Capacitors

Derivation:

1. Electric field between plates: $E = \frac{Q}{\varepsilon_0 A}$
2. Potential difference: $V = Ed = \frac{Qd}{\varepsilon_0 A}$
3. Capacitance: $C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}$

Physical significance:

• $\varepsilon_0$: permittivity of vacuum (fundamental constant)
• A: area of plates (↑A → ↑C, more surface for charge)
• d: separation between plates (↑d → ↓C, weaker interaction)

The formula shows capacitance increases with plate area and decreases with plate separation.

Step 1: Find equivalent capacitance of parallel section $C_{parallel} = C_2 + C_3$

Step 2: Find total equivalent capacitance (series combination) $\frac{1}{C_{equivalent}} = \frac{1}{C_1} + \frac{1}{C_{parallel}}$

$\frac{1}{C_{equivalent}} = \frac{1}{C_1} + \frac{1}{C_2 + C_3}$

$C_{equivalent} = \frac{C_1(C_2 + C_3)}{C_1 + C_2 + C_3}$

This is a simplified formula for this specific configuration.

To calculate charge distributions in this mixed network:

1. Establish charge variables:

   • Let Q = charge supplied to positive terminal
   • Let Q₁ = charge on middle capacitor (C₂)
   • Then (Q-Q₁) = charge on rightmost capacitor (C₃)

2. Apply potential analysis:

   • Set potential at P as V and at N as 0
   • Let intermediate potential at junction be V₁

3. Apply capacitor equations:

   • For C₁: Q = C₁(V-V₁)
   • For C₂: Q₁ = C₂V₁
   • For C₃: (Q-Q₁) = C₃V₁

4. Solve the system:

   • From C₂ and C₃ equations: Q = (C₂+C₃)V₁
   • From C₁ equation: V-V₁ = Q/C₁
   • Combining: Q/C₁ + Q/(C₂+C₃) = V
   • Therefore: Q = V·[C₁(C₂+C₃)/(C₁+C₂+C₃)]

5. Individual charges can then be determined by substitution.

The induced charge magnitude is:

$Q_p = Q\left(1-\frac{1}{K}\right)$

Derivation:

• The resultant field in the dielectric: $E = \frac{E_0}{K} = \frac{Q}{\varepsilon_0 AK}$
• The field from free charge: $E_0 = \frac{Q}{A\varepsilon_0}$
• The field from induced charge: $E_p = \frac{Q_p}{A\varepsilon_0}$
• The resultant field: $E = E_0 - E_p = \frac{Q-Q_p}{A\varepsilon_0}$

Equating the two expressions for $E$: $\frac{Q-Q_p}{A\varepsilon_0} = \frac{Q}{\varepsilon_0 AK}$

Solving for $Q_p$ yields the result.

For a capacitor partially filled with dielectric:

$C = \frac{K\varepsilon_0 A}{Kd - x(K-1)}$

Derivation:

• The system is equivalent to two capacitors in series:
  • $C_1 = \frac{K\varepsilon_0 A}{x}$ (dielectric region)
  • $C_2 = \frac{\varepsilon_0 A}{d-x}$ (air gap region)
• For series capacitors: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$
• $\frac{1}{C} = \frac{x}{K\varepsilon_0 A} + \frac{d-x}{\varepsilon_0 A} = \frac{Kd-x(K-1)}{K\varepsilon_0 A}$
• Therefore: $C = \frac{K\varepsilon_0 A}{Kd-x(K-1)}$

As $x$ approaches $d$, this approaches $\frac{K\varepsilon_0 A}{d}$ (fully filled capacitor). As $x$ approaches 0, this approaches $\frac{\varepsilon_0 A}{d}$ (empty capacitor).

