Electrostatics — JEE Mains & Advanced
Coulomb's Law
Force between two point charges:
F = kq₁q₂/r²
• k = 1/4πε₀ = 9 × 10⁹ N·m²/C²
• ε₀ = 8.85 × 10⁻¹² C²/N·m² (permittivity of free space)
• Force is along line joining charges
• Like charges repel, unlike attract
Electric Field
Force per unit positive test charge
E = F/q₀ = kq/r²
Unit: N/C or V/m
Due to point charge: E = kq/r² (radially outward for +ve)
Due to infinite line charge: E = λ/2πε₀r
Due to infinite plane sheet: E = σ/2ε₀
Electric Field Lines
• Start from positive, end at negative
• Never intersect
• Perpendicular to conductor surface
• Closer lines = stronger field
Gauss's Law ← JEE Favourite
φ = Q_enclosed/ε₀
φ = ∮ E·dA (total flux through closed surface)
Applications:
1. Spherical shell: E = kQ/r² (outside), E = 0 (inside)
2. Solid sphere: E = kQr/R³ (inside, r<R), E = kQ/r² (outside)
3. Infinite line: E = λ/2πε₀r
4. Infinite plane: E = σ/2ε₀
Electric Potential
Work done per unit charge to bring test charge from infinity to that point
V = kq/r (for point charge)
• Scalar quantity
• Unit: Volt (V) = J/C
• E = -dV/dr (field = negative gradient of potential)
Potential at center of charged ring: V = kQ/√(R²+0) = kQ/R
Potential Energy
U = kq₁q₂/r (system of two charges)
For assembling n charges: U = k × Σ(qᵢqⱼ/rᵢⱼ) for all pairs
Capacitors
Capacitance: C = Q/V
Unit: Farad (F)
Parallel plate capacitor: C = ε₀A/d
With dielectric: C = Kε₀A/d (K = dielectric constant)
Combinations
Series: 1/C = 1/C₁ + 1/C₂ + 1/C₃
Parallel: C = C₁ + C₂ + C₃
Energy stored: U = ½CV² = Q²/2C = ½QV
Conductor in Electric Field
• E = 0 inside conductor
• All charge on surface
• E perpendicular to surface
• Potential same throughout conductor
JEE Important Points
1. E inside hollow charged sphere = 0 (by Gauss's law)
2. Electric field is discontinuous at surface of charged conductor (σ/ε₀)
3. Potential is continuous everywhere
4. Work done moving charge on equipotential surface = 0
5. Electric field always perpendicular to equipotential surfaces
Formula Sheet
``
F = kq₁q₂/r² [Coulomb's law]
E = F/q₀ = kq/r² [Electric field]
V = kq/r [Potential]
U = kq₁q₂/r [Potential energy]
C = ε₀A/d [Capacitance parallel plate]
U = ½CV² [Energy stored]
Gauss: φ = Q_enc/ε₀
``
PYQ Solutions Approach
Q: Find electric field at point P due to uniformly charged ring at distance x from center.
Solution: E = kQx/(R²+x²)^(3/2) [Component along axis adds, perpendicular cancels]
Q: Three capacitors 2μF, 3μF, 6μF in series. Find equivalent.
1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1 → C = 1μF