Formula
Energy stored in capacitor
U = (1/2) C V² = (1/2) Q V = Q²/(2C). Energy density u = (1/2) ε₀ E² (J/m³) in the field volume.
-- NCERT Class 12 Physics, Ch. 2, p. 75Energy in Capacitor
8 MCQs1 revision card9-step worked example
Source: NCERT ElectrostaticsPYQ coverage: NEET 2022, 2026Official key: NTA-verifiedLast updated: 21 Sep 2026
Energy in Capacitor, explained for NEET
The trap: writing the stored energy as U = QV. The correct result carries a factor of one-half: U = ½QV. The ½ is not a detail you can drop — QV is exactly twice the stored energy.
Where the ½ comes from. NCERT Class 12 Physics Part I, Chapter 2, Section 2.15 builds the charge up bit by bit. When the conductors already carry Q′ and −Q′, the potential difference is only V′ = Q′/C, so moving a further small charge δQ′ costs δW = (Q′/C) δQ′ (Eq. 2.68, page 74). The potential difference grows from zero to its final value while charging, so the total work is not "final V times Q". Integrating gives (Eq. 2.69, page 74):
U = Q²/2C = ½CV² = ½QV
Work is done externally because, at every stage, the conductor receiving positive charge is at the higher potential. Since the electrostatic force is conservative, this work is stored as potential energy and does not depend on how the charge was built up (page 74). When the capacitor discharges, the energy is released.
Where the energy sits. NCERT views it as stored in the electric field between the plates; for a parallel plate capacitor this leads to an energy density u = ½ε₀E² (Eq. 2.73, page 75).
Bridge to NEET. Pick the form that matches the data. Given C and V, use ½CV². Given Q and C, use Q²/2C. Given Q and V, use ½QV. Two common slips: dropping the ½, and forgetting to square (V in ½CV², Q in Q²/2C). A unit slip — leaving µF or pF unconverted — shifts the answer by powers of ten.
Watch out: in NCERT Example 2.10 (pages 75–76), a charged capacitor shares its charge with an identical uncharged one. No charge is lost, yet the total energy halves. The missing energy leaves as heat and electromagnetic radiation during the transient current — do not assume energy is conserved inside the capacitors.
Can you answer these Energy in 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
A capacitor of capacitance C carries charge Q with potential difference V between its plates. Which expression gives the energy stored in it?
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.69 (page 74): U = Q²/2C = ½CV² = ½QV.
Why A is wrong: A is wrong because QV drops the factor ½; it is twice the stored energy, since the potential difference rises from zero during charging.
Why B is wrong: B is wrong because Q²/C drops the 2 in the denominator; the correct form is Q²/2C, so this is again twice the stored energy.
Why D is wrong: D is wrong because C²V has the wrong dimensions for energy: C²V equals C × Q, not an energy; the voltage must be squared, not the capacitance.
MCQ 2Easy RecallPractice
According to NCERT, where can the potential energy of a charged capacitor be viewed as stored?
Show answer and why every option is right or wrong
Answer: B. NCERT Class 12 Physics Part I, Chapter 2, Section 2.15 (page 74) views the capacitor's potential energy as stored in the electric field between the plates, leading to the energy density ½ε₀E² (Eq. 2.73, page 75).
Why A is wrong: A is wrong because the charges are at rest in electrostatics; the stored energy is potential energy of the charge configuration, not kinetic energy.
Why C is wrong: C is wrong because NCERT places the energy where the field exists — the region between the plates of volume Ad — not outside the plates.
Why D is wrong: D is wrong because the energy belongs to the whole configuration of +Q and −Q (the field between them), not to one plate.
MCQ 3Easy RecallPractice
NCERT states that the energy U = Q²/2C of a capacitor does not depend on the manner in which its charge was built up. What reason does NCERT give?
Show answer and why every option is right or wrong
Answer: D. NCERT Class 12 Physics Part I, Chapter 2, Section 2.15 (page 74): because the electrostatic force is conservative, the charging work is stored as potential energy, and for the same reason the result is independent of how the charge is built up.
Why A is wrong: A is wrong because the potential difference is Q′/C at each stage, so it grows from zero as charge accumulates; that growth is where the ½ comes from.
Why B is wrong: B is wrong because NCERT says work is done externally, since the conductor receiving positive charge is always at the higher potential.
Why C is wrong: C is wrong because the small steps are only the method used to calculate the work; they are not the reason the result is path-independent.
