Capacitance & Dielectrics
Practice JEE Main Capacitance & Dielectrics PYQs online. Parallel plate capacitors, energy density u = ½ ε₀ E², dielectric slab insertion & solutions.
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Capacitance & Dielectrics:
- Parallel Plate Capacitor: C = ε₀ A / d
- Series and Parallel Combinations of Capacitors
- Dielectric Slab Insertion (Battery Connected vs Disconnected)
- Stored Electrostatic Energy U = ½ C V² = Q² / (2C)
- Energy Density in Electric Field u = ½ ε₀ E²
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Capacitors questions in under 45 seconds:
Battery Disconnected (Q = const): Inserting dielectric increases C to KC, decreases V to V/K, and decreases energy to U/K.
Battery Connected (V = const): Inserting dielectric increases C to KC, increases charge to KQ, and increases energy to KU.
Attractive Force between plates: F = Q² / (2 ε₀ A).
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: 8C
- Step 1: Original capacitance with air: C = ε₀ A / d.
- Step 2: New separation is d' = d / 2 and dielectric constant is K = 4.
- Step 3: New capacitance is C' = K ε₀ A / d' = 4 ε₀ A / (d/2).
- Step 4: C' = 4 × 2 × (ε₀ A / d) = 8 C.
⚡ Speed Tip: Capacitance formula: C' = K (d / d') C = 4 × (1 / 0.5) C = 8C.
💡 View Step-by-Step Solution & Answer
Correct Answer: 8 V
- Step 1: In series combination, equivalent capacitance is C_eq = (C₁ C₂) / (C₁ + C₂) = (3 × 6) / (3 + 6) = 18 / 9 = 2 μF.
- Step 2: Total charge supplied by battery: Q = C_eq × V = 2 μF × 12 V = 24 μC.
- Step 3: In series, both capacitors carry the same charge Q = 24 μC.
- Step 4: Potential difference across 3 μF capacitor: V₁ = Q / C₁ = 24 μC / 3 μF = 8 V.
⚡ Speed Tip: Voltage divider in series capacitors: V₁ = V × C₂ / (C₁ + C₂) = 12 × 6 / 9 = 8 V.
💡 View Step-by-Step Solution & Answer
Correct Answer: Decrease to U / K
- Step 1: When battery is disconnected, charge Q on the capacitor remains strictly constant.
- Step 2: Stored electrostatic energy can be written as U = Q² / (2C).
- Step 3: After inserting dielectric slab of constant K, capacitance increases to C' = K C.
- Step 4: New energy is U' = Q² / (2 C') = Q² / (2 K C) = U / K. Thus energy decreases by factor K.
⚡ Speed Tip: Battery disconnected => Q = const => U = Q²/(2C) => U' = U/K. (Battery connected => V = const => U' = K U).
❓ Frequently Asked Questions (FAQs)
What is the efficiency of charging a capacitor by a battery?
Exactly 50%! Battery does work W = C V², capacitor stores U = ½ C V², and the remaining ½ C V² is always dissipated as heat.