Effects of inserting a dielectric with constant K:

1. Capacitance change:

   • C = KC₀ (where C₀ is the original capacitance)
   • C₀ = ε₀A/d
   • C = Kε₀A/d

2. Relationship between induced charge (Qp) and free charge (Q):

   • Qp = Q(1-1/K)

3. Physical meaning:

   • When K=1 (vacuum): Qp=0 (no induced charge)
   • As K increases: Qp approaches Q
   • The induced charge partially neutralizes the electric field
   • This reduces the potential difference, allowing more charge to be stored at the same voltage

This explains why materials with high dielectric constants (like ceramics) are valuable in capacitor manufacturing - they dramatically increase charge storage capacity.

When a point charge q is placed in an infinite dielectric medium:

1. The charge induces polarization in the surrounding dielectric material
2. This polarization creates bound charges in the medium
3. The bound charge on the cavity surface equals: $q_p = -q\left(1-\frac{1}{K}\right)$
4. The effective charge becomes: $q_{effective} = \frac{q}{K}$

The electric field is reduced because:

• The polarized molecules in the dielectric align their dipoles to oppose the original field
• This creates an internal field (E_p) that opposes the applied field (E_0)
• The resultant field is reduced by a factor of K: $E = \frac{E_0}{K}$

Example: A 1 nC charge in a medium with K=4 will produce the same electric field as a 0.25 nC charge would in vacuum.

• Electric field strength just outside a conductor is E = σ/ε₀ (directly proportional to charge density)
• Near pointed regions (small radius of curvature):
  • Electric field is intensified
  • May exceed dielectric breakdown threshold of surrounding medium (typically air)
  • Results in corona discharge

Applications:

• Lightning rods: Pointed tips create strong fields that ionize air and provide path for discharge
• Van de Graaff generators: Use sharp needle points for charge transfer
• Electrostatic precipitators: Sharp electrodes create strong fields to charge and collect particles
• Limitation in high-voltage equipment: Rounded edges prevent unwanted discharge

Key Principles for Equivalent Capacitance

1. Parallel Capacitors:

   • Capacitances add directly: $C_{eq} = C_1 + C_2 + ... + C_n$
   • Physical interpretation: Total plate area increases

2. Series Capacitors:

   • Reciprocals add: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}$
   • Alternative form: $C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2}$ (for two capacitors)
   • Physical interpretation: Effective plate separation increases

Step-by-Step Problem-Solving Approach:

1. Identify Sub-Circuits:

   • Locate pure series or parallel arrangements
   • Work from innermost to outermost combinations

2. Reduce Systematically:

   • Replace each identified arrangement with its equivalent capacitance
   • Continue until the entire circuit is reduced to a single equivalent capacitor

Example Applications:

Example 1: Two capacitors (10 μF, 20 μF) in parallel, combined in series with a 30 μF capacitor:

• Parallel combination: 10 μF + 20 μF = 30 μF
• Series combination: $\frac{1}{C_{eq}} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30}$
• Final result: $C_{eq} = 15$ μF

Example 2: 12 μF and 6 μF in series (one branch), in parallel with 2 μF:

• Series branch: $\frac{1}{C_{series}} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} = \frac{1}{4}$ → 4 μF
• Parallel combination: 4 μF + 2 μF = 6 μF

Note: Always verify your answer makes physical sense - the equivalent capacitance of series capacitors is always less than the smallest individual capacitor, while parallel combinations are always larger than any individual component.

For a point charge $q$ in an infinite homogeneous dielectric medium with dielectric constant $K$:

$E = \frac{q}{4\pi\varepsilon_0 Kr^2}$

Key characteristics:

• Field magnitude is reduced by factor $K$ compared to vacuum
• Field remains radially directed (away from positive charge)
• Field lines maintain their radial pattern but with reduced density

Physical mechanism behind the reduction:

1. The dielectric medium becomes polarized in response to the charge
2. Polarized molecules align their dipoles to partially counteract the original field
3. This creates an induced charge $-q(1-\frac{1}{K})$ on the cavity surface
4. The charge effectively becomes $\frac{q}{K}$ due to this polarization effect

Example: In water ($K \approx 80$), the electric field from a point charge is 80 times weaker than in vacuum at the same distance.

Note: This reduction only applies in infinite dielectric media; boundary conditions modify the field pattern in finite dielectrics.