MCQ 4Direct ApplicationPractice
A 25 µF capacitor is charged to a potential difference of 12 V. How much energy is stored in it?
Show answer and why every option is right or wrong
Answer: D. U = ½CV² = ½ × 25 × 10⁻⁶ F × (12 V)² = ½ × 25 × 10⁻⁶ × 144 = 1.8 × 10⁻³ J (NCERT Chapter 2, Eq. 2.69, page 74).
Why A is wrong: A is wrong because 3.6 × 10⁻³ J is CV² = 25 × 10⁻⁶ × 144, which drops the factor ½.
Why B is wrong: B is wrong because 1.5 × 10⁻⁴ J is ½CV = ½ × 25 × 10⁻⁶ × 12, which forgets to square the voltage.
Why C is wrong: C is wrong because 1.8 × 10³ J comes from ½ × 25 × 144 with the capacitance left as 25 instead of converting 25 µF to 25 × 10⁻⁶ F.
MCQ 5Direct ApplicationPractice
A 3.0 µF capacitor holds a charge of 6.0 × 10⁻⁴ C. What energy is stored in it?
Show answer and why every option is right or wrong
Answer: A. U = Q²/2C = (6.0 × 10⁻⁴)² / (2 × 3.0 × 10⁻⁶) = 3.6 × 10⁻⁷ / 6.0 × 10⁻⁶ = 6.0 × 10⁻² J (NCERT Chapter 2, Eq. 2.69, page 74).
Why B is wrong: B is wrong because 1.2 × 10⁻¹ J is Q²/C = 3.6 × 10⁻⁷ / 3.0 × 10⁻⁶, which drops the 2 in the denominator.
Why C is wrong: C is wrong because 9.0 × 10⁻¹⁰ J is ½QC = ½ × 6.0 × 10⁻⁴ × 3.0 × 10⁻⁶, which multiplies by C where Q² must be divided by 2C.
Why D is wrong: D is wrong because 1.0 × 10² is Q/2C = 6.0 × 10⁻⁴ / 6.0 × 10⁻⁶, which forgets to square Q; that number is half the voltage in volts, not an energy.
MCQ 6Direct ApplicationPractice
A capacitor carries a charge of 8.0 × 10⁻⁶ C when the potential difference across it is 25 V. What energy is stored?
Show answer and why every option is right or wrong
Answer: B. U = ½QV = ½ × 8.0 × 10⁻⁶ C × 25 V = 1.0 × 10⁻⁴ J (NCERT Chapter 2, Eq. 2.69, page 74).
Why A is wrong: A is wrong because 2.0 × 10⁻⁴ J is QV = 8.0 × 10⁻⁶ × 25, which drops the factor ½ — the trap this lesson names.
Why C is wrong: C is wrong because 8.0 × 10⁻¹⁰ J is ½Q²V = ½ × 6.4 × 10⁻¹¹ × 25, which squares Q in the ½QV form; only Q²/2C has Q squared.
Why D is wrong: D is wrong because 2.5 × 10⁻³ J is ½QV² = ½ × 8.0 × 10⁻⁶ × 625, which squares V in the ½QV form; only ½CV² has V squared.
MCQ 7CalculationPractice
A 2.0 µF capacitor stores 1.0 × 10⁻² J of energy. What charge does it carry?
Show answer and why every option is right or wrong
Answer: A. From U = ½CV², V² = 2U/C = 2 × 1.0 × 10⁻² / 2.0 × 10⁻⁶ = 1.0 × 10⁴ V², so V = 1.0 × 10² V. Then Q = CV = 2.0 × 10⁻⁶ × 1.0 × 10² = 2.0 × 10⁻⁴ C. (Check: Q = √(2UC) = √(4.0 × 10⁻⁸) = 2.0 × 10⁻⁴ C.)
Why B is wrong: B is wrong because 1.4 × 10⁻⁴ C is √(UC) = √(2.0 × 10⁻⁸); it uses U = Q²/C, dropping the ½ of Eq. 2.69.
Why C is wrong: C is wrong because 4.0 × 10⁻⁸ C is 2UC, which is Q², not Q; the square root was never taken.
Why D is wrong: D is wrong because 2.0 × 10⁻² C is C × V² = 2.0 × 10⁻⁶ × 1.0 × 10⁴; it stops at V² and multiplies by C instead of first taking the square root to get V.
MCQ 8CalculationPractice
The potential difference across a capacitor of fixed capacitance is raised from V to 3V. By what factors do its charge and its stored energy change?
