Capacitors & Parallel Plate Capacitance

What is a Capacitor?

Think of a capacitor as a 'bucket' for electric charge. While a battery provides a steady flow of energy, a capacitor is designed to store charge and release it almost instantly. This is why capacitors are essential in camera flashes and power supply filters.

A capacitor is a system of two conductors separated by an insulator (dielectric). Its primary function is to store electric charge and electrical potential energy.

Technically, a capacitor is a system of two conductors separated by an insulator (dielectric). The two conductors usually have charges $Q$ and $-Q$, with a potential difference $V$ between them.

Capacitance ($C$)

Experiments show that the charge $Q$ stored is directly proportional to the potential difference $V$ applied across the conductors: $$Q \propto V \implies Q = CV$$

Here, $C$ is the Capacitance of the capacitor. It represents the ability of the system to store charge per unit volt.

Capacitance tells us how much charge a conductor or capacitor can store per unit potential difference.

• SI Unit: Farad (F). $1 \text{ F} = 1 \text{ Coulomb/Volt}$.
• Practical Units: Since 1 Farad is a very large unit, we use microfarads ($1 \mu\text{F} = 10^{-6}$ F), nanofarads ($1 \text{nF} = 10^{-9}$ F), or picofarads ($1 \text{pF} = 10^{-12}$ F).
• Factors affecting $C$: Capacitance depends ONLY on the geometry (shape, size, separation) of the conductors and the nature of the medium between them. It does not depend on $Q$ or $V$ themselves.

The Parallel Plate Capacitor

This is the most common type of capacitor, consisting of two large plane parallel conducting plates of area $A$, separated by a small distance $d$.

1. Electric Field between Plates: Let the surface charge density on the plates be $\sigma = Q/A$. Using Gauss's Law, the electric field in the region between the plates (away from the edges) is: $$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$$ Outside the plates, the fields from the two plates cancel each other out, making the net field zero.

2. Potential Difference ($V$): For a uniform electric field, the potential difference between the plates is the product of the field and the distance: $$V = E \cdot d$$ Substituting the expression for $E$: $$V = \left( \frac{Q}{A\epsilon_0} \right) d$$

3. Capacitance ($C$): Using the definition $C = Q/V$: $$C = \frac{Q}{(Qd / A\epsilon_0)}$$ $$C = \frac{\epsilon_0 A}{d}$$

Conclusion: The capacitance of a parallel plate capacitor depends only on its geometry (Area $A$ and separation $d$) and the medium between the plates. It is independent of the charge $Q$ or potential $V$ applied.

Effect of Dielectric on Capacitance

When a dielectric slab of dielectric constant $K$ is completely filled in the space between the plates:

1. The electric field is reduced to $E' = E_0/K$.
2. The potential difference is reduced to $V' = E' \cdot d = (E_0 d) / K = V_0 / K$.
3. The new capacitance $C'$ becomes: $$C' = \frac{Q}{V'} = \frac{Q}{(V_0 / K)} = K \left( \frac{Q}{V_0} \right)$$ $$C' = K C_0$$

Key Insight: Inserting a dielectric increases the capacitance by a factor of $K$. This allows the capacitor to store more charge at the same potential difference.

🧠 Memory Capsule

• The Bucket Analogy: $Q = CV$. $C$ is the size of the bucket.
• Geometry Rule: For parallel plates, $C = \frac{\epsilon_0 A}{d}$. Area $\uparrow \implies C \uparrow$, Distance $\downarrow \implies C \uparrow$.
• Dielectric Power: $C$ increases $K$ times when a dielectric is added.
• Units: $1 \mu F = 10^{-6}$ F, $1 pF = 10^{-12}$ F.
• Independence: $C$ is constant for a given capacitor; changing $V$ just changes the amount of charge $Q$ it holds, not the capacitance $C$ itself.

Example 1: Basic Definition

A capacitor is connected to a 12 V battery and stores 24 $\mu C$ of charge. Find its capacitance.

Solution:

1. Given: $Q = 24 \times 10^{-6}$ C, $V = 12$ V.
2. Formula: $$C = \frac{Q}{V}$$
3. Calculation: $$C = \frac{24 \times 10^{-6}}{12} = 2 \times 10^{-6}\text{ F}$$
4. Final Answer: $$C = 2\,\mu\text{F}$$

Example 2: Basic Parallel Plate Calculation

A parallel plate capacitor has plate area $6 \times 10^{-3} \text{ m}^2$ and the distance between the plates is 3 mm. Calculate its capacitance.

Solution:

1. Given: $A = 6 \times 10^{-3} \text{ m}^2$, $d = 3 \times 10^{-3} \text{ m}$.
2. Formula: $$C = \frac{\epsilon_0 A}{d}$$
3. Substitute: $$C = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}$$
4. Calculation: $$C = 8.854 \times 10^{-12} \times 2 = 17.708 \times 10^{-12}\text{ F}$$
5. Final Answer: $$C \approx 17.7\text{ pF}$$

Example 3: Charge for a Given Potential

How much charge is stored in a 10 pF capacitor when connected to a 50 V supply?

