1. Chapter at a Glance
Electrostatics is the study of charges at rest. This chapter explains what electric charge is and how two charges push or pull each other (Coulomb's law). It then introduces the electric field, field lines, the electric dipole, electric flux, and Gauss's law, which gives the field of symmetric charge distributions very quickly. NEET usually asks two questions from this chapter, often mixed with the next one (potential and capacitance).
- Coulomb force, the superposition principle and the null point between two charges.
- Electric field of a point charge, a dipole (axial and equatorial), a line charge and a sheet.
- Torque on a dipole in a uniform field.
- Electric flux and Gauss's law (charge enclosed in a closed surface).
- Properties of field lines and of conductors.
2. Electric Charge and Its Properties
- Charge is of two kinds: positive and negative. Like charges repel; unlike charges attract.
- Additivity: the total charge of a system is the algebraic sum of its charges (it behaves like a scalar).
- Conservation: charge can be neither created nor destroyed; it only moves from one body to another.
- Quantisation: every charge is an integer multiple of the elementary charge, q = ne, with e = 1.6 × 10⁻¹⁹ C.
- Rubbing a glass rod with silk transfers electrons from glass to silk; the rod becomes positive and the silk negative.
- Conductors let charges move freely (metals); insulators do not (glass, plastic). Earthing (grounding) shares charge with the Earth.
- Charging by induction: bringing a charged rod near a conductor separates its charges; earthing it and then removing the earth and the rod leaves it with the opposite charge.
1 C is a very large charge: it equals the charge of about 6.25 × 10¹⁸ electrons. Everyday static charges are in nanocoulombs (nC) or microcoulombs (μC).
3. Coulomb's Law and Superposition
- The force acts along the line joining the charges, and it obeys Newton's third law.
- In a medium of dielectric constant K, the force becomes F/K (it is weaker in water, K ≈ 80).
- Superposition: the net force on a charge is the vector sum of the forces due to all other charges, each calculated as if the others were absent.
- When two identical conducting spheres touch, they share their total charge equally.
For two like charges q₁ and q₂ a distance d apart, the field is zero between them at a distance x from q₁, where x = d√q₁/(√q₁ + √q₂). For unlike charges, the null point lies outside, beyond the smaller charge.
4. Electric Field and Field Lines
- The field points away from a positive charge and towards a negative charge.
- A charge q in a field E feels a force F = qE; an electron moves opposite to the field.
Properties of field lines
- They start on positive charges and end on negative charges (or at infinity).
- The tangent at any point gives the direction of E; two lines never cross.
- Where lines are crowded, the field is strong; parallel, equally spaced lines mean a uniform field.
- Electrostatic field lines never form closed loops.
5. Electric Dipole
Two equal and opposite charges ±q separated by a small distance 2a form a dipole. Its dipole moment is p = q × 2a, directed from −q to +q (unit C m).
| Situation | Result | NEET Note |
|---|---|---|
| Field on the axis (r ≫ a) | E = 2kp/r³ | Along p |
| Field on the equatorial line | E = kp/r³ | Opposite to p; half the axial value |
| Torque in a uniform field | τ = pE sinθ | Maximum at θ = 90° |
| Net force in a uniform field | F = 0 | Only rotation, no translation |
| Net force in a non-uniform field | F ≠ 0 | Dipole also moves |
| Potential energy | U = −pE cosθ | Stable at θ = 0, unstable at θ = 180° |
A dipole field falls off as 1/r³, faster than the 1/r² field of a single charge, because the fields of the two opposite charges almost cancel far away.
6. Electric Flux and Gauss's Law
- Flux is a scalar measured in N m² C⁻¹. Only the charge inside the closed surface decides the total flux; charges outside contribute zero net flux.
- The shape and size of the closed surface do not matter.
- A charge q at the centre of a cube gives flux q/(6ε₀) through each face.
| Charge distribution | Field | NEET Note |
|---|---|---|
| Infinite line charge (λ) | E = λ/(2πε₀r) = 2kλ/r | E ∝ 1/r |
| Infinite plane sheet (σ) | E = σ/(2ε₀) | Independent of distance |
| Two oppositely charged sheets | E = σ/ε₀ between, 0 outside | Basis of the parallel-plate capacitor |
| Thin spherical shell (outside, r > R) | E = kQ/r² | Acts like a point charge at the centre |
| Thin spherical shell (inside, r < R) | E = 0 | Electrostatic shielding |
- Inside a conductor in electrostatic equilibrium, E = 0 and any excess charge sits on the outer surface.
- Just outside a charged conductor, E = σ/ε₀, perpendicular to the surface.
7. Formula Master Table
| Quantity | Formula |
|---|---|
| Quantisation | q = ne |
| Coulomb's law | F = kq₁q₂/r² |
| Force in a medium | Fm = F/K |
| Field of a point charge | E = kq/r² |
| Dipole moment | p = q(2a) |
| Dipole field (axial / equatorial) | 2kp/r³ / kp/r³ |
| Torque on a dipole | τ = pE sinθ |
| Electric flux | Φ = EA cosθ |
| Gauss's law | Φ = qin/ε₀ |
| Line charge | E = 2kλ/r |
| Plane sheet | E = σ/2ε₀ |
| Near a conductor | E = σ/ε₀ |
8. Master Summary Table
| Concept | Key idea | NEET Note |
|---|---|---|
| Quantisation | q = ne | Charge 2.4 × 10⁻¹⁹ C is impossible |
| Conservation | Total charge constant | Used after touching spheres |
| Coulomb force | ∝ q₁q₂/r² | Halve r ⇒ force × 4 |
| Field lines | + to −, never cross | No closed loops |
| Dipole | E ∝ 1/r³ | Axial field = 2 × equatorial |
| Dipole in uniform E | F = 0, τ = pE sinθ | Stable when p ∥ E |
| Gauss's law | Φ = qin/ε₀ | Outside charges give zero net flux |
| Conductor | E = 0 inside | Charge on outer surface |
9. Mind Map
- Quantised
- Conserved
- Additive
- Coulomb's law
- Superposition
- Medium: F/K
- E = F/q
- Field lines
- Dipole
- Flux
- Line, sheet, shell
- Conductors
10. Tips and Tricks
- Always put charges in coulombs and distances in metres before using k = 9 × 10⁹.
- For flux questions, first list which charges are inside the surface; ignore the rest.
- A dipole in a uniform field needs work W = pE(1 − cosθ) to turn it from θ = 0 to θ.
- For touching spheres: add the charges with sign, then divide equally.
- Taking the dipole length as a instead of 2a in p = q(2a).
- Thinking a charge outside a closed surface changes its total flux.
- Forgetting that the field of an infinite sheet does not depend on distance.
- Assuming the net force on a dipole is zero in a non-uniform field.
11. Important One-liners
- The SI unit of charge is the coulomb; e = 1.6 × 10⁻¹⁹ C.
- Coulomb's law is an inverse-square law, like Newton's law of gravitation, but electric forces can attract or repel.
- The electric force between two protons is about 10³⁶ times their gravitational attraction.
- Electric field lines are always perpendicular to the surface of a conductor.
- The field inside a hollow charged conducting sphere is zero.
- The dipole moment points from the negative to the positive charge.
- A dipole in a uniform field is in stable equilibrium when p is along E.
- The unit of electric flux is N m² C⁻¹ (or V m).
12. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take k = 9 × 10⁹ N m² C⁻² and e = 1.6 × 10⁻¹⁹ C. Click to reveal each answer and working.
Average level
Charges of +5 μC and −4 μC are 20 cm apart in air. Find the force between them and state its nature.
Show answer & solution
Given: q₁ = 5 × 10⁻⁶ C, q₂ = 4 × 10⁻⁶ C (magnitude), r = 0.2 m. Concept: F = kq₁q₂/r².
- F = 9 × 10⁹ × 5 × 10⁻⁶ × 4 × 10⁻⁶ / (0