On this page
The formula ½CV² tells you how much energy a capacitor stores, but not where. NCERT's answer is that it sits in the electric field between the plates, and that idea works for any arrangement of charges. This post derives the energy density, checks it against ½CV², and uses it in places where ½CV² would not help.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "energy stored in a capacitor", the JEE (Advanced) 2026 brochure lists "Energy stored in a capacitor", and the NEET (UG) 2026 syllabus has the same line. Energy density itself is not named in those lists, but it is derived in section 2.15 of the NCERT Class 12 chapter, Electrostatic Potential and Capacitance (2026-27 reprint). The CBSE 2026-27 curriculum marks energy stored in a capacitor "no derivation, formulae only". Check the documents for your year.
Where does ½ε₀E² come from?
NCERT starts from a parallel plate capacitor with plate area A, separation d and surface charge density σ. The stored energy is Q²/2C, with Q = Aσ and C = ε₀A/d, so
U = (Aσ)² × d / (2ε₀A).
The field between the plates is E = σ/ε₀. Substituting gives U = ½ε₀E² × Ad. Since Ad is the volume between the plates, where the field exists, the energy per unit volume is
u = ½ε₀E².
NCERT adds that although it was derived for a parallel plate capacitor, the result is very general and holds for the field of any configuration of charges. Its unit is J/m³.
Worked example 1: a capacitor, two ways
Plates of area 0.010 m² are 2.0 mm apart and connected to 500 V. Take ε₀ = 8.854 × 10⁻¹² F/m, the CODATA value from NIST.
From the field. E = V/d = 500 / 2.0 × 10⁻³ = 2.5 × 10⁵ V/m. Then u = ½ × 8.854 × 10⁻¹² × (2.5 × 10⁵)² ≈ 0.277 J/m³. Volume = 0.010 × 2.0 × 10⁻³ = 2.0 × 10⁻⁵ m³. U = 0.277 × 2.0 × 10⁻⁵ ≈ 5.53 × 10⁻⁶ J.
From the capacitor. C = ε₀A/d = 8.854 × 10⁻¹² × 0.010 / 2.0 × 10⁻³ ≈ 4.43 × 10⁻¹¹ F. U = ½CV² = ½ × 4.43 × 10⁻¹¹ × 500² ≈ 5.53 × 10⁻⁶ J.
The two agree, about 5.5 microjoules.
Worked example 2: the most energy air can hold
NCERT gives the dielectric strength of air as about 3 × 10⁶ V/m; beyond that, air breaks down. The largest energy density in air is then
u = ½ × 8.854 × 10⁻¹² × (3 × 10⁶)² ≈ 40 J/m³.
That is tiny: a full cubic metre of air at the edge of breakdown holds about 40 J. It explains why air-gap capacitors store little energy, and why practical capacitors use dielectrics that can take stronger fields.
Worked example 3: the field energy of a charged sphere
A conducting sphere of radius R = 0.10 m carries Q = 1.0 × 10⁻⁸ C. Outside it, E = Q/(4πε₀r²); inside, E = 0. The field is not uniform, so add up thin shells of radius r and thickness dr, each of volume 4πr² dr:
dU = ½ε₀ [Q/(4πε₀r²)]² × 4πr² dr = Q² dr / (8πε₀r²).
Integrating from R to infinity gives U = Q²/(8πε₀R).
Numbers: U = (1.0 × 10⁻⁸)² / (8π × 8.854 × 10⁻¹² × 0.10) ≈ 4.49 × 10⁻⁶ J.
Check. The sphere's potential is V = Q/(4πε₀R) ≈ 899 V, and ½QV = ½ × 1.0 × 10⁻⁸ × 899 ≈ 4.49 × 10⁻⁶ J. Same answer.
A bonus. The energy between R and a radius r is a fraction 1 − R/r of the total. At r = 2R that is one half: half the sphere's field energy lies within one radius of its surface.
Which mistakes cost marks?
Forgetting the ½. u = ε₀E² double-counts.
Multiplying by the wrong volume. Use only the region where the field exists; inside a conductor E = 0.
Treating a non-uniform field as uniform. Integrate shell by shell.
Mixing up mm and m in E = V/d. A 1000-fold error in E is a million-fold error in u.
Our posts on capacitor and dielectric problems and electrostatic pressure on a conductor use this energy density in other ways.
A practice routine
Rederive u = ½ε₀E² from Q²/2C on one side of a page.
Redo Example 1 with the plates moved to 4.0 mm at fixed V. Predict first: E halves, u falls to a quarter, volume doubles, so U halves.
Repeat Example 3 for R = 0.20 m and check with ½QV.
Mix in two capacitor questions you got wrong before and solve them with the energy density method.
If integrating shell by shell is the hard part, the Fermi Concept Library covers every concept in physics and mathematics with its own lesson, so you can revise integration before returning to the sphere.
Tonight, check Example 1 both ways without looking at the working.
