: D is wrong because 2.5 × 10⁻⁶ C comes from reading 10 pF as 10 × 10⁻⁹ F (nanofarads): 5.0 × (1.0 × 10⁻⁸) × 50; 1 pF = 10⁻¹² F, so 10 pF = 1.0 × 10⁻¹¹ F.
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Parallel Plate Capacitor: quick recall before you leave
How do you solve a Parallel Plate Capacitor question? A worked example
- 1
Given
(NCERT Class 12 Physics Part I, Chapter 2, Example 2.8, page 71.) A parallel plate capacitor has plate separation d. A slab of material of dielectric constant K, with the same area as the plates and thickness (3/4)d, is inserted between the plates. The charge on the plates is unchanged.
- 2
Required
How the capacitance changes when the slab is inserted, i.e. C in terms of the original capacitance C₀.
- 3
Concept
Before insertion the field between the plates is E₀ = V₀/d. With the free charge Q₀ unchanged, the field in the air part of the gap stays E₀, while inside the dielectric it is reduced to E₀/K. The potential difference is the sum of field × distance over the two parts, and the capacitance is Q₀/V.
- 4
Formula
V = E × (distance) for each uniform-field region; C = Q/V; C₀ = Q₀/V₀.
- 5
Substitution
Air thickness = d − (3/4)d = (1/4)d; dielectric thickness = (3/4)d.
V = E₀ (1/4)d + (E₀/K)(3/4)d - 6
Calculation
V = E₀d (1/4 + 3/(4K)) = V₀ (K + 3)/(4K)
The potential difference decreases by the factor (K + 3)/4K while Q₀ is unchanged, so
C = Q₀/V = (4K/(K + 3)) × Q₀/V₀ = (4K/(K + 3)) C₀
The numbers 3/4, 1/4 and 4 are exact fractions from the geometry of the problem; they do not contribute to any significant-figure count, and the answer is symbolic.
Check: with K = 1 (no real dielectric) the factor 4K/(K + 3) = 1, so C = C₀, as it must be. - 7
Final answer
C = 4K C₀/(K + 3); the capacitance increases by the factor 4K/(K + 3).
- 8
Common trap
Writing C = KC₀ because "a dielectric was inserted." That result needs the dielectric to fill the whole gap. Here one quarter of the gap is still air, so the increase is smaller: for any K > 1, 4K/(K + 3) is less than K. A second slip is multiplying the whole potential difference by 1/K instead of reducing only the part across the slab.
- 9
Similar NEET-style question
"A slab of dielectric constant 2.0, with the same area as the plates, fills half the gap d of a parallel plate capacitor of capacitance C₀ (charge on the plates unchanged). Find the new capacitance."
Strategy: V = E₀(d/2) + (E₀/2.0)(d/2) = (3/4)V₀, so C = (4/3)C₀ ≈ 1.3C₀. Do not write 2.0C₀ — the slab fills only half the gap.
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What to remember before solving Parallel Plate Capacitor questions
Which Parallel Plate Capacitor formulas do you need for NEET?
1 formula — click to collapse
Parallel plate capacitor
Capacitance of parallel-plate capacitor. eps_r = 1 for vacuum/air.
| Symbol | Quantity | SI Unit |
|---|---|---|
| C | capacitance | F |
| A | plate area | m^2 |
| d | plate separation | m |
| eps_r | relative permittivity | - |
Valid when
- Plate separation << plate dimensions (uniform field)
- Uniform dielectric
More in Electrostatics: 3 exam traps and mistakes · 10 formulas · 5 question patterns from its other lessons.
Parallel Plate Capacitor questions from past NEET papers
3 questions from NEET 2020, 2024, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
11.33
22.66
32.33
41.60
NTA Answer: Option 2(final)
NEET 2024Revised key
1A, B and E only
2A, C and E only
3B, D and E only
4A, B and C only
NTA Answer: Option 2(revised_final)
NEET 2020
15.00 C² N⁻¹ m⁻²
20.44×10⁻¹³ C² N⁻¹ m⁻²
31.77×10⁻¹² C² N⁻¹ m⁻²
40.44×10⁻¹⁰ C² N⁻¹ m⁻²
NTA Answer: Option 4(final)
How does NEET ask about Parallel Plate Capacitor?
1 recurring pattern from past papers — click to collapse
Capacitance change when dielectric inserted: C = K*C_0. Find new C, V, or E given fixed Q vs fixed V.
Multi StepMedium
Common distractors
forgets K multiplier
Uses C = eps_0 A/d ignoring dielectric
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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