Capacitors

Energy density derivation:

1. Energy stored in capacitor: $U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{\varepsilon_0 A}{d}V^2$
2. Volume between plates: $Volume = Ad$
3. Energy density: $u = \frac{U}{Volume} = \frac{1}{2}\varepsilon_0\frac{V^2}{d^2}$
4. Since $E = \frac{V}{d}$, energy density: $u = \frac{1}{2}\varepsilon_0 E^2$

Physical significance:

• Energy is stored in the electric field itself, not on the plates
• Energy density is proportional to the square of field strength
• This formula applies to any electric field, not just in capacitors
• The concept shows that empty space containing an electric field contains energy

Derivation:

1. Electric field due to positive plate: $E_+ = \frac{Q}{2A\varepsilon_0}$
2. Force on negative plate: $F = -QE_+ = -Q \cdot \frac{Q}{2A\varepsilon_0} = -\frac{Q^2}{2A\varepsilon_0}$
3. Magnitude of force: $F = \frac{Q^2}{2A\varepsilon_0}$

Physical implications:

• The force is attractive (opposite plates attract)
• Force increases with the square of charge ($Q^2$)
• Force decreases with increasing plate area (A)
• The charged plates always pull toward each other
• This attraction explains why work must be done to separate capacitor plates

In a capacitor:

• The positive plate carries charge +Q
• The negative plate carries charge -Q
• The total charge of the entire capacitor system is zero (+Q + (-Q) = 0)

Key points:

• "Charge on a capacitor" refers specifically to the magnitude |Q| on either plate
• The charges are equal in magnitude but opposite in sign
• When connected to a voltage source, electrons flow from one plate to the other
• The charge separation creates the potential difference between plates
• The plates attract each other due to their opposite charges

This maintains electrical neutrality while allowing charge storage through separation.

The capacitance of a cylindrical capacitor is:

$C = \frac{2\pi\epsilon_0 l}{\ln(R_2/R_1)}$

Where:

• ε₀ is the permittivity of free space
• l is the length of the cylinders
• R₁ is the radius of the inner cylinder
• R₂ is the radius of the outer cylinder
• ln(R₂/R₁) is the natural logarithm of the ratio of radii

This formula applies when the cylinders are long compared to their separation distance.

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.

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.

Corona discharge:

• Electrical discharge that occurs around pointed regions of a charged conductor
• Air becomes ionized and partially conductive near these regions
• Often produces a visible glow around pointed ends
• Charge can "leak" from the conductor into the air

Factors influencing corona discharge:

1. Radius of curvature: More likely at portions with smaller radius (pointed ends)
2. Charge density relationship: $\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1}$ (smaller radius → higher charge density)
3. Electric field at surface: $E = \frac{\sigma}{\varepsilon_0}$ (higher charge density → stronger field)
4. Proximity to dielectric breakdown: Occurs when field exceeds dielectric strength of air (≈3 kV/mm)

Applications: Lightning rods utilize this principle to safely discharge atmospheric electricity through pointed conductors.

When a dielectric material is inserted between capacitor plates:

1. Three distinct electric fields exist:

   • E₀: The original field due to the free charges (+Q, -Q) on the capacitor plates
   • Ep: The polarization field in the opposite direction due to induced charges
   • E: The resultant field (reduced in magnitude)

2. The relationship between these fields:

   • E = E₀/K (where K is the dielectric constant)
   • E is in the same direction as E₀ but with reduced magnitude

3. The induced charges (±Qp) on the dielectric surfaces partially neutralize the effect of the free charges, reducing the net electric field within the dielectric.

This reduction in electric field is why the potential difference decreases while the capacitance increases by a factor of K.

The maximum voltage is limited by dielectric breakdown of the surrounding medium (typically air):

• When the electric field at the sphere's surface exceeds the dielectric strength of air (~3×10^6 V/m), ionization occurs
• For a conducting sphere: V = ER, where V is potential, E is electric field, and R is radius
• For a 1-meter radius sphere in air, maximum voltage ≈ 3×10^6 volts

Methods to increase maximum voltage:

1. Increase sphere radius (V is directly proportional to R)
2. Place the generator in a highly evacuated chamber (vacuum has higher dielectric strength)
3. Use insulating gas with higher dielectric strength than air
4. Improve insulation of the support structure
5. Round all edges to minimize local field concentrations

The ultimate theoretical limit relates to the relationship between electric field, potential, and geometry (V = ER for a sphere).

The potential at point D is 16V, calculated as follows:

1. First, determine the voltage across the series branch (which equals the battery voltage):

   • V_series = 24V

2. In series capacitors, voltage divides inversely proportionally to capacitance:

   • V₁/V₂ = C₂/C₁

3. Voltage across the 6 μF capacitor:

   • V_6μF = (96 μC)/(6 μF) = 16V

4. By voltage division in the series branch:

   • V_12μF = 24V - 16V = 8V

5. Since point B is at zero potential, and the 6 μF capacitor has 16V across it, point D must be at:

   • V_D = V_B + V_6μF = 0V + 16V = 16V

This demonstrates how voltage divides across series capacitors and how potential can be determined at any point in a capacitive circuit.