Electrostatics
Electric Charges and Fields · Electrostatic Potential and Capacitance
Build the field–potential–energy chain, apply Gauss’s law only when symmetry permits it, and connect capacitance to geometry and dielectric response.
Electrostatics · corrected and expanded
Official coverage, visual understanding, formula recall, worked problems, common-error repairs and all explained MCQs.
Official syllabus coverage
16 marks with Unit IINothing in the current theory list is hidden or silently omitted.
- ✓Charge, conservation, Coulomb’s law, multiple charges, superposition and continuous charge distributions
- ✓Electric field, field lines, point-charge and dipole fields, and torque on a dipole
- ✓Electric flux and Gauss’s theorem for a long wire, infinite plane sheet and thin spherical shell
- ✓Potential and potential difference due to a point charge, dipole and system of charges
- ✓Equipotential surfaces and potential energy of two charges and an electric dipole
- ✓Conductors, insulators, free and bound charges, dielectrics and polarisation
- ✓Capacitance, series/parallel combinations, parallel plates with/without dielectric, and stored energy
What you should be able to do
- Use Coulomb’s law and superposition in scalar and vector form.
- Choose a Gaussian surface from symmetry and calculate flux.
- Solve potential, potential-energy and capacitor-combination problems.
Core understanding
- Charge is conserved and quantised: q = ne.
- Electric field is force per unit positive test charge; electric potential is work per unit charge.
- In electrostatic equilibrium the field inside a conductor is zero and excess charge lies on its surface.
- A dielectric increases capacitance by reducing the effective field for a given free charge.
- For a continuous distribution, replace the charge sum by an integral: dq = λdl, σdA or ρdV according to geometry.
- An electric dipole has moment p = q(2a), directed from negative to positive charge; in a uniform field τ = p × E.
- Gauss-law results to recall: E = λ/(2πε₀r) for a long line, E = σ/(2ε₀) for an infinite sheet, and a charged spherical shell has E = 0 inside and kQ/r² outside.
- On an equipotential surface ΔV = 0, so electrostatic work along the surface is zero and E is normal to it.
- Series capacitors carry equal charge; parallel capacitors share equal potential difference.
Formula system
F = (1/4πε₀) q₁q₂/r²E = F/q₀; Φ = ∮E·dA∮E·dA = q_enclosed/ε₀V = (1/4πε₀) q/r; U = qVC = Q/V; C_parallel plate = ε₀K A/dU_capacitor = ½CV² = Q²/2CWorked problem studio
A 4 μF capacitor is charged to 12 V. Find its charge and stored energy.
- Use Q = CV.
- Use U = ½CV² with capacitance in farads.
AnswerQ = 48 μC and U = 2.88 × 10⁻⁴ J.
A dielectric of K = 3 fully enters an isolated charged capacitor. What happens to C, V and U?
- Isolated means free charge Q stays fixed.
- C′ = KC, then V′ = Q/C′ and U′ = Q²/(2C′).
AnswerC triples; V and stored energy each become one-third.
Corrections and exam traps
CorrectedKeep r² in the force magnitude and use the unit vector separately; do not mix displacement and distance.
CorrectedPotential is a scalar but electric field is a vector.
CorrectedA Gaussian surface is a calculation surface, not a physical boundary.