Electrostatics for NEET

Electrostatics covers two NCERT chapters: electric charges and fields, and electrostatic potential and capacitance. Together they are among the heavier units in Class 12 physics, and they set up everything that follows in current electricity and magnetism.

NEET questions here come in three shapes: a superposition problem (add fields or forces from several charges), a Gauss's law shortcut (field of a sheet, a wire or a shell), or a capacitor circuit (combinations, energy, what happens when a dielectric goes in). Get the capacitor "battery connected or disconnected" table right and you will save a lot of marks.

Charge and Coulomb's law

Electric charge has three basic properties:

• It is quantised: q = ne, where e = 1.6 × 10⁻¹⁹ C.
• It is conserved: it can be transferred but not created or destroyed.
• It is additive: the total charge is the algebraic sum.

Coulomb's law: F = kq₁q₂/r², with k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².

In a medium of dielectric constant K, the force falls to F/K. For several charges, the net force is the vector sum of the individual forces (superposition).

Electric field and dipoles

The field of a point charge is E = kq/r², pointing away from a positive charge.

Field lines start on positive charges and end on negative ones, never cross, and are perpendicular to a conductor's surface. Where they crowd together, the field is stronger.

An electric dipole is a pair of charges +q and −q separated by 2a. Its dipole moment is p = q × 2a, directed from −q to +q.

A dipole's field falls as 1/r³, faster than a single charge's 1/r².

Gauss's law

The electric flux through a closed surface equals the enclosed charge divided by ε₀:

Φ = q_enclosed / ε₀

Charges outside the surface contribute zero net flux. With a symmetric surface, this gives fields in one line:

Potential and potential energy

The potential of a point charge is V = kq/r. Potential is a scalar, so potentials from several charges simply add, with their signs.

• The field and potential are linked by E = −dV/dr. The field points towards decreasing potential.
• Equipotential surfaces are perpendicular to field lines. No work is done moving a charge along one.
• The work done moving a charge q from A to B is q(V_B − V_A), independent of the path.
• Dipole potential: V = kp cos θ / r². It is zero everywhere on the equatorial line.
• A charged shell has potential kQ/R everywhere inside, equal to its surface value, even though the field inside is zero.

The potential energy of two point charges is U = kq₁q₂/r. For a system of several charges, add this for every pair.

Conductors

In electrostatic equilibrium, the field inside a conductor is zero, all excess charge sits on the outer surface, the whole conductor is at one potential, and the field just outside is perpendicular to the surface. Charge concentrates at sharp points, which is why lightning conductors are pointed.

Capacitors

C = Q/V, measured in farads.

• Parallel-plate capacitor: C = ε₀A/d. Filled with a dielectric: C = Kε₀A/d.
• A slab of thickness t and constant K between the plates: C = ε₀A / (d − t + t/K). A metal slab behaves like K → ∞: C = ε₀A/(d − t).
• Series: 1/C = 1/C₁ + 1/C₂ + … (same charge on each).
• Parallel: C = C₁ + C₂ + … (same voltage across each).
• Energy: U = ½CV² = Q²/2C = ½QV. The energy density in the field is ½ε₀E².

The most tested idea is what happens when a dielectric of constant K fills the gap:

Charge sharing. When a charged capacitor is joined to another, charge flows until the voltages match. The common potential is V = (C₁V₁ + C₂V₂)/(C₁ + C₂). Some energy is always lost as heat in the wires, equal to C₁C₂(V₁ − V₂)² / 2(C₁ + C₂).

Worked numericals

Example 1: where the field is zero

Charges of +1 μC and +4 μC are 30 cm apart. Where between them is the net field zero?

• Let the point be x from the 1 μC charge. Then k(1)/x² = k(4)/(0.3 − x)².
• Taking square roots: (0.3 − x)/x = 2, so x = 0.1 m, that is 10 cm from the smaller charge.

For like charges, the zero-field point is between them and nearer the smaller charge.

