Chapter 3 · Class 11-12 & JEE/NEET

Capacitors

41 topics in 13 parts

Full lessonClass 12 Physics · Electrostatic Potential and Capacitance Practice180 practice questions

Key formulas

Capacitance of Parallel Plate Capacitor
$C_0 = \frac{\epsilon_0 A}{d}, \quad C_{dielectric} = \frac{K \epsilon_0 A}{d} = K C_0$
Series & Parallel Combinations
$\frac{1}{C_{series}} = \sum \frac{1}{C_i}, \quad C_{parallel} = \sum C_i$
Stored Electrostatic Energy
$U = \frac{1}{2} C V^2 = \frac{Q^2}{2C} = \frac{1}{2} Q V, \quad u_E = \frac{1}{2}\epsilon_0 E^2$
Force Between Plates
$F = \frac{Q^2}{2\epsilon_0 A} = \frac{1}{2} Q E$

Parallel-plate capacitance Capacitance Parallel-Plate Capacitance Parallel-Plate Capacitor with Dielectric Capacitance of an Isolated Sphere Spherical Capacitor

Key ideas

Parallel Plate Capacitor with Electric Field Lines & Dielectric Polarization

Exam Worked Example

Problem: A parallel plate capacitor of capacitance 10 μF is charged to 100 V and then disconnected from the battery. A dielectric slab of dielectric constant K = 4 is inserted between the plates. Find the new potential difference and the change in stored energy.

● Step 1: Initial charge Q = C * V = 10 μF * 100 V = 1000 μC. Initial energy U_i = 1/2 C V² = 0.5 * 10e-6 * 10000 = 0.05 Joules.

● Step 2: Since battery is disconnected, charge Q remains constant at 1000 μC. New capacitance C' = K * C = 4 * 10 = 40 μF.

● Step 3: New potential difference V' = Q / C' = 1000 μC / 40 μF = 25 Volts.

● Step 4: New stored energy U_f = Q² / (2 C') = (1000e-6)² / (2 * 40e-6) = 0.0125 Joules. Energy lost as heat = 0.05 - 0.0125 = 0.0375 J.

4. Systematic Comparison & Parameter Classifications

Parameter	Battery Remains CONNECTED (V = Const)	Battery is DISCONNECTED (Q = Const)
Charge (Q)	Increases: Q = K * Q0	Remains Constant: Q = Q0
Potential Diff (V)	Remains Constant: V = V0	Decreases: V = V0 / K
Capacitance (C)	Increases: C = K * C0	Increases: C = K * C0
Electric Field (E)	Remains Constant: E = E0	Decreases: E = E0 / K
Stored Energy (U)	Increases: U = K * U0	Decreases: U = U0 / K

Checkpoint

1. When a dielectric slab (K > 1) is inserted into a capacitor while the battery remains connected, what happens to the electric field between the plates?

All 41 topics in Capacitors

Capacitance
- Concept & Definition
- Isolated Conductor
- System of Conductors
Parallel Plate Capacitor
- Basic Capacitance
- Electric Field & Charge Distribution
- Multi-Plate Capacitors
- Variable Geometry
Capacitance of a Spherical Conductor or Capacitor
- Isolated Spherical Conductor
- Concentric Spherical Capacitor
Cylindrical Capacitor
- Coaxial Cylinders
Dielectrics in Capacitors
- Dielectric Slabs
- Effective Capacitance with Dielectrics
- Effect at Constant Voltage
- Effect at Constant Charge
- Polarisation & Induced Charge
- Dielectric Strength
Combination of Capacitors
- Series Combination
- Parallel Combination
- Series-Parallel Networks
- Infinite Networks
Kirchhoff's Rules for Capacitors
- Steady-State DC Circuits
- Kirchhoff's Loop Rule
- Nodal Analysis
- Wheatstone Bridge
- Extended Wheatstone Bridge
Charging & Discharging Through Resistance
- RC Transient
- Time Constant
- Transient Current & Voltage
LC Oscillations
- Energy Exchange in LC Circuits
Sharing of Charge
- Charge Redistribution
- Charge Flow
- Energy Loss on Reconnection
Energy Stored in a Charged Conductor or Capacitor
- Energy of a Capacitor
- Alternative Method
- Energy Density
- Energy Dissipation as Heat
Force Between the Plates of a Parallel Plate Capacitor
- Force Between Plates
- Work Done & Change in Plate Separation
- Spring-Coupled Capacitors
Force on Dielectric Slab
- Constant-Voltage Insertion
- Constant-Charge Insertion