Capacitance and Dielectrics: Concepts, Calculations, and Applications

Capacitance and Dielectrics

Capacitance

Capacitance is a measure of a system's ability to store electric charge. It is defined for a pair of conductors carrying equal and opposite charges, with a potential difference between them. The device that stores this charge is called a capacitor.

• Definition: The capacitance C is the ratio of the charge q on one conductor to the potential difference V between the conductors.

$q = CV$

• SI Unit: The unit of capacitance is the farad (F), where $1\ \mathrm{F} = 1\ \mathrm{C}/\mathrm{V}$.

• Permittivity of Free Space: $\epsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2) = 8.85 \times 10^{-12}\ \mathrm{F}/\mathrm{m}$

Calculating Capacitance for Common Geometries

The capacitance depends on the geometry of the conductors and the medium between them.

• Parallel-Plate Capacitor: Two parallel plates of area A separated by distance d:

$C = \epsilon_0 \frac{A}{d}$

• Cylindrical Capacitor: Two coaxial cylinders, inner radius a, outer radius b, length L:

$C = 2\pi\epsilon_0 \frac{L}{\ln(b/a)}$

• Spherical Capacitor: Two concentric spheres, inner radius a, outer radius b:

$C = 4\pi\epsilon_0 \frac{ab}{b-a}$

Capacitors in Parallel and Series

Capacitors can be combined to achieve desired capacitance values. The rules for combining capacitors depend on whether they are connected in parallel or in series.

• Parallel Combination: The equivalent capacitance is the sum of individual capacitances.

$C_{\mathrm{eq}} = \sum_i C_i$

• Series Combination: The reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances.

$\frac{1}{C_{\mathrm{eq}}} = \sum_i \frac{1}{C_i}$

Energy Stored in a Capacitor

When a capacitor is charged, it stores energy in the electric field between its plates. The energy U stored is given by:

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

• Energy Density: The energy per unit volume in the electric field is:

$u = \frac{1}{2}\epsilon_0 E^2$

Capacitors and Dielectrics

Inserting an insulating material (dielectric) between the plates of a capacitor increases its capacitance by a factor called the dielectric constant κ:

$C = \kappa C_{\mathrm{air}}$

• Dielectric Constant (κ): A dimensionless number characterizing the material's ability to increase capacitance.

Worked Examples

Unit Consistency for Permittivity

• Show that $\epsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$ is equivalent to $8.85 \times 10^{-12}\ \mathrm{F}/\mathrm{m}$.

$1\ \mathrm{F}/\mathrm{m} = 1\ \mathrm{C}/(\mathrm{V}\cdot\mathrm{m}) = 1\ \mathrm{C}/(\mathrm{N}\cdot\mathrm{m}/\mathrm{C}\cdot\mathrm{m}) = 1\ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$

Example: Charging a Capacitor

• A 25 μF capacitor is connected to a 120 V battery. The charge stored is:

$q = CV = (25 \times 10^{-6}\ \mathrm{F})(120\ \mathrm{V}) = 3.0 \times 10^{-3}\ \mathrm{C} = 3.0\ \mathrm{mC}$

Example: Capacitance of a Parallel-Plate Capacitor

• Plates of radius 8.2 cm, separation 1.3 mm:

$C = \epsilon_0 \frac{\pi r^2}{d} = (8.85 \times 10^{-12}\ \mathrm{F}/\mathrm{m}) \frac{\pi (8.2 \times 10^{-2}\ \mathrm{m})^2}{1.3 \times 10^{-3}\ \mathrm{m}} = 1.4 \times 10^{-10}\ \mathrm{F} = 140\ \mathrm{pF}$

• Charge for 120 V applied:

$q = CV = (1.4 \times 10^{-10}\ \mathrm{F})(120\ \mathrm{V}) = 1.7 \times 10^{-8}\ \mathrm{C} = 17\ \mathrm{nC}$

Example: Plate Separation for 1 F Capacitance

• Area = 1.00 m2, C = 1.00 F. Find d:

$d = \frac{\epsilon_0 A}{C} = \frac{8.85 \times 10^{-12}\ \mathrm{F}/\mathrm{m} \times 1.00\ \mathrm{m}^2}{1.00\ \mathrm{F}} = 8.85 \times 10^{-12}\ \mathrm{m}$

• This separation is much smaller than atomic dimensions, so such a capacitor is not physically realizable.

Example: Spherical Capacitor Volume

• Given C = 2.0 μF, outer sphere radius is twice the inner: b = 2a.

• Capacitance formula:

$C = 4\pi\epsilon_0 \frac{ab}{b-a} = 8\pi\epsilon_0 a$

• Solve for a:

$a = \frac{C}{8\pi\epsilon_0} = \frac{2.0 \times 10^{-6}\ \mathrm{F}}{8\pi \times 8.85 \times 10^{-12}\ \mathrm{F}/\mathrm{m}} = 9.0 \times 10^3\ \mathrm{m}$ $b = 2a = 1.8 \times 10^4\ \mathrm{m}$

• Volume between spheres:

$V_{\mathrm{enc}} = \frac{4}{3}\pi (b^3 - a^3) = \frac{4}{3}\pi ((2a)^3 - a^3) = \frac{4}{3}\pi (7a^3) = 2.1 \times 10^{13}\ \mathrm{m}^3$

Example: Equivalent Capacitance (Mixed Series and Parallel)

• Given: C1 = 10.0 μF, C2 = 5.00 μF (series), C3 = 4.00 μF (parallel with series combination).

Step 1: Combine series capacitors:

$\frac{1}{C_{\mathrm{eq,series}}} = \frac{1}{10.0\ \mu\mathrm{F}} + \frac{1}{5.00\ \mu\mathrm{F}} = 0.300\ \mu\mathrm{F}^{-1}$ $C_{\mathrm{eq,series}} = 3.33\ \mu\mathrm{F}$

Step 2: Add parallel capacitor:

$C_{\mathrm{eq,total}} = 3.33\ \mu\mathrm{F} + 4.00\ \mu\mathrm{F} = 7.33\ \mu\mathrm{F}$

Example: Number of Capacitors Needed

• How many 1.00 μF capacitors in parallel are needed to store 1.00 C at 110 V?

$n = \frac{q_{\mathrm{tot}}}{CV} = \frac{1.00\ \mathrm{C}}{(1.00 \times 10^{-6}\ \mathrm{F})(110\ \mathrm{V})} = 9.09 \times 10^3$

• So, 9090 capacitors are required.

Summary Table: Capacitance Formulas for Common Geometries

Geometry	Capacitance Formula	Parameters
Parallel-Plate	$C = \epsilon_0 \frac{A}{d}$	Area $A$, separation $d$
Cylindrical	$C = 2\pi\epsilon_0 \frac{L}{\ln(b/a)}$	Length $L$, radii $a$, $b$
Spherical	$C = 4\pi\epsilon_0 \frac{ab}{b-a}$	Radii $a$, $b$

Additional info: The notes include practical limitations (e.g., atomic scale limits for plate separation), and the effect of dielectrics is summarized but not deeply derived. For more advanced study, see derivations of energy density and dielectric breakdown.