Capacitance and Dielectrics

Capacitance and Dielectrics

1: Capacitance

• Definition & Concept

  • Consider a parallel plate capacitor without any medium between the plates. The electric field (E) within the capacitor is calculated as:
    $E = \frac{|σ|}{ε_0}$

  • The potential difference (ΔV) across the plates can be determined using the formula:
    $\Delta V = E d = \left(\frac{|σ|}{ε_0}\right) d$

  • Setting the potential of the negative plate as 0 V and the positive plate as V, we find:
    $V = \frac{σ}{ε_0} d$

  • Expressing σ in terms of charge (Q) and area (A), we have:
    $σ = \frac{Q}{A}$

  • Consequently, the potential at the positive plate is:
    $V = \frac{Q d}{A ε_0}$

  • Rearranging this, we find the charge (Q) on the positive plate:
    $Q = \left(\frac{A ε_0}{d}\right)V$

  • Capacitance (C) is defined as:
    $C = \frac{ε_0 A}{d}$

  • Hence, the relationship between charge and potential for a capacitor is given by:
    $Q = C V$

2: Capacitance Characteristics

• Dependence of Capacitance

  • From the capacitance formula:
    $C = \frac{ε_0 A}{d}$

  • Capacitance depends solely on the geometry of the capacitor and changes only if either the plate area (A) or the distance (d) between the plates changes.

• Units of Capacitance

  • The units of capacitance stem from the relation $Q = C V$ (thus, $C = Q/V$):

  $[C] = \frac{[Q]}{[V]} = \frac{1 C}{1 V} = 1 F$

  • Capacitance is termed so because it sets the charge capacity of a capacitor. With a constant voltage, increasing capacitance allows more charge (Q) to be held: $Q = C V$.

• Dynamics of Charging

  • When a potential difference is applied across an initially neutral capacitor, an electric field forms, causing electrons to transfer from one plate to another through a circuit. The process stops at equilibrium when the charge satisfies:
    $Q = C V$.

• Isolation of Capacitor

  • If a capacitor is disconnected from the battery, the charge remains constant. If the plate distance changes, this alters capacitance which affects voltage inversely (since $V = \frac{Q}{C}$). Since $E = \frac{V}{d}$, any alteration in distance translates to a proportional change in voltage.

3: Proportions with Capacitance

• Proportional Relationships

  • Given the formula
    $C = \frac{ε_0 A}{d}$, capacitance shows proportional behavior:

  • Capacitance is proportional to the area of plates (A) and inversely proportional to the distance (d):

  1. If A increases by 4 and d decreases by 2:

     • Change in capacitance:
       $C' = 4 * 2 = 8 ext{ (8 times increase)}$

  2. If A increases by 3 and d increases by 4:

     • Change in capacitance:
       $C' = \frac{3}{4} = 0.75$

  3. If A decreases by 5 and d decreases by 10:

     • Change in capacitance:
       $C' = \frac{10}{5} = 2$

4: Capacitor Example Setup

• Sequence of Capacitor Changes:

  • The setup shows the changes in a capacitor as various quantities are measured during different states:

  1. State 1:

     • Quantities at Position 1: Q, C, V, E

     • Plate distance (d) = 0.2 m; Voltage = 10 V; Area = 0.5 m²

  2. State 2:

     • Distance increased to 0.4 m

     • Quantities are Q', C', V', and E'

  3. State 3:

     • Battery removed; Capacitor disconnected

  4. State 4:

     • Plates moved closer to 0.1 m; Quantities are Q'', C'', V'', E''

5: Calculations for Capacitor at Positions

Position 1:

• Parameters and Calculations:

  • Capacitance Calculation:
    $C = \frac{ε_0 A}{d} = \frac{(8.85 \times 10^{-12} C^2/(N m^2))(0.5 m^2)}{0.2 m} = 2.21 \times 10^{-11} F$

  • Voltage across capacitor:
    $V = 10 V$

  • Charge on plates:
    $Q = C V = (2.21 \times 10^{-11} F)(10 V) = 2.21 \times 10^{-10} C$

