# 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

- Normally consists of two conducting objects; plates, sheets
- Two main use cases
- Storing energy for later use; e.x. camera flash
- Block surges of charge and energy to protect circuits

- The amount of charge .$Q$ acquired by each plate is proportional to
- .$V$: The potential difference of the two plates (Volts)
- .$C$: 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

- Assign an arbitrary charge .$\pm q$ to the two plates.
- Using Gauss’s law or other techniques, calculate the electric field between these two plates
- From that electric field, calculate the potential difference between the plates, .$V = -\int \vec E \cdot d \vec s$
- 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.

- The net effect of connecting capacitors in parallel is to increase the capacitance

# 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

- E.x. lets find the energy stored in a parallel-plate capacitor in terms of the electric field

# 24.5 Dielectrics #

- Dielectrics are the insulating material sheet placed between conductors
- They serve to
- 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
- Allow the plates to be placed closer together without touching, allowing an increased capacitance because the thickness .$d$ is smaller
- 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 .$\varepsilon = K \varepsilon_0$