19.5 Capacitors and Dielectrics

Capacitor Fundamentals

Capacitance and charge storage

A capacitor stores electric charge and energy by maintaining an electric field between two conducting plates. The quantity that describes how much charge a capacitor can hold per volt applied is called capacitance ($C$), defined as:

$C = \frac{Q}{V}$

Capacitance is measured in farads (F), where 1 F = 1 C/V. One farad is actually a huge amount of capacitance, so you'll typically see values in microfarads ($\mu$F) or picofarads (pF).

Here's how a capacitor actually works:

• Two conducting plates are separated by an insulating material called a dielectric (this can be air, paper, plastic, ceramic, etc.).
• When you connect a voltage source, positive charge builds up on one plate and negative charge on the other.
• The opposite charges on the two plates attract each other, but the dielectric prevents charge from flowing directly between them.
• This charge separation creates a potential difference (voltage) across the plates and stores energy in the electric field between them.

Capacitance of parallel plate capacitors

For a parallel plate capacitor, the capacitance depends on three things: plate area ($A$), plate separation ($d$), and the permittivity of free space ($\varepsilon_0$):

$C = \frac{\varepsilon_0 A}{d}$

where $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m.

Two key relationships to remember:

• Capacitance is directly proportional to plate area. A larger area gives more room for charge to spread out, so more charge can be stored at the same voltage.
• Capacitance is inversely proportional to plate separation. Bringing the plates closer together strengthens the electric field between them, which increases the capacitor's ability to hold charge.

Dielectrics and Capacitor Configurations

Dielectrics and capacitance increase

A dielectric is an insulating material placed between the plates of a capacitor. Common examples include air, paper, plastic, and ceramic. Inserting a dielectric does several useful things:

• It increases capacitance by a factor called the dielectric constant ($\kappa$), which is always greater than 1. The modified formula becomes:

$C = \kappa \frac{\varepsilon_0 A}{d}$

• It prevents electrical breakdown (sparking) between the plates, allowing you to apply higher voltages without discharge.
• It increases the total energy the capacitor can store for a given plate area and separation.

Why does this work? When a dielectric is placed in an electric field, its molecules polarize, meaning the positive and negative charges within each molecule shift slightly. This polarization partially cancels the electric field between the plates, which reduces the voltage for a given charge. Since $C = Q/V$, a lower $V$ for the same $Q$ means higher capacitance.

Problem-solving for capacitor configurations

The charge stored in any capacitor is found from:

$Q = CV$

where $Q$ is charge in coulombs, $C$ is capacitance in farads, and $V$ is voltage in volts.

Series capacitors (connected end-to-end along a single path):

1. The total capacitance is less than the smallest individual capacitance.
2. Use the reciprocal formula: $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + ...$
3. Each capacitor holds the same charge ($Q$ is equal across all).
4. Voltage divides among the capacitors, with smaller capacitors getting a larger share of the voltage (since $V = Q/C$).

Parallel capacitors (connected across the same two nodes):

1. Total capacitance is the sum: $C_{total} = C_1 + C_2 + ...$
2. Each capacitor has the same voltage (equal to the source voltage).
3. Charge divides among capacitors in proportion to their capacitances (larger $C$ stores more $Q$).

Energy Storage in Capacitors

Capacitors store potential energy in the electric field between their plates. The energy stored is:

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

You can also write this in two equivalent forms using $Q = CV$:

$U = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}QV$

All three expressions give the same result; which one you use depends on what quantities you're given.

The $V^2$ in the first form is worth paying attention to: doubling the voltage across a capacitor quadruples the stored energy. This is why capacitors charged to high voltages can store (and release) significant amounts of energy.