Notes can be found as interactive webpage at

24: Capacitance, Dielectrics, Electric Energy Storage

24.1 Capacitors #

  • Capacitors are devices that store an electric charge
    • Normally consists of two conducting objects; plates, sheets
      • When a voltage is applied, the two plates become charged: one positive, one negative
    • Conductors are placed near one another, but not touching
      • This distance is typically due to an insulator between sheets
      • Capacitors are typically rolled so that they take up less room
  • Two main use cases
    1. Storing energy for later use; e.x. camera flash
    2. Block surges of charge and energy to protect circuits
  • The amount of charge .$Q$ acquired by each plate is proportional to
    1. .$V$: The potential difference of the two plates (Volts)
    2. .$C$: The constant capacitance of the capacitor (Coulombs per volt, farad) $$Q = CV$$

24.2 Determination of Capacitance #

  • In the real world, capacitance is determined experimentally by using the prior equations
  • For ideal cases where the sheets are separated by a vacuum or air, however, we can use the following equations
  • For a parallel-plate capacitor where .$A$ is the area of each plate and .$d$ is the distance between plates: $$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$
    • We also know this because .$E = \sigma / \varepsilon_0$ and .$\sigma = Q/A$
  • Since .$V = \int E\ dl = \frac{Qd}{\varepsilon_0 A}$, we can relate it to .$C$ as $$C = \frac{Q}{V} = \varepsilon_0 \cdot \frac{A}{d}$$

Capacitance-finding strategy

  1. Assign an arbitrary charge .$\pm q$ to the two plates.
  2. Using Gauss’s law or other techniques, calculate the electric field between these two plates
  3. From that electric field, calculate the potential difference between the plates, .$V = -\int \vec E \cdot d \vec s$
  4. Calculate the capacitance using .$C = q/V$. The arbitrary charge .$q$ from (1) should cancel out.

24.3 Capacitors in Series and Parallel #

Series #

  • The current/charge on each capacitor has the same magnitude: $$Q = Q_1 = Q_2 = \dots$$
  • The total voltage across all capacitors is sumo of the voltage drops of the individual components: $$V = V_1 + V_2 + \dots = I(R_1 + R_2 + \dots)$$
  • And since .$V = Q/C$, capacitance is then $$\frac{Q}{C_\text{eq}} = \frac{Q}{C_1} + \frac{Q}{C_2} + \dots \Longrightarrow \frac{1}{C_\text{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots $$
    • Notice that the equivalence capacitance is smaller than the smallest contributing capacitance

Parallel #

  • The total current/charge is the sum of the currents flowing through each component $$Q = Q_1 + Q_2 + \dots = V (R^{-1}_1 + R^{-1}_2 + \dots)$$
  • Voltage (potential difference) is the same across all paths/capacitors $$V = V_1 = V_2 = \dots$$
  • Therefore, we can use .$V = Q/C$ to write the equivalent capacitance as $$Q = C_1 V + C_2 V + \dots$$ $$Q = C_\text{eq} V = (C_1 + \dots)V \Longrightarrow C_\text{eq} = C_1 + \dots$$
    • The net effect of connecting capacitors in parallel is to increase the capacitance
      • Makes sense: We’re essentially increasing area of the plates
    • The overall working voltage is always limited by the smallest working voltage of an individual capacitor.

24.4 Storage of Electric Energy #

  • The energy stored in a capacitor is equal to the work done to charge it.
    • Initially, an uncharged capacitor requires no work to move the first few bits of charge
    • As more charge is stored, more work is needed to add more charge of the same sign because of the electric repulsion
    • That is, the more charge already on a plate, the more work required to add additional charge
  • Since we know .$dW = V\ dq$ and .$V = q/C$, we can write the work needed to store charge .$Q$ as $$W = \int_0^Q V\ dq = \frac{1}{C}\int_0^Q q \ dq = \frac{1}{2} \frac{Q^2}{C}$$
  • Since .$U = W$ and .$Q = CV$, we can write the energy stored in a capacitor with charges .$+Q$ and .$-Q$ on its two conductors as $$U = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2}CV^2 = \frac{1}{2}QV$$
  • It is useful to think of the energy stored in a capacitor as being stored in the electric field between the plates.
    • E.x. lets find the energy stored in a parallel-plate capacitor in terms of the electric field
      • We know for two close parallel plates we can find the potential difference as .$V = Ed$ where .$d$ is distance between plates
      • We also know .$C = \varepsilon_0 A/d$ for parallel plate capacitors, thus we can write $$U = \frac{1}{2}CV^2 = \frac{1}{2}\bigg(\frac{\varepsilon_0 A}{d}\bigg)(E^2 d^2) = \frac{1}{2} \varepsilon_0 E^2 Ad$$
      • We can recognize .$Ad$ as the volume between the plates where .$E$ exists
      • If we divide both sides of by this volume, we can an equation for the energy density .$u$: $$u = \frac{\text{energy}}{\text{volume}} = \frac{1}{2}\varepsilon_0 E^2$$
    • Thus, electric energy stored per unit volume in any region of space is proportional to the square of the electric field
    • We proved this with parallel plates, but this can be shown for any region with an electric field

24.5 Dielectrics #

  • Dielectrics are the insulating material sheet placed between conductors
  • They serve to
    1. Because they don’t break down, they allow electric charge to flow as easily as air so higher voltages can be applied without charge passing across the gap
    2. Allow the plates to be placed closer together without touching, allowing an increased capacitance because the thickness .$d$ is smaller
    3. Dielectrics increase the capacitance by a factor .$K$ (known as the dielectric constant) $$C = KC_0$$
      • .$C_0$ is the capacitance when the space is a vacuum/air
      • .$C$ is the capacitance with the dielectric filling the space
  • For parallel-plate capacitors, we use .$C = Q/V = \varepsilon_0 A/d$ and .$C = KC_0$ $$C = K \varepsilon_0 \frac{A}{d}$$
  • Energy density also changes with a dielectric as $$u = \frac{1}{2}K \varepsilon_0 E^2 = \frac{1}{2}\varepsilon E^2$$
  • Likewise, .$E$ and .$V$ are both also altered:
    • With no dielectric, the field is .$E_0 = \frac{V_0}{d}$ where .$V_0$ is the potential difference
    • If the capacitor is isolated (i.e. not connected to a battery) so that the charge stays constant, potential difference drops: .$V = V_0/K$
    • Therefore, .$E = \frac{V}{d} = \frac{V_0}{Kd} = \frac{E_0}{K}$
    • .$\varepsilon$ is the permittivity of the dielectric material defined as .$\varepsilon = K \varepsilon_0$