Question

Derive the expression for the capacitance of a parallel plate capacitor (with and without dielectric slab).

Hint:

Remember: - Capacitance increases with area and dielectric constant. - Capacitance decreases with plate separation. - Dielectric reduces electric field and increases charge storing ability.

Answer 1

Concept: Capacitance is defined as the charge stored per unit potential difference: \[ C = \frac{Q}{V} \] For a parallel plate capacitor:

• Plate area = \( A \)
• Separation between plates = \( d \)
• Medium between plates affects capacitance

The derivation is based on electric field and potential difference between plates.
Part 1: Capacitance of Parallel Plate Capacitor (Without Dielectric)
Step 1: Electric field between plates Surface charge density: \[ \sigma = \frac{Q}{A} \] Electric field due to one plate: \[ E = \frac{\sigma}{2\varepsilon_0} \] Between two oppositely charged plates: \[ E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A\varepsilon_0} \]
Step 2: Potential difference \[ V = Ed = \frac{Qd}{A\varepsilon_0} \]
Step 3: Capacitance \[ C = \frac{Q}{V} = \frac{Q}{\frac{Qd}{A\varepsilon_0}} \] \[ \boxed{C = \frac{\varepsilon_0 A}{d}} \] This is the capacitance in vacuum or air.
Part 2: Capacitance with Dielectric Slab Filling Entire Space Let dielectric constant be \( K \). When dielectric is inserted:

• Electric field reduces by factor \( K \)
• Permittivity becomes \( \varepsilon = K\varepsilon_0 \)

New electric field: \[ E = \frac{E_0}{K} \] Potential difference: \[ V = \frac{Ed}{K} \] Capacitance: \[ C = \frac{Q}{V} = \frac{K\varepsilon_0 A}{d} \] \[ \boxed{C = \frac{K\varepsilon_0 A}{d}} \] Thus, dielectric increases capacitance by factor \( K \).
Part 3: Dielectric Slab Partially Filling the Capacitor Let:

• Thickness of dielectric slab = \( t \)
• Remaining air gap = \( d - t \)

This behaves like two capacitors in series:

• One with dielectric
• One with air

Capacitances of parts: \[ C_1 = \frac{K\varepsilon_0 A}{t}, \quad C_2 = \frac{\varepsilon_0 A}{d - t} \] For series combination: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \] \[ \frac{1}{C} = \frac{t}{K\varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A} \] Taking common factor: \[ \frac{1}{C} = \frac{1}{\varepsilon_0 A} \left( \frac{t}{K} + d - t \right) \] \[ C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \] \[ \boxed{C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}} \]
Special Cases:

• If \( t = 0 \): \( C = \frac{\varepsilon_0 A}{d} \) (air)
• If \( t = d \): \( C = \frac{K\varepsilon_0 A}{d} \) (fully filled)

Final Results:

• Without dielectric: \( C = \dfrac{\varepsilon_0 A}{d} \)
• Fully filled dielectric: \( C = \dfrac{K\varepsilon_0 A}{d} \)
• Partially filled: \( C = \dfrac{\varepsilon_0 A}{d - t + \frac{t}{K}} \)