Electric Potential and Capacitance: Parallel Plate Capacitors and Dielectrics

Electric Potential

The Parallel Plate Capacitor

The parallel plate capacitor consists of two conductive plates separated by a distance, with an electric field established between them when a potential difference is applied. The field is uniform between the plates, except near the edges.

• Charge density ($\sigma$): $\sigma = \frac{q}{A}$, where $q$ is charge and $A$ is plate area.

• Electric field ($E$): $E = \frac{q}{\varepsilon_0 A} = \frac{\sigma}{\varepsilon_0}$

• Permittivity of free space ($\varepsilon_0$): $8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2$

Equipotential Surfaces and Their Relation to the Electric Field

Equipotential surfaces are regions where the electric potential is constant. The electric field is perpendicular to these surfaces and relates to the potential difference and separation:

• Electric field from potential gradient: $E = -\frac{\Delta V}{\Delta s}$

• For parallel plate capacitors: $E = \frac{V}{d}$ or $V = Ed$

Capacitors and Dielectrics

Dielectrics

A dielectric is an electrically insulating material placed between capacitor plates to increase capacitance and prevent breakdown. Advantages include:

• Reduced breakdown (sparking) compared to air

• Allows plates to be closer together

• Increases ability to store charge

Relation Between Charge and Potential Difference for a Capacitor

The charge stored on a capacitor is proportional to the potential difference:

• $q = CV$

• Capacitance ($C$): property determining how much charge can be stored for a given voltage

• SI Unit: coulomb/volt = farad (F)

The Dielectric Constant

The dielectric constant ($\kappa$) quantifies how much a dielectric increases capacitance:

• $\kappa = \frac{E_0}{E}$, where $E_0$ is field without dielectric, $E$ is field with dielectric

• $\kappa \geq 1$

Table: Dielectric Constants of Common Substances

Substance	Dielectric Constant, $\kappa$
Vacuum	1
Air	1.00054
Teflon	2.1
Benzene	2.28
Paper (royal gray)	3.3
Ruby mica	5.4
Neoprene rubber	6.7
Methyl alcohol	33.6
Water	80.4

Capacitance of a Parallel Plate Capacitor

For a parallel plate capacitor filled with a dielectric:

• $C = \frac{\kappa \varepsilon_0 A}{d}$

• Capacitance increases with increasing $\kappa$, increasing $A$, and decreasing $d$

Dielectric Breakdown

If the electric field exceeds a material's dielectric strength, the dielectric breaks down and conducts electricity.

Substance	Dielectric Strength (V/m)
Mica	$100 \times 10^6$
Air	$3 \times 10^6$

Example: Computer Keyboard

Capacitance changes in a keyboard are detected electronically. When a key is pressed, the plate separation decreases, increasing capacitance.

• Key not pressed: $C = 0.589 \times 10^{-12}$ F

• Key pressed: $C = 19.6 \times 10^{-12}$ F

• Change: $\Delta C = 19.0$ pF

Example: Plate Size for 1 F Capacitor

• For $d = 1$ mm in air: $A = 1.13 \times 10^6$ m$^2$ ($\sqrt{A} = 1.1$ km)

• For $d = 1$ μm with paper: $A = 3.42 \times 10^4$ m$^2$ ($\sqrt{A} = 185$ m)

Conceptual Example: Effect of Dielectric with Constant Charge

When a dielectric is inserted after disconnecting the battery (constant charge):

• Voltage across plates decreases

• $V = V_0/\kappa$ for $\kappa > 1$

Conceptual Example: Effect of Dielectric with Constant Voltage

When a dielectric is inserted while the battery remains connected (constant voltage):

• Charge on plates increases

• $q = \kappa q_0$ for $\kappa > 1$

Energy Storage in a Capacitor

Energy Stored

The energy stored in a capacitor is given by:

• $Energy = \frac{1}{2} qV = \frac{1}{2} CV^2 = \frac{q^2}{2C}$

Energy Density

The energy density in the electric field of a capacitor is:

• $Energy\ density = \frac{Energy}{Volume} = \frac{1}{2} \kappa \varepsilon_0 E^2$

Example: Energy Stored in a Capacitor

• For $C = 10$ F, $V = 3$ V: $Energy = \frac{1}{2} (10)(3^2) = 45$ J

• Equivalent to gravitational potential energy: $GPE = mgh = (4.6)(9.8)(1) = 45$ J

Example: Energy Density Near a Point Charge

• For $q = 40$ μC at $r = 2.0$ cm: $E = \frac{kq}{r^2} = 8.99 \times 10^8$ V/m

• Energy density: $\frac{1}{2}(8.85 \times 10^{-12})(8.99 \times 10^8)^2 = 3.6 \times 10^6$ J/m$^3$

• Energy in shell of thickness 1 mm: $Volume = 5.0 \times 10^{-6}$ m$^3$, $Energy = 18$ J

Additional info: These notes cover the essential concepts of electric potential, parallel plate capacitors, dielectrics, capacitance, energy storage, and related examples, suitable for college-level physics students.