Capacitors can be built with two conductors, called plates, separated by an insulating medium.

Capacitance:

Capacitance is the ability to store electrical charges. It is defined as the ratio between the amount of charges \(Q\) of the positive plate and the \(\mathbb{V}_{AB}\) between plate \(A\) and \(B\), so \(C = \frac{Q}{\mathbb{V}_{AB}}.\) But, for simplicity, we can just write $$C = \frac{Q}{\mathbb{V}}.$$ The capacitance of a conductor depends on its geometry and the material between conductors. The \(IS\) unit of capacitance is Faraday, which is Coulomb per volt ( \([F] = \frac{C}{V} \) ).

Example 1

Example 2

Isolated Spherical Conductor:

The capacitance for an isolated spherical conductor is: $$C = 4\pi \epsilon R $$ Where \(R\) is the radius of the spherical conductor and \(\epsilon\) is the electric permittivity of the medium.

Flat and Parallel Plates

The capacitance of a capacitor made of flat parallel plates, in vacuum, is directly proportional to the area of the plates and inversely proportional to the distance between them, $$C = \epsilon_0 \frac{A}{d}$$ Where \(A\) is the useful area of the plates, \(d\) is the separation between the plates and \(\epsilon_0\) is the vacuum electric permittivity.

Example 3

Surface Charge Density \((\sigma)\)

For an area \(A\) with a charge amount \(Q\) evenly distributed on its surface, the surface density of electric charge is: $$\sigma = \frac{Q}{A}.$$

Capacitors Association

Serial Association

The charge on each capacitor is the same, and the \(\mathbb{V}\) in the \(i\)-th capacitor is given by: $$\mathbb{V}_i = \frac{Q}{C_i}$$ Such that, $$\mathbb{V} = \mathbb{V}_1 + \mathbb{V}_2 + \ldots + \mathbb{V}_n$$ and dividing both sides by \(Q\) we obtain: $$\frac{1}{C_{eq}} = \frac{1}{C_{1}} +\frac{1}{C_{2}} + \ldots+ \frac{1}{C_{n}}$$

Parallel Association

The \(\mathbb{V}\) in each capacitor is the same, and the load on the \(i\)th capacitor is given by \(Q_i = \mathbb{V} C_i\) or \(C_i=\frac{Q_i}{\mathbb{V}}\) . The total load is $$Q = Q_1 + Q_2 + \ldots + Q_n,$$ and dividing both sides of the above equation by \(\mathbb{V}\), we obtain $$\frac{Q}{\mathbb{V}} = \frac{Q_1}{\mathbb{V}} + \frac{Q_2}{\mathbb{V}} + \ldots + \frac{Q_n}{\mathbb{V}},$$ and thus, $$C_{eq} = C_{1} + C_{2} + \ldots + C_{n}.$$

Effect of a Dielectric

A non-conductive material is also called dielectric. When introducing a dielectric between the conductors of a capacitor, the electric field is reduced and the capacitance \(C\) increases by a factor \(K\), the dielectric constant. This means: \(C = K C_0\) , were \(C_0\) is the capacitance without the dielectric. The field in the dielectric is reduced because the dipole moments of molecules (either permanent or induced) tend to align with the field and generate another field that opposes the former. The dipole moment aligned with the field is proportional to the intensity of the driving field.

Utilities of a dielectric:

• It increases the capacitance of a capacitor.
• It increases the dielectric strength.
• It provides mechanical separation between the conductors.

Energy Stored in a Capacitor

The potential energy stored, \(U\), in an electric field of a capacitor with capacitance \(C\) , with a charge \(Q\) and subjected to a potential difference \(\mathbb{V}\), is given by: $$U = \frac{C\mathbb{V}^2}{2} = \frac{Q^2}{2C} = \frac{q\mathbb{V}}{2} $$