Electrostatics

Covers NCERT Class 12 Physics Part I, Chapters 1 & 2 — Electric Charges and Fields and Electrostatic Potential and Capacitance. Consistently one of the highest-weightage chapters in NEET Physics.

Foundation Advanced High-Yield Revision
Advanced NTA favourite Very high weightage

⭐ NTA-favourite subtopics in this chapter

Electric dipole field (axial & equatorial), Gauss's law applications (infinite sheet, uniformly charged sphere, cylindrical symmetry), electrostatic potential energy of a charge system, equipotential surfaces & their properties, capacitors in series/parallel with dielectric slabs, and energy stored in a capacitor / energy density. These sub-areas account for a disproportionate share of repeat-style NEET questions from this chapter.

Exam-oriented notes

NCERT-tagged theory, simplified

Each point is tagged to the NCERT section it comes from (section numbers are stable across editions — check the matching heading in your own copy for the full derivation).

NCERT 1.5

Charge is quantised (q = ne) and conserved in an isolated system — both properties are directly testable as assertion-reason or one-line factual MCQs.

NCERT 1.6

Coulomb's Law: F = kq₁q₂/r², with k = 1/4πε₀ ≈ 9×10⁹ N m² C⁻². Vector form matters for problems with multiple charges at angles — always resolve components before adding.

NCERT 1.8

Electric field E = F/q₀ (test charge → 0). For a point charge, E = kq/r², directed away from positive charge, towards negative.

NCERT 1.11

Dipole moment p = q×2a (from −q to +q). Axial field E ≈ 2kp/r³; equatorial field E ≈ kp/r³ (direction opposite to p). The factor-of-2 difference between axial and equatorial is a classic trap in options.

NCERT 1.12

Torque on a dipole in uniform field: τ = pE sinθ = p×E. Net force is zero in a uniform field but non-zero in a non-uniform field — a frequent assertion-reason pairing.

NCERT 1.14–1.15

Gauss's law: Φ = q_enclosed/ε₀. Use symmetry to pick the Gaussian surface — sphere for point/spherical charge, cylinder for line charge, pillbox for infinite sheet. Field due to infinite sheet: E = σ/2ε₀ (independent of distance); for a conductor surface: E = σ/ε₀.

NCERT 2.2–2.3

Potential V = kq/r for a point charge; potential is a scalar, so contributions from multiple charges simply add algebraically (unlike field, which is a vector sum).

NCERT 2.6

Equipotential surfaces are always perpendicular to field lines; no work is done moving a charge along one; surfaces are closer where the field is stronger.

NCERT 2.7

Potential energy of a two-charge system: U = kq₁q₂/r. For a system of three or more charges, sum U over every unique pair — a common source of missed terms under time pressure.

NCERT 2.9

Inside a conductor, E = 0 and the entire conductor (including its surface) is an equipotential region; charge resides only on the outer surface.

NCERT 2.11–2.12

Capacitance C = Q/V. Parallel plate capacitor: C = ε₀A/d (vacuum). Introducing a dielectric of constant K anywhere between the plates always increases capacitance.

NCERT 2.14

Series: 1/C_eq = Σ(1/Cᵢ) — equivalent capacitance is always less than the smallest individual capacitor. Parallel: C_eq = ΣCᵢ — always greater than the largest.

NCERT 2.15

Energy stored: U = ½CV² = ½QV = Q²/2C. Energy density in the field: u = ½ε₀E². When a battery stays connected, V is constant during a change; when disconnected, Q is constant instead — this single distinction resolves most "before/after inserting dielectric" questions.

Printable

Formula sheet

Electrostatics — Chapters 1 & 2

Coulomb's law
F = kq₁q₂ / r²,  k = 9×10⁹ N·m²/C²
Electric field (point charge)
E = kq / r²
Dipole moment
p = q · (2a)
Axial field of dipole
E_axial = 2kp / r³
Equatorial field of dipole
E_eq = kp / r³
Torque on dipole
τ = pE sinθ
Field: infinite sheet
E = σ / 2ε₀
Field: charged conductor surface
E = σ / ε₀
Potential (point charge)
V = kq / r
Potential energy (2 charges)
U = kq₁q₂ / r
Capacitance (parallel plate)
C = Kε₀A / d
Series combination
1/C_eq = 1/C₁ + 1/C₂ + …
Parallel combination
C_eq = C₁ + C₂ + …
Energy stored
U = ½CV² = Q²/2C
Energy density
u = ½ε₀E²
Key constant
ε₀ = 8.85×10⁻¹² C²N⁻¹m⁻²
PYQ-style practice

Practice set, NEET pattern

Original questions modelled on recurring NEET question types for this chapter — not verbatim reproductions of any official paper. Each includes a worked solution.

Q1. Two point charges +2q and −q are placed 2a apart. At what point on the line joining them (outside, on the −q side) is the electric field zero?
  • At distance a from −q
  • At distance a(1+√2) from −q
  • At distance 2a from −q
  • Field is never zero outside the segment on this side
Show solution
Field zero must occur closer to the smaller-magnitude charge, outside the segment, on the −q side. Setting k(2q)/(x+2a)² = kq/x² and solving gives x = a(1+√2) from −q. Answer: B
Q2. A dipole of moment p is placed in a uniform electric field E, initially anti-parallel to E. Work done to rotate it to be parallel to E is:
  • 0
  • pE
  • 2pE
  • −2pE
Show solution
W = U_f − U_i = (−pE cos0°) − (−pE cos180°) = −pE − pE = −2pE. Since the field does positive work rotating the dipole into alignment, the work done by an external agent against the field is +2pE, but work done by the field is −2pE — read the question stem carefully. Taking "work done to rotate it" as work done by the field: Answer: D
Q3. A hollow conducting sphere carries charge Q on its surface. The electric field at the centre of the sphere is:
  • kQ/R²
  • kQ/R
  • Zero
  • Infinite
Show solution
By symmetry/Gauss's law, field inside a uniformly charged hollow conducting sphere is zero everywhere inside, including the centre. Answer: C
Q4. Three capacitors of 2 μF, 3 μF and 6 μF are connected in series across a 12 V battery. The equivalent capacitance is:
  • 11 μF
  • 1 μF
  • 6 μF
  • 0.5 μF
Show solution
1/C_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 → C_eq = 1 μF. Answer: B
Q5. A parallel plate capacitor is charged and then isolated from the battery. A dielectric slab (K = 4) is now inserted, filling the gap. Which quantity remains unchanged?
  • Capacitance
  • Potential difference
  • Charge on the plates
  • Energy stored
Show solution
Isolated (battery disconnected) ⇒ charge Q cannot change. C increases by factor K, so V = Q/C decreases, and U = Q²/2C decreases. Answer: C
Q6. The SI unit of electric flux is:
  • N C⁻¹
  • N m² C⁻¹
  • C m⁻²
  • N m C⁻¹
Show solution
Φ = E·A, so units are (N C⁻¹)(m²) = N m² C⁻¹. Answer: B
Daily Practice Problems

DPP — Electrostatics (15 questions)

Scaled from direct NCERT application (Q1–Q6) to mixed NEET-level difficulty (Q7–Q15). Attempt in 18 minutes, then check the answer key.

1.State the SI unit of permittivity of free space, ε₀.
2.Two charges of +5 μC and −5 μC are 10 cm apart. Find the force between them.
3.Define electric field intensity and give its SI unit.
4.A capacitor of 4 μF is charged to 100 V. Find the charge and energy stored.
5.Why is the electric field inside a conductor always zero in electrostatic equilibrium?
6.State Gauss's law and write the expression for flux through a closed surface.
7.Two identical spheres with charges +8 μC and −2 μC are touched together and separated. Find the new charge on each.
8.An electric dipole of moment 4×10⁻⁹ C·m is placed in a field of 5×10⁴ N/C at 30° to the field. Find the torque on it.
9.Find the potential at the centre of a square of side a, with equal charges +q at each corner.
10.Three capacitors 2 μF, 4 μF, 6 μF are connected in parallel. Find the equivalent capacitance and total charge at 10 V.
11.A parallel plate capacitor's capacitance doubles when a dielectric slab of thickness equal to half the gap is inserted. Find K.
12.Two point charges repel each other with force F. If the distance is doubled and each charge is halved, find the new force in terms of F.
13.Derive, in outline, why equipotential surfaces around a point charge are concentric spheres.
14.A capacitor network has 3 μF and 6 μF in series, and this combination in parallel with 2 μF. Find equivalent capacitance.
15.Assertion: Work done in moving a charge on an equipotential surface is zero. Reason: Electric field is always perpendicular to an equipotential surface. Judge the assertion-reason pair.
Show answer key
1. C²N⁻¹m⁻²  |  2. 22.5 N (attractive)  |  3. F/q₀, N/C  |  4. Q = 4×10⁻⁴ C, U = 0.02 J  |  5. Free charges rearrange until net internal field cancels the external field  |  6. Φ = q_enc/ε₀  |  7. +3 μC each  |  8. τ = 1×10⁻⁴ N·m  |  9. V = 4kq/(a/√2) = 4√2 kq/a  |  10. 12 μF, 120 μC  |  11. K = 2  |  12. F/16  |  13. V = kq/r is constant only when r is constant ⇒ locus is a sphere  |  14. 4 μF  |  15. Both true, reason correctly explains assertion
Printable mnemonic chart

Sign & trap chart for Electrostatics

🧠 "SIP the Signs" — quick recall chart

SituationBattery stays connectedBattery disconnected
What's constant?V (voltage)Q (charge)
Insert dielectric KC↑, Q↑, U↑C↑, V↓, U↓
Increase plate separation dC↓, Q↓C↓, V↑

Memory hook: "Still connected → V is Still" (Still/Still/V) vs "Isolated → charge is Imprisoned" (Isolated/Imprisoned/Q). Two S-words pair with V, two I-words pair with Q.

🧠 Axial vs equatorial dipole field

"Axial is Active" — axial field is twice as strong (2kp/r³) and points along p; equatorial is the "quiet" one (kp/r³) and points opposite to p.