When charges are moved, work must be done, since they are forced to follow a potential difference. The work required to charge a capacitor is then stored as energy in the capacitor’s electric field.
To calculate this more easily, imagine moving individual electrons from the negatively charged plate across the insulating gap to the positively charged plate. With each additional charge, the force needed increases because the potential difference grows.
The infinitesimal work needed for a small added charge $dQ$ is:
\[
dW = U(Q)\, dQ
\]
Using the definition of capacitance
\[
C = \frac{Q}{U}
\]
we can substitute for $U(Q)$ and integrate over all charges from $0$ to $Q_0$:
\[
E_\mathrm{el} = \int_0^{Q_0} dW
= \frac{1}{C} \int_0^{Q_0} Q\, dQ
= \frac{1}{2}\frac{Q_0^2}{C}
\]
Since $Q_0 = CU_0$, this can be expressed in terms of the applied voltage:
\[
\boxed{E_\mathrm{el} = \tfrac{1}{2} C U_0^2}
\]
Thus, the energy stored in the capacitor grows quadratically with the voltage and linearly with the capacitance.
This energy is released again as work when the capacitor discharges.
Applications:
- Defibrillators use this principle: a battery is stepped up to several kilovolts using a DC-DC converter.
A large capacitor is charged, which can then deliver a powerful pulse in milliseconds — much higher than the battery could provide directly. - Camera flash units also use capacitors. The stored energy is released in a very short time, producing an intense flash of light.
Example:
A defibrillator capacitor with $C = 200\,\mu\text{F}$ is charged to $U_0 = 4\,\text{kV}$.
How much energy is stored?
Solution:
\[
E = \tfrac{1}{2} C U_0^2 = \tfrac{1}{2} \cdot 200 \times 10^{-6} \cdot (4000)^2
\]
\[
E \approx 1.6 \,\text{kJ}
\]
The capacitor stores about 1.6 kJ of energy.
The d of the differentials are written in the wrong font.
Posted by supermeisi on 2025-09-26 16:24:54