  • Electric Field:
    $E = \frac{V}{d} = \frac{10 V}{0.2 m} = 50 V/m$

Position 2:

• Changes After Increasing Distance

  • Upon increasing plate distance from 0.2 m to 0.4 m:

  • Capacitance decreases:
    $C' = \frac{1}{2} C = \frac{1}{2} (2.21 \times 10^{-11} F) = 1.11 \times 10^{-11} F$

  • Voltage remains constant:
    $V' = 10 V$

  • Charge changes:
    $Q' = \frac{1}{2} Q = \frac{1}{2} (2.21 \times 10^{-10} C) = 1.11 \times 10^{-10} C$

  • Electric field decreases:
    $E' = \frac{1}{2} E = \frac{1}{2} (50 V/m) = 25 V/m$

Position 4:

• After Removing Battery

  • Charge remains constant at:
    $Q'' = Q' = 1.11 \times 10^{-10} C$

  • Reducing distance to 0.1 m increases capacitance:
    $C'' = 4 C' = 4(1.11 \times 10^{-11} F) = 4.44 \times 10^{-11} F$

  • Voltage decreases:
    $V'' = \frac{1}{4} V' = \frac{1}{4} (10 V) = 2.5 V$

  • Electric field remains unchanged:
    $E'' = \frac{V''}{d} = \frac{2.5 V}{0.1 m} = 25 V/m$

6: Energy Stored by a Capacitor

• Energy Principles:

  • Energy is required to move charges, which is also the energy stored in a capacitor charged to $Q = C V$:

  • General formula:
    $U = \frac{1}{2} \frac{Q^2}{C}$

  • Rewritten in terms of voltage:
    $U = \frac{1}{2} \frac{(C V)^2}{C} = \frac{1}{2} C V^2$

  • Another representation:
    $U = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} Q V$

  • All variations yield the same energy based on the provided parameters, with preferred equations dependent on known values.

7: Insulators (Dipoles)

• Characteristics of Insulators:

  • In insulators, outer electrons are bound to their atomic nuclei, unlike conductors where they can flow freely.

  • Neutral atoms exhibit a positive charge at the nucleus and negative charge from surrounding electrons, forming electric dipoles: a +q charge at the center and –q charge at a distance.

8: Dipoles in External Electric Fields

• Behavior in Electric Fields:

  • Without an external electric field, dipoles orient randomly. In contrast, when exposed to an electric field (E), dipoles experience forces leading to an alignment along the field's direction.

  • The resulting dipole field (E′) opposes the applied electric field (E).

9: Dielectrics in Capacitors

• Effect of Dielectrics:

  • When a dielectric material (like plastic) is placed within an isolated charged capacitor, dipoles align, creating an opposing electric field (E′) to the capacitor’s original field (E).

  • This alignment reduces the effective electric field, resulting in a decrease in voltage (V) across the plates while keeping distance (d) constant:
    $V = E d$

  • Since $Q = C V$ remains constant, a decrease in V necessitates an increase in capacitance (C) by the factor of the dielectric constant (κ):
    $C = κ C<em>0 = \frac{κ ε</em>0 A}{d}$ where C₀ is capacitance without dielectric.

10: Energy Density for Electric Field

• Energy Density Formulation:

  • Stored energy can also be expressed as:
    $U = \frac{1}{2} C V^2$, combined with $V = Ed$ gives:
    $U = \frac{1}{2} C (E d)^2 = \frac{1}{2} C E^2 d^2$

  • Incorporating capacitance formula gives:
    $U = \frac{1}{2} κ ε_0 E^2 d A$

  • By substituting volume representation we find:
    $U = \frac{1}{2} κ ε_0 E^2 ext{(Volume)}$

  • Further divided by volume yields energy density:
    $u = \frac{U}{Vol} = \frac{1}{2} κ ε_0 E^2$. This density applies universally to electric fields, implying that an electric field in space possesses associated energy, defined per volume.

• Units of Energy Density:

  • The unit of energy density is given as:
    $[u] = \frac{[U]}{[Vol]} = \frac{J}{m^3}$