Solution:

1. Given: $C = 10 \times 10^{-12}$ F, $V = 50$ V.
2. Formula: $$Q = CV$$
3. Calculation: $$Q = 10 \times 10^{-12} \times 50 = 500 \times 10^{-12}\text{ C}$$
4. Final Answer: $$Q = 500\text{ pC} = 0.5\text{ nC}$$

Example 4: Inserting a Dielectric with Battery Connected

If a mica sheet ($K = 6$) is inserted between the plates of a capacitor of capacitance 17.7 pF while the voltage remains constant, find the new capacitance and charge when the applied voltage is 100 V.

Solution:

1. New capacitance: $$C' = KC = 6 \times 17.7\text{ pF} = 106.2\text{ pF}$$
2. Voltage remains constant: $$V = 100\text{ V}$$
3. New charge: $$Q' = C'V = 106.2 \times 10^{-12} \times 100 = 10.62 \times 10^{-9}\text{ C}$$
4. Final Answer: $$C' = 106.2\text{ pF}, \quad Q' = 10.62\text{ nC}$$

Example 5: Isolated Capacitor and Dielectric

A capacitor is charged to 100 V and then disconnected from the battery. A dielectric of constant $K = 5$ is inserted. What is the new potential difference?

Solution:

1. Battery disconnected: charge remains constant.
2. New capacitance: $$C' = 5C$$
3. Using $$V = \frac{Q}{C}$$, the new potential becomes: $$V' = \frac{Q}{C'} = \frac{Q}{5C} = \frac{V}{5}$$
4. Calculation: $$V' = \frac{100}{5} = 20\text{ V}$$
5. Final Answer: $$V' = 20\text{ V}$$

Example 6: Ratio Analysis

If the area of the plates of a parallel plate capacitor is doubled and the separation is halved, what happens to the capacitance?

Solution:

1. Formula: $$C = \frac{\epsilon_0 A}{d}$$
2. New values: $$A' = 2A, \quad d' = \frac{d}{2}$$
3. New capacitance: $$C' = \frac{\epsilon_0 (2A)}{d/2} = 4\frac{\epsilon_0 A}{d}$$
4. Final Answer: $$C' = 4C$$

Example 7: Electric Field Intensity

A capacitor is charged to 200 V. If the plate separation is 2 mm, find the electric field between the plates.

Solution:

1. Given: $$V = 200\text{ V}, \quad d = 2 \times 10^{-3}\text{ m}$$
2. Formula: $$E = \frac{V}{d}$$
3. Calculation: $$E = \frac{200}{2 \times 10^{-3}} = 10^5\text{ V/m}$$
4. Final Answer: $$E = 1 \times 10^5\text{ V/m}$$

Example 8: Force Between Plates (Advanced)

A parallel plate capacitor has charge $Q$ and plate area $A$. Find the force of attraction between the plates.

Solution:

1. Field due to one plate: $$E_{\text{one}} = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2A\epsilon_0}$$
2. Force on the other plate: $$F = Q E_{\text{one}}$$
3. Substitute: $$F = Q \cdot \frac{Q}{2A\epsilon_0} = \frac{Q^2}{2A\epsilon_0}$$
4. Final Answer: $$F = \frac{Q^2}{2A\epsilon_0}$$

Example 9: Capacitance of an Isolated Sphere (Advanced)

Calculate the capacitance of an isolated spherical conductor of radius $R$.

Solution:

1. Potential of sphere: $$V = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$
2. Capacitance: $$C = \frac{Q}{V} = 4\pi\epsilon_0 R$$
3. Final Answer: $$C = 4\pi\epsilon_0 R$$

Example 10: Earth’s Capacitance (Advanced)

Estimate the capacitance of the Earth, taking its radius as $6400$ km.

Solution:

1. Formula: $$C = 4\pi\epsilon_0 R$$
2. Substitute: $$R = 6.4 \times 10^6\text{ m}, \quad 4\pi\epsilon_0 = \frac{1}{9 \times 10^9}$$
3. Calculation: $$C = \frac{6.4 \times 10^6}{9 \times 10^9} = 0.711 \times 10^{-3}\text{ F}$$
4. Final Answer: $$C \approx 711\,\mu\text{F}$$

Example 11: Effect of Increasing Plate Separation with Battery Connected

A parallel plate air capacitor is connected to a battery. If the distance between plates is increased, what happens to the charge on the plates?

Solution:

1. Battery connected: potential difference remains constant.

2. Capacitance formula: $C = \frac{\epsilon_0 A}{d}$

3. If $d$ increases, then $C$ decreases.

4. Since $Q = CV$ and $V$ is constant, $Q$ decreases.

5. Final Answer: The charge on the plates decreases.