Deriving the Stored Energy
Consider two initially uncharged conductors, and imagine transferring positive charge bit by bit from conductor 2 to conductor 1, so that conductor 1 ends with charge $Q$ and conductor 2 with $-Q$. At any intermediate stage the conductors carry $+Q'$ and $-Q'$, and the potential difference between them is $V' = Q'/C$, where $C$ is the capacitance. Crucially, this potential difference is not the final $V$ — it grows as charge accumulates.
The work done in transferring a further infinitesimal charge $\mathrm{d}Q'$ across this instantaneous potential difference is
$$ \mathrm{d}W = V'\,\mathrm{d}Q' = \frac{Q'}{C}\,\mathrm{d}Q'. $$
Integrating from $0$ to the final charge $Q$ gives the total work, which — because the electrostatic force is conservative — is stored as potential energy:
$$ W = \int_0^{Q} \frac{Q'}{C}\,\mathrm{d}Q' = \frac{Q^2}{2C}. $$
The stored energy equals the shaded triangular area under the linear $Q$ versus $V$ line. Because $Q = CV$ is a straight line through the origin, the area is $\tfrac{1}{2}(\text{base})(\text{height}) = \tfrac{1}{2}QV$ — the factor $\tfrac{1}{2}$ is the average potential during charging.
NIOS arrives at the same answer through a shortcut: since the potential rises linearly from $0$ to $V$, the average potential difference during charging is $V/2$, and the work is simply (charge) × (average potential) $= Q \cdot V/2 = \tfrac{1}{2}QV$. Both routes give the identical result, and both make the origin of the factor $\tfrac{1}{2}$ explicit.
The Three Equivalent Forms
Using the defining relation $Q = CV$, the stored energy can be written in three algebraically identical ways. They describe the same physical quantity; the only reason to prefer one over another is to match the variables that stay constant in a given problem.
| Form | Expression | Use it when… |
|---|---|---|
| In terms of Q and V | U = ½ QV | Both charge and voltage of the state are known |
| In terms of C and V | U = ½ CV² | Voltage is fixed — capacitor stays connected to the battery |
| In terms of Q and C | U = ½ Q²/C | Charge is fixed — capacitor disconnected after charging |
The SI unit of $U$ is the joule. Note the structural parallels: $\tfrac{1}{2}CV^2$ mirrors the kinetic energy form $\tfrac{1}{2}mv^2$ (capacitance plays the role of mass, voltage of velocity), while $\tfrac{1}{2}Q^2/C$ mirrors spring energy $\tfrac{1}{2}kx^2$ with $1/C$ as the stiffness. These analogies help recall which variable goes in the numerator.
The missing factor of ½
A common error is writing $U = CV^2$ or $U = QV$, dropping the $\tfrac{1}{2}$. This comes from forgetting that the plates are not at the full potential $V$ during charging — the potential builds up from zero. The $\tfrac{1}{2}$ is the average potential and is mandatory.
Energy of a charged capacitor always carries the factor ½: $U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \tfrac{1}{2}Q^2/C$. Without it, the answer is exactly double the correct value.
Energy Stored in the Field
Because the electrostatic force is conservative, the stored energy is independent of how the charge configuration was built; it depends only on the final state. NCERT then offers a deeper interpretation: the energy is not located on the plates but in the electric field occupying the gap between them. For a parallel-plate capacitor of plate area $A$ and separation $d$, start from $U = \dfrac{Q^2}{2C}$ with $C = \dfrac{\varepsilon_0 A}{d}$ and $Q = \sigma A$:
$$ U = \frac{(\sigma A)^2}{2}\cdot \frac{d}{\varepsilon_0 A} = \frac{\sigma^2}{2\varepsilon_0}\,(A d). $$
Using the field between the plates $E = \sigma/\varepsilon_0$, this rearranges into
$$ U = \frac{1}{2}\,\varepsilon_0 E^2 \,(A d), $$
where $Ad$ is exactly the volume of the region between the plates — the region where the field exists. The stored energy is the field-energy density multiplied by the volume the field occupies, which justifies viewing the energy as residing in the field itself.
Energy must be tracked carefully when capacitors are joined in networks. See combination of capacitors for series and parallel energy distribution.
Energy Density
Dividing the field energy by the volume gives the energy density $u$ — the electrostatic energy stored per unit volume of space:
$$ u = \frac{U}{A d} = \frac{1}{2}\,\varepsilon_0 E^2 \quad (\text{vacuum}). $$
NCERT stresses that although this is derived for a parallel-plate capacitor, the result is general: it holds for the electric field of any charge configuration. When the gap is filled with a dielectric of constant $K$, the permittivity becomes $K\varepsilon_0$ and the density generalises to $u = \tfrac{1}{2}K\varepsilon_0 E^2$.
Between the plates the uniform field $E = \sigma/\varepsilon_0$ fills the volume $Ad$. Each cubic metre of this region holds energy $u = \tfrac{1}{2}\varepsilon_0 E^2$; outside the plates the field (ideally) vanishes and stores no energy.
Energy density has its own ½
The energy density is $\tfrac{1}{2}\varepsilon_0 E^2$, not $\varepsilon_0 E^2$ and not $\tfrac{1}{2}\varepsilon_0 E$. Examiners also test whether you multiply by the correct volume: total energy $= u \times (\text{volume of field})$, which for a parallel-plate capacitor is $Ad$, not $A$ or $d$ alone.
Vacuum: $u = \tfrac{1}{2}\varepsilon_0 E^2$. With dielectric $K$: $u = \tfrac{1}{2}K\varepsilon_0 E^2$. Total energy is density times the field volume.
Charge Sharing and Energy Loss
A capacitor $C_1$ charged to potential $V_1$ is connected (after the battery is removed) to a second capacitor $C_2$ at potential $V_2$. Charge flows until both reach a common potential $V$. By charge conservation, the total charge is unchanged:
$$ V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}. $$
The total energy after sharing, $\tfrac{1}{2}(C_1 + C_2)V^2$, is always less than the initial energy $\tfrac{1}{2}C_1 V_1^2 + \tfrac{1}{2}C_2 V_2^2$ whenever $V_1 \ne V_2$. The energy lost is
$$ \Delta U = \frac{C_1 C_2}{2(C_1 + C_2)}\,(V_1 - V_2)^2. $$
This dissipation occurs during the transient current that flows while the system settles, lost as heat in the connecting wires and as electromagnetic radiation. Remarkably, the expression contains no resistance: the same energy is lost regardless of how thin or thick the connecting wire is. NCERT poses precisely this puzzle in Example 2.10 — "Where has the remaining energy gone?" — and answers it with the transient heat-and-radiation mechanism.
A charged $C_1$ (at $V_1$) is connected in parallel to $C_2$ (at $V_2$). Charge redistributes to a common potential $V$; the difference between the initial and final energies is dissipated.
Charge sharing never conserves energy
Students often assume energy is conserved because charge is conserved. It is not. Charge is conserved, but energy is always lost (as long as the two capacitors start at different potentials). The identical-capacitor case — two equal $C$, one charged to $V$ and one uncharged — loses exactly half the energy, since the final energy is $\tfrac{1}{2}(2C)(V/2)^2 = \tfrac{1}{4}CV^2$ against an initial $\tfrac{1}{2}CV^2$.
Charge is conserved; energy is dissipated. Use $\Delta U = \dfrac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$ — the loss is independent of wire resistance.
Energy Change on Inserting a Dielectric
Inserting a dielectric of constant $K$ raises the capacitance to $KC$, but the effect on stored energy depends entirely on what is held fixed. The two cases give opposite results, a distinction examiners exploit repeatedly.
| Condition | What is constant | Best form | Energy change |
|---|---|---|---|
| Battery disconnected | Charge Q | U = ½ Q²/C | Falls to U/K (C rises, Q fixed) |
| Battery connected | Voltage V | U = ½ CV² | Rises to KU (C rises, V fixed) |
With the battery disconnected the charge is trapped, so $U = Q^2/2C$ decreases as $C$ grows — the field does positive work pulling the slab in, drawing energy out of the capacitor. With the battery connected the voltage is clamped, so $U = \tfrac{1}{2}CV^2$ increases; here the battery supplies extra charge and energy. The full treatment, including the force on the slab, is developed in effect of dielectric on capacitance.
A 900 pF capacitor is charged by a 100 V battery, then disconnected and connected to an identical uncharged 900 pF capacitor. Find the initial and final energies.
Initial charge $Q = CV = 900\times10^{-12}\times100 = 9\times10^{-8}\,\text{C}$. Initial energy $= \tfrac{1}{2}QV = \tfrac{1}{2}\times 9\times10^{-8}\times100 = 4.5\times10^{-6}\,\text{J}$.
After sharing, charge conservation gives $Q' = Q/2$ on each, so the common voltage is $V' = V/2 = 50\,\text{V}$. Final energy $= \tfrac{1}{2}(C_1+C_2)V'^2 = \tfrac{1}{2}\times1800\times10^{-12}\times50^2 = 2.25\times10^{-6}\,\text{J}$ — exactly half the initial value. The missing $2.25\times10^{-6}\,\text{J}$ is dissipated in the transient. (NCERT Example 2.10.)
Energy Stored in a Capacitor at a glance
- Derived by integrating $\mathrm{d}W = (Q'/C)\,\mathrm{d}Q'$ from 0 to $Q$, giving $U = Q^2/2C$.
- Three equivalent forms: $U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \tfrac{1}{2}Q^2/C$ — the factor ½ is mandatory.
- Energy resides in the field; density $u = \tfrac{1}{2}\varepsilon_0 E^2$ (vacuum), $\tfrac{1}{2}K\varepsilon_0 E^2$ with a dielectric.
- Total field energy $= u \times (\text{field volume}) = \tfrac{1}{2}\varepsilon_0 E^2 (Ad)$ for a parallel-plate capacitor.
- Charge sharing conserves charge but loses energy: $\Delta U = C_1 C_2 (V_1-V_2)^2 / 2(C_1+C_2)$, independent of wire resistance.
- Dielectric insertion: energy falls to $U/K$ at constant $Q$; rises to $KU$ at constant $V$.