19.7 Energy Stored in Capacitors

Applications and Energy Storage of Capacitors

Capacitors store electrical energy and release it when needed. They show up everywhere in electronics, from smoothing out power supply noise to delivering life-saving shocks in defibrillators. The physics behind energy storage in capacitors connects directly to the relationship between charge, voltage, and capacitance.

Applications of Capacitors in Electronics

Capacitors serve several distinct roles depending on how they're wired into a circuit:

• Filtering and smoothing
  • Remove ripples and noise from power supplies (voltage regulators)
  • Smooth out voltage fluctuations in audio circuits (amplifiers)
• Energy storage
  • Provide short-term power during outages or switching (computers, clocks)
  • Store energy in camera flash units for rapid discharge
• Timing and oscillation
  • Paired with resistors in RC circuits to create time delays (timers, alarms)
  • Set the frequency of oscillation in resonant circuits (radio tuners)
• Coupling and decoupling
  • Block DC signals while allowing AC signals to pass, called coupling (audio amplifiers)
  • Prevent high-frequency noise from reaching sensitive components, called decoupling (microprocessors)
• Tuning and filtering
  • Adjust the frequency response of a circuit (bandpass filters)
  • Filter out unwanted frequencies in radio and TV receivers (channel selectors)

Energy Calculation for Capacitors

When you charge a capacitor, you're doing work by moving charge from one plate to the other against an increasing voltage. The first bit of charge moves easily because the voltage is still near zero. As more charge builds up, each additional bit requires more work because the voltage across the plates keeps rising. This is why the energy formula has a factor of $\frac{1}{2}$ rather than just $CV^2$.

The energy stored in a capacitor is:

$E = \frac{1}{2}CV^2$

• $E$ = energy stored (joules, J)
• $C$ = capacitance (farads, F)
• $V$ = voltage across the capacitor (volts, V)

Two proportionality relationships are worth remembering:

• Energy is directly proportional to capacitance. Doubling $C$ doubles the stored energy.
• Energy is proportional to voltage squared. Doubling $V$ quadruples the stored energy. This is the one students often forget on exams.

Example calculation:

1. You have a capacitor with $C = 10 \, \mu F$ and $V = 5 \, V$.
2. Substitute into the formula: $E = \frac{1}{2}(10 \times 10^{-6} \, F)(5 \, V)^2$
3. Calculate: $E = \frac{1}{2}(10 \times 10^{-6})(25) = 125 \times 10^{-6} \, J = 125 \, \mu J$

Two equivalent forms of the energy equation are also useful. Since $Q = CV$, you can substitute to get:

$E = \frac{Q^2}{2C} \quad \text{or} \quad E = \frac{1}{2}QV$

Use whichever form matches the quantities you're given in a problem.

Capacitors in Medical Defibrillators

A defibrillator delivers a controlled electric shock to restore normal heart rhythm during cardiac arrest (specifically ventricular fibrillation, where the heart quivers instead of pumping). Capacitors make this possible.

Here's how the process works:

1. The capacitor is charged to a high voltage, typically around 3000 V.
2. When the paddles are placed on the patient's chest and triggered, the capacitor discharges its stored energy in a very short burst (around 10 ms).
3. The electrical pulse depolarizes all the heart muscle cells at once, giving the heart's natural pacemaker a chance to re-establish a normal rhythm.

The energy delivered is carefully controlled, typically between 200 and 360 J. Too little won't reset the heart; too much risks tissue damage.

Capacitors are well-suited for this job because they can store a large amount of energy in a compact space, release it in milliseconds, and be recharged within seconds for repeat shocks if needed.

Quick check: A defibrillator capacitor stores 360 J at 3000 V. What's the capacitance? Rearranging $E = \frac{1}{2}CV^2$ gives $C = \frac{2E}{V^2} = \frac{2(360)}{(3000)^2} = 80 \times 10^{-6} \, F = 80 \, \mu F$. That's a physically small component delivering a huge energy pulse.

Capacitor Physics and Electrostatics

A parallel plate capacitor consists of two conductive plates separated by a dielectric (insulating) material. Its capacitance depends on three things: the plate area $A$, the separation distance $d$, and the permittivity $\varepsilon$ of the dielectric between the plates.

The charge on the plates creates a uniform electric field in the gap between them. The voltage across the plates is related to stored charge by $V = \frac{Q}{C}$, so for a given capacitance, more stored charge means a higher voltage and more stored energy.