Example 2: dipole in a field

A dipole has charges ±2 μC separated by 2 cm, in a uniform field of 5 × 10⁴ N C⁻¹.

• p = 2 × 10⁻⁶ × 0.02 = 4 × 10⁻⁸ C m.
• Torque at 30°: τ = pE sin 30° = 4 × 10⁻⁸ × 5 × 10⁴ × 0.5 = 1 × 10⁻³ N m.
• Work to turn it from θ = 0 to 180°: W = U(180°) − U(0°) = pE − (−pE) = 2pE = 4 × 10⁻³ J.

Example 3: capacitors in series

A 2 μF and a 3 μF capacitor are in series across 100 V.

• C = (2 × 3)/(2 + 3) = 1.2 μF.
• Charge on each: Q = 1.2 × 10⁻⁶ × 100 = 120 μC.
• Voltages: 120/2 = 60 V and 120/3 = 40 V. The smaller capacitor takes the larger voltage.
• Energy: ½ × 1.2 × 10⁻⁶ × 100² = 6 × 10⁻³ J.

Example 4: charge sharing and energy loss

A 10 μF capacitor charged to 100 V is connected to an identical uncharged capacitor.

• Common voltage: (10 × 100 + 0)/20 = 50 V.
• Initial energy: ½ × 10⁻⁵ × 100² = 0.05 J.
• Final energy: ½ × (2 × 10⁻⁵) × 50² = 0.025 J.
• Energy lost as heat = 0.025 J, half the original. The formula gives the same: (10 × 10 × 10⁻¹²)(100)² / (2 × 20 × 10⁻⁶) = 0.025 J.

Practice MCQs

1. Two charges repel with force F. Each charge is doubled at the same separation. The new force is: (a) 4F (b) 2F (c) F (d) F/4
2. Three 3 μF capacitors in series have a total capacitance of: (a) 9 μF (b) 3 μF (c) 1 μF (d) 1/3 μF
3. The electric field inside a uniformly charged thin spherical shell is: (a) kQ/R² (b) kQ/r² (c) infinite (d) zero
4. A charge q sits at the centre of a cube. The flux through one face is: (a) q/ε₀ (b) q/(6ε₀) (c) q/(8ε₀) (d) zero
5. The work done in moving 2 μC between two points on the same equipotential surface at 50 V is: (a) 100 μJ (b) 50 μJ (c) 25 μJ (d) zero
6. An air capacitor stays connected to a battery while a dielectric of K = 3 fills it. The energy stored becomes: (a) one-third (b) the same (c) three times (d) nine times
7. Four charges +q sit at the corners of a square of side a. At the centre, E and V are: (a) 0 and 0 (b) 0 and 4√2 kq/a (c) 4√2 kq/a² and 0 (d) both non-zero
8. At the same large distance, the ratio of a dipole's axial field to its equatorial field is: (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 4 : 1

Answers

1. (a) F ∝ q₁q₂, so 2 × 2 = 4.
2. (c) 1/C = 1/3 + 1/3 + 1/3 = 1.
3. (d) A Gaussian surface inside encloses no charge.
4. (b) Total q/ε₀ shared equally by six faces.
5. (d) ΔV = 0 on an equipotential.
6. (c) With V fixed, U = ½CV² rises with C, by K = 3.
7. (b) The fields cancel in pairs. Each charge is a/√2 from the centre, so V = 4kq/(a/√2) = 4√2 kq/a.
8. (c) 2kp/r³ against kp/r³.

What to do next

• Make one table with the field and potential of a point charge, a dipole, a line, a sheet and a shell, inside and outside.
• Learn the dielectric table (battery connected or removed) until you can rewrite it without errors.
• Solve five superposition questions with charges on a square or triangle, drawing every vector.
• Carry on to current electricity for NEET, where potential difference drives current.

For the numeric habits that stop unit slips in μC and μF, see NEET physics numericals.