Energy Stored in a Charged Capacitor
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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.
Definition: Energy Stored in a Capacitor
The work done in the transfer of charge q between the two plates of a capacitor, which gets stored in the form of potential energy of the system, is called the energy stored in a capacitor.
Formula: Energy Stored / Work Done in a Capacitor
W = \[\frac {1}{2}\]qV
U = \[\frac {Q^2}{2C}\] = \[\frac {1}{2}\]QV = \[\frac {1}{2}\]CV 2
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:
V = \[\frac {q}{C}\]
If an additional small charge dq is supplied, the small work done is:
dW = V dq = \[\frac {q}{C}\]dq
Total work done in charging the capacitor from 0 to the final charge Q is:
W = \[\int_0^Q\frac{q}{C}dq\]
W = \[\frac{1}{C}\int_0^Qqdq=\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q\]
W = \[\frac{Q^2}{2C}\]
Since this work is stored as energy:
U = \[\frac{Q^2}{2C}\]
Using Q = CV, the equivalent forms become:
U = \[\frac{Q^2}{2C}=\frac{1}{2}QV=\frac{1}{2}CV^2\]
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.
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.
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?
At constant voltage, energy U = \[\frac {1}{2}\]CV 2 , so energy ∝ capacitance.
Steps in simple words
With air: use given C = 3 × 10⁻⁹ F and V = 400 V. U air = \[\frac {1}{2}\]CV 2 = 24 × 10 −5 J.
With dielectric: new capacitance C′ = kC = 3 × 3 × 10 −9 = 9 × 10 −9 F. U dielectric = \[\frac {1}{2}\]C′V 2 = 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.
(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.
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 = \[\frac {1}{2}\]CV 2 = \[\frac {1}{2}\]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.
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.
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.
Shaalaa.com | Capacitor and Capacitance part 19 (Energy Stored in Capacitors, Energy density)
Related Questions VIEW ALL [45]
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An air-filled parallel plate capacitor has a uniform electric field `overset(->)("E")` in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is ______ (∈ 0 = permittivity of free space)
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CBSE Previous Year Question Papers With Solutions for Class 10
Maharashtra State Board Previous Year Question Papers With Solutions for Class 12 Arts
Maharashtra State Board Previous Year Question Papers With Solutions for Class 12 Commerce
Maharashtra State Board Previous Year Question Papers With Solutions for Class 12 Science
Maharashtra State Board Previous Year Question Papers With Solutions for Class 10
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 12 Arts
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 12 Commerce
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 12 Science
CISCE ICSE / ISC Board Previous Year Question Papers With Solutions for Class 10
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