Electrostatics Class 11 Notes
Coulomb's law, electric field, potential, capacitance, dielectrics, and Gauss's law.
Electrostatics — Detailed Notes
Electrostatics is an important chapter in Physics and is frequently tested in both conceptual and application-based questions. Students should first understand the core definition, then connect the topic with real-life observations and exam patterns.
Electrostatics studies electric charges at rest, the forces between them (Coulomb's law), and the associated electric fields and potentials. In school and entrance exams, questions usually check your conceptual clarity, step-wise logic, and ability to avoid common mistakes.
To prepare effectively, break Electrostatics into smaller sub-parts: definition, laws/rules, examples, formulas, and revision questions. After theory, solve short questions, then move to mixed-level numericals or application prompts.
A smart revision strategy is to maintain a one-page summary for Electrostatics. Include important terms, two solved examples, and last-minute checkpoints before exams.
Key Exam Points
- Coulomb's law: F = kq₁q₂/r² where k = 9×10⁹ N·m²/C². Force is along line joining charges.
- Electric field E = F/q₀ = kQ/r² (N/C or V/m); direction away from +, towards -.
- Electric potential V = kQ/r (scalar); relationship: E = -dV/dr.
- Gauss's law: ∮E·dA = Q_enc/ε₀ — total electric flux through closed surface equals enclosed charge/ε₀.
- Capacitance C = Q/V; parallel plate: C = ε₀A/d; energy stored: U = ½CV² = Q²/2C.
- Dielectric constant K reduces E and increases C: C = Kε₀A/d.
Important Formula / Rule
F = kq₁q₂/r² | E = kQ/r² | V = kQ/r | C = ε₀A/d | U = ½CV² | k = 1/(4πε₀) = 9×10⁹
What You Will Learn in Electrostatics
Electrostatics studies electric charges at rest, the forces between them (Coulomb's law), and the associated electric fields and potentials.
- Coulomb's law: F = kq₁q₂/r² where k = 9×10⁹ N·m²/C². Force is along line joining charges.
- Electric field E = F/q₀ = kQ/r² (N/C or V/m); direction away from +, towards -.
- Electric potential V = kQ/r (scalar); relationship: E = -dV/dr.
- Gauss's law: ∮E·dA = Q_enc/ε₀ — total electric flux through closed surface equals enclosed charge/ε₀.
- Capacitance C = Q/V; parallel plate: C = ε₀A/d; energy stored: U = ½CV² = Q²/2C.
- Dielectric constant K reduces E and increases C: C = Kε₀A/d.
Key Formulas
F = kq₁q₂/r²E = kQ/r²V = kQ/rC = ε₀A/dU = ½CV²k = 1/(4πε₀) = 9×10⁹
Example
Two charges of +2μC and -2μC separated by 0.3 m attract each other with F = 9×10⁹×4×10⁻¹²/0.09 = 0.4 N.
Solved Numerical Example
Two point charges q₁=+4μC and q₂=+9μC are 0.3 m apart. Find where electric field is zero. Let distance from q₁ be x. E₁=E₂: k×4/(x²) = k×9/(0.3-x)². 4(0.3-x)² = 9x² → 2(0.3-x) = 3x → 0.6-2x=3x → x=0.12 m from q₁.
Expected Exam Questions — Electrostatics
Q1.State Gauss's Law and give one application.
Q2.A 4μF capacitor is connected to 100V. Find charge and energy stored.
Q3.What is the effect of inserting a dielectric in a capacitor? Explain.
🔘 MCQ Practice — Electrostatics
MCQ 1.The SI unit of electric field intensity is:
✓ Correct Answer: N/C or V/m
MCQ 2.Equipotential surfaces are always:
✓ Correct Answer: Perpendicular to electric field
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