Capacitance

What and why?

They are used in various circuits such as frequency filters, televisions, motherboards, and even your cellphones! We used them in lab to create touchscreens.

Definition

Capacitors are made by separating two parallel plates by a dielectric (nonconductive material). They are used to store charge across the two plates, and the resulting electric field between the plates stores a voltage potential.

Capacitance

Capacitors are measured by their ability to store a charge, and how much potential is required to store said charge. This measurement is called capacitance, and their units are Farads (F), where $1 \text{ Farad} = \frac{1\text{ Coulomb}}{1\text{ Volt}}$.

Below is an important relationship between capacitance, charge, and voltage:

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

The specific material used for the dielectric is important because it affects the field, and thus capacitance, of the capacitor. The dielectric's effect is described by its permittivity constant, $\varepsilon$, which is a material property measured in Farads per meter.

Capacitor Diagram

Capacitance is described in terms of the dielectric plates' area, their permittivity constant, and the distance between them:

$C= \int_0^l \int_0^w \frac{\varepsilon}{d} \,dw\,dl = \frac{\varepsilon lw}{d}=\frac{\varepsilon A}{d}$

*For this class, we will only really consider the parallel plate model and the integration won't be required.

In the next section, we will analyze the effects of stacking capacitors on top of each other or putting them side by side.

Capacitors In Circuits

In circuits, it is common to have two capacitors in series, parallel, or both! Luckily, we can fall back on our equation for capacitance to find the effects of having multiple capacitors in different configurations.

Parallel

Capacitors in Parallel

When two capacitors, $C_1$ and $C_2$, are in parallel, we know that they have the same voltage ($V$) across them. Consider each parallel capacitor as contributing to the overall area of the equivalent capacitor. If we refer back to the equation for capacitance, we know that the areas of capacitors are directly proportional to their capacitance, thus creating an additive property:

$C_{eq} = C_1 + C_2$

This relationship will always hold across different permittivities, areas, and distances between plates as long as the capacitors are in parallel, because the relationship comes from equal voltage being supplied to the capacitors. This tells us that the equivalent capacitance across capacitors in parallel is the sum of all parallel capacitors' capacitances.

Now let's examine the charge on parallel capacitors:

$Q_1 = C_1V, \quad Q_2 = C_2V$

$Q_{eq} = C_{eq}V = \left(C_1 + C_2\right)V = C_1V + C_2V = Q_1 + Q_2$

Thus, we see that the equivalent charge across capacitors in parallel is the sum of all parallel capacitors' charges.

Series

Capacitors in Series

When two capacitors are in series, they have the same charge ($Q$) across their plates. Consider the outermost plates to be part of one equivalent capacitor; the distances between the plates are, therefore, additive. Referring back to our equation for capacitance, distances are inversely proportional to capacitance. This means our equation for equivalent capacitance in series will look quite similar to that for parallel resistors:

$C_{eq} = \left(\frac{1}{C_1} + \frac{1}{C_2}\right)^{-1} = \frac{C_1C_2}{C_1 + C_2} = C_1||C_2$

This can be abbreviated by the shorthand parallel operator. However, it is important to note that the parallel operator does not necessarily mean the components are in parallel.

Now, let's examine the voltage across capacitors in series:

$V_1 = \frac{Q}{C_1},\quad V_2 = \frac{Q}{C_2}$

$V_{eq} = \frac{Q}{C_{eq}} = \frac{Q}{\frac{C_1C_2}{C_1 + C_2}} = \frac{Q\left(C_1 + C_2\right)}{C_1C_2} = \frac{Q}{C_1} + \frac{Q}{C_2} = V_1 + V_2$

Therefore, the equivalent voltage of capacitors in series is equal to the sum of all present voltages.

Steady State

While capacitors introduce a time element to a circuit, for this class we will primarily look at them in their steady state. A capacitor's steady state is the point when a it is eitcompletely charged or discharged. Intuitively, a capacitor is discharged when it has $Q = 0$:

Discharged

$Q = CV \implies V = 0$

When discharged, a capacitor will begin charging when connected to a voltage supplier. Below, we see a circuit that will allow a capacitor, $C$, to become fully discharged:

Discharged C

Charged

So how do we find the maximum charge for a capacitor? When connected to a voltage supply, the capacitor will max out at the total voltage it is supplied:

Charging C

In this example, our capacitor will charge to the voltage being supplied from the voltage source: $Q = CV$

In the following example, we introduce a voltage divider:

Now, since we have a voltage divider, the capacitor can only charge to the relative voltage across $R_2$: $Q = C \left( V \cdot \frac{R_2}{R_1 + R_2} \right)$

At steady state, charged capacitors are treated as open circuits.

Energy

In the process of charging a capacitor, we are storing energy. The energy stored can be found by integrating our equation for charge, and using the relationship between current and voltage:

$Q = CV, \quad I = \frac{dQ}{dt} = C \frac{dV}{dt}$

$P(t) = IV = CV \frac{dV}{dt}$

$E = \int_0^\infty P(t) = \frac{1}{2} CV^2$

Practical Applications

Capacitors can be used to add a time element to circuits. We used this element during the imaging module in lab to smooth out the light sensors' readings.

They can also be used to measure a variable distance and react to the changes in these differences, like in the capacitive touchscreen module, or in the homework, where a train modulated its height using giant capacitors.

Example

We have provided a video walkthrough of a capacitance problem below:

The problem: