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.

Spherical capacitor (concentric spheres): $C = \frac{4\pi\varepsilon_0 R_1 R_2}{R_2 - R_1}$

Where:

• $R_1$ = radius of inner sphere
• $R_2$ = radius of outer sphere
• $\varepsilon_0$ = permittivity of vacuum

For an isolated sphere ($R_2 \to \infty$): $C = 4\pi\varepsilon_0 R_1$

Physical interpretation: The capacitance of an isolated sphere depends only on its radius and is directly proportional to it.

If both cylinders are enlarged while maintaining the same ratio R₂/R₁:

• The capacitance will increase proportionally to the scaling factor
• This occurs because the capacitance formula $C = \frac{2\pi\epsilon_0 l}{\ln(R_2/R_1)}$ depends on the logarithm of the ratio, not the absolute sizes

For example, if both R₁ and R₂ are doubled:

• The ratio R₂/R₁ remains the same
• ln(R₂/R₁) remains unchanged
• The effective area increases, increasing capacitance

This proportional scaling differs from a parallel plate capacitor, where capacitance remains unchanged if plate separation and area increase proportionally.

Derivation using charge conservation:

1. Starting conditions:

   • All capacitors have equal charge Q
   • Total potential difference: V = V₁ + V₂ + V₃ + ...

2. For each capacitor:

   • V₁ = Q/C₁
   • V₂ = Q/C₂
   • V₃ = Q/C₃

3. Substitute into total potential:

   • V = Q/C₁ + Q/C₂ + Q/C₃ + ...
   • V = Q(1/C₁ + 1/C₂ + 1/C₃ + ...)

4. By definition, equivalent capacitance C_eq relates to total charge and potential:

   • Q = C_eq × V

5. Therefore:

   • Q/V = C_eq = 1/(1/C₁ + 1/C₂ + 1/C₃ + ...)

6. Or:

   • 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + ...

This formula works because the charge conservation principle requires each capacitor to hold identical charge, while the potential differences distribute according to each capacitor's individual capacitance.

The displacement vector $\vec{D}$ is:

$\vec{D} = \varepsilon_0\vec{E} + \vec{P}$

Where:

• $\vec{E}$ is the electric field
• $\vec{P}$ is the polarization vector (dipole moment per unit volume)
• For homogeneous, isotropic dielectrics: $\vec{D} = \varepsilon_0K\vec{E}$

Alternative form of Gauss's law using $\vec{D}$: $\oint \vec{D} \cdot d\vec{S} = Q_{free}$

Key properties:

• Only counts free charges (not bound/polarization charges)
• Simplifies calculations in dielectrics
• For homogeneous dielectrics: $\oint K\vec{E} \cdot d\vec{S} = \frac{Q_{free}}{\varepsilon_0}$
• In vacuum ($K=1$), reduces to the standard form of Gauss's law

This formulation separates free charges from polarization effects.

Relationship between dielectric constant and polarization:

• Polarization vector: $\vec{P}$ represents dipole moment per unit volume
• For homogeneous, isotropic dielectrics: $\vec{P} = \varepsilon_0(K-1)\vec{E}$
• Field due to polarization: $\vec{E}_p = -\frac{\vec{P}}{\varepsilon_0}$
• Net electric field: $\vec{E} = \vec{E}_0 + \vec{E}_p = \vec{E}_0 - \frac{\vec{P}}{\varepsilon_0}$

Substituting:

• $\vec{E} = \vec{E}_0 - \frac{\varepsilon_0(K-1)\vec{E}}{\varepsilon_0} = \vec{E}_0 - (K-1)\vec{E}$
• This gives: $K\vec{E} = \vec{E}_0$ or $\vec{E} = \frac{\vec{E}_0}{K}$

The dielectric constant determines how much the material polarizes in response to an applied field, which then determines how much the internal field is reduced from the applied field.

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.

• Charge density on non-uniform conductors varies inversely with the radius of curvature of the surface
• Highest charge density occurs at points with smallest radius of curvature (sharp points, edges)
• Mathematical relationship: σ₁/σ₂ = r₂/r₁ (charge densities are inversely proportional to radii)
• This occurs because all points on a conductor must maintain the same electrical potential
• Example: On a teardrop-shaped conductor, the pointed end will have significantly higher charge density than the rounded end

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.