Electrostatics
Electrostatics in the Tamil Nadu HSE Second Year syllabus deals with charges at rest: Coulomb's law, electric field and potential, the dipole, Gauss's law and capacitors. Numericals dominate, so be fluent with vector directions, unit prefixes and dielectric effects.
Core rules
- Coulomb force in free space: F = (1/4πε₀)(q₁q₂/r²), where 1/4πε₀ = 9×10⁹ N m² C⁻². In a medium of dielectric constant K the force becomes F/K.
- Electric field E = F/q₀ is a vector, pointing away from positive charge and towards negative charge. Field lines never cross and meet conductors at right angles.
- Potential V = W/q is a scalar and E = −dV/dr. Work done in moving a charge along an equipotential surface is zero.
- Dipole moment p = q(2a) points from −q to +q. In a uniform field, torque τ = pE sin θ and potential energy U = −pE cos θ.
- Gauss's law gives E = σ/ε₀ just outside a charged conductor and E = σ/2ε₀ near an infinite sheet.
- Capacitance C = Q/V; parallel-plate C = ε₀KA/d; energy U = ½CV² = Q²/2C. Capacitors in series add as reciprocals, in parallel directly.
Worked example
Charges +4 μC and −4 μC are 0.20 m apart. Find the field at the midpoint. Each charge is 0.10 m away. For the positive charge, E₁ = 9×10⁹ × 4×10⁻⁶/(0.10)² = 3.6×10⁶ N C⁻¹, directed away from it, i.e. towards the negative charge. The negative charge produces an equal field also pointing towards itself. The two add: E = 7.2×10⁶ N C⁻¹, directed from +4 μC towards −4 μC.
Traps
Mixing μC (10⁻⁶) with cm (10⁻²); forgetting that midpoint fields of opposite charges add rather than cancel; applying E = V/d to a non-uniform field; assuming a dielectric raises capacitance without lowering the field; and treating capacitor combinations like resistor combinations.