Electrostatics – Chapter 2 (IIT–JEE)
Page 2: Parallel Plate Capacitor
1️⃣ What Is a Parallel Plate Capacitor?
A parallel plate capacitor consists of:
- Two large conducting plates
- Each plate has area A
- Separation between plates = d
- Air or vacuum between plates
✔ Plates are assumed very large compared to separation ✔ Edge effects are neglected (JEE assumption)
2️⃣ Charge Distribution
When the capacitor is charged:
- +Q appears on one plate
- −Q appears on the other plate
Charge resides only on the inner surfaces of plates.
⚠️ JEE Point: No charge exists inside the conductor.
3️⃣ Electric Field Between Plates
Electric field due to a single charged plate:
E = σ / (2ε₀)
Where σ = Q/A is surface charge density.
For two oppositely charged plates, fields add up:
E = σ / ε₀
4️⃣ Potential Difference Between Plates
Potential difference is given by:
V = E × d
Substituting value of E:
V = (σ d) / ε₀
5️⃣ Derivation of Capacitance
Capacitance is defined as:
C = Q / V
Substituting Q = σA and V = (σd)/ε₀:
C = ε₀ A / d
6️⃣ Important Observations (JEE Focus)
- Capacitance ∝ Area (A)
- Capacitance ∝ 1 / Separation (d)
- Independent of charge Q
- Independent of voltage V
⚠️ JEE Trap: Increasing charge increases voltage, not capacitance.
7️⃣ Validity Conditions
- d ≪ √A
- Plates are parallel
- No dielectric (vacuum/air)
🔑 Page-2 Summary
✔ Parallel plate capacitor is the base model ✔ Capacitance = ε₀A / d ✔ Geometry decides capacitance ✔ Core formula for all future problems
📌 Master this page → 70% capacitor problems become easy.
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