Show answer and why every option is right or wrong
Answer: C. With C fixed, Q = CV so the charge triples. U = ½CV² so the energy rises by 3² = 9. Cross-check with ½QV: both Q and V triple, so U rises by 3 × 3 = 9 (NCERT Chapter 2, Eq. 2.69, page 74).
Why A is wrong: A is wrong because it treats the energy as proportional to V; in ½CV² the voltage is squared, so tripling V multiplies U by 9, not 3.
Why B is wrong: B is wrong because the charge Q = CV is linear in V at fixed C; it triples, it does not rise ninefold.
Why D is wrong: D is wrong because it holds Q fixed while V changes; at fixed C, Q = CV must triple along with V, so ½QV rises by 9, not 3.
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Energy in Capacitor: quick recall before you leave
How do you solve a Energy in Capacitor question? A worked example
- 1
Given
NCERT Class 12 Physics Part I, Chapter 2, Example 2.10 (pages 75–76).
(a) A 900 pF capacitor (C = 9.0 × 10⁻¹⁰ F) is charged by a 100 V battery (V = 1.0 × 10² V).
(b) The capacitor is then disconnected from the battery and connected to another, uncharged, 900 pF capacitor. - 2
Required
(a) The electrostatic energy stored by the first capacitor. (b) The electrostatic energy stored by the two-capacitor system in the steady state.
- 3
Concept
(a) Energy of a charged capacitor, Eq. 2.69. (b) After connection, the positive plates are at one common potential and the negative plates at another, so both capacitors share one potential difference V′. Charge is conserved and the capacitors are identical, so the charge Q divides equally: Q′ = Q/2 on each.
- 4
Formula
Q = CV; U = ½CV² = ½QV (Eq. 2.69). For (b): Q′ = CV′ = Q/2, so V′ = V/2, and total energy = 2 × ½Q′V′.
- 5
Substitution
(a) Q = 9.0 × 10⁻¹⁰ F × 1.0 × 10² V; U = ½ × Q × 1.0 × 10² V
(b) Total energy = 2 × ½ × (Q/2) × (V/2) = ¼QV - 6
Calculation
(a) Q = 9 × 10⁻⁸ C; U = ½ × 9 × 10⁻⁸ C × 100 V = 4.5 × 10⁻⁶ J
(b) Total energy = ¼ × 9 × 10⁻⁸ C × 100 V = 2.25 × 10⁻⁶ J
The ½, the ¼, and the factor 2 (for two capacitors) are exact numbers and do not affect the significant-figure count. The answers are quoted as NCERT prints them. - 7
Final answer
(a) 4.5 × 10⁻⁶ J; (b) 2.25 × 10⁻⁶ J — half the initial energy.
- 8
Common trap
Assuming the total energy stays 4.5 × 10⁻⁶ J because no charge is lost. Charge is conserved, but energy in the capacitors is not: NCERT notes that during the transient period a current flows from the first capacitor to the second, and energy is lost as heat and electromagnetic radiation. A second slip is keeping V = 100 V across each capacitor in (b); sharing the charge halves the potential difference.
- 9
Similar NEET-style question
"A 5.0 µF capacitor is charged to 12 V, disconnected from the supply, and connected to an identical uncharged 5.0 µF capacitor. How much energy is lost?"
Strategy: Initial U = ½ × 5.0 × 10⁻⁶ × 12² = 3.6 × 10⁻⁴ J. After sharing, V′ = 6.0 V on each, so total = 2 × ½ × 5.0 × 10⁻⁶ × 6.0² = 1.8 × 10⁻⁴ J. Energy lost = 1.8 × 10⁻⁴ J (half the initial energy).
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What to remember before solving Energy in Capacitor questions
Which Energy in Capacitor formulas do you need for NEET?
1 formula — click to collapse
Energy stored in capacitor
Energy stored as work to charge the capacitor.
| Symbol | Quantity | SI Unit |
|---|---|---|
| C | capacitance | F |
| V | voltage | V |
| Q | charge | C |
Valid when
- Linear capacitor
- Static charging or quasi-static
More in Electrostatics: 3 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Energy in Capacitor questions from past NEET papers
2 questions from NEET 2022, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
12/3
21/3
31/2
41/6
NTA Answer: Option 4(final)
NEET 2022
11.5 × 10⁻⁶ J
24.5 × 10⁻⁶ J
33.25 × 10⁻⁶ J
42.25 × 10⁻⁶ J
NTA Answer: Option 4(final)
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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