Energy Stored in a Charged Capacitor

A capacitor stores electrical potential energy when it is charged. This energy is associated with the electric field set up between its plates, not merely with charge “sitting” on the plates.

A useful analogy is a water tank being filled against gravity: as the water level rises, further filling requires more work. In the same way, as a capacitor charges, the potential difference rises, so adding more charge requires increasing work.

W = \[\frac {1}{2}\]qV

OR

U = \[\frac {Q^2}{2C}\] ​= \[\frac {1​}{2}\]QV = \[\frac {1}{2}\]​CV²

SI unit: Joule (J)

Stepwise Derivation

Consider a capacitor of capacitance C. At some intermediate stage of charging, let the charge on it be q. Then the potential difference across it is:

If an additional small charge dq is supplied, the small work done is:

Total work done in charging the capacitor from 0 to the final charge Q is:

Since this work is stored as energy:

Using Q = CV, the equivalent forms become:

The energy stored in a charged capacitor is actually stored in the electric field between its plates. For a parallel-plate capacitor, the energy density of the electric field is:

where u is energy per unit volume, and E is electric field intensity.

Meaning

• Energy density indicates how much electrical energy is stored per unit volume of the field.
• This is an important bridge between electrostatics and field-based physics.

Question idea:

A parallel-plate capacitor (3 × 10⁻⁹ F) is connected to 400 V.
A dielectric slab (K = 3, thickness 3 cm) completely fills the space. The voltage is kept constant. What is the change in energy when the slab is removed?

Key idea:

At constant voltage, energy U = \[\frac {1}{2}\]CV², so energy ∝ capacitance.

Steps in simple words

• With air: use given C = 3 × 10⁻⁹ F and V = 400 V.
  U_air = \[\frac {1}{2}\]CV² = 24 × 10⁻⁵ J.
• With dielectric: new capacitance C′ = kC = 3 × 3 × 10⁻⁹ = 9 × 10⁻⁹ F.
  U_dielectric = \[\frac {1}{2}\]C′V² = 72 × 10⁻⁵ J.
• Change in energy = U_dielectric − U_air = (72 − 24) × 10⁻⁵ = 48 × 10⁻⁵ J.

Conclusion in one line:

When the dielectric is present, energy is higher by 48 × 10⁻⁵ J at constant voltage.

Question idea:

(a) A 900 pF capacitor is charged by 100 V. Find the stored energy.
(b) Then it is disconnected and connected to another 900 pF capacitor (uncharged). Find a new total energy.

Part (a)

• Capacitance C = 900 pF = 900 × 10⁻¹² F, voltage V = 100 V.
• Charge: Q = CV = 900 × 10⁻¹² × 100 = 9 × 10⁻⁸ C.
• Energy: U = \[\frac {1}{2}\]CV² = \[\frac {1}{2}\]QV.
  U = \[\frac {1}{2}\] × 9 × 10⁻⁸ × 100 = 4.5 × 10⁻⁶ J.

Simple meaning:

This is how much energy the single capacitor stores when charged to 100 V.

Part (b)

• Now two equal capacitors (both 900 pF) are connected, so the final charges are shared equally.
• Let the final potential be V′. Then each capacitor has Q′ = CV′.
• Total charge is conserved: Q_total = Q. So each gets Q′ = Q/2.
• Therefore V′ = Q′/C = (Q/2)/C = V/2.
• Energy in each capacitor: \[\frac {1}{2}\]Q′V′. Total energy = 2 × \[\frac {1}{2}\]Q′V′ = Q′V′ = \[\frac {1}{4}\]QV.
  Numerically: U_final = 2.25 × 10⁻⁶ J.

Key point:

Final energy is half of the initial: 4.5 × 10⁻⁶ → 2.25 × 10⁻⁶ J. The “missing” energy is lost as heat and electromagnetic radiation when charge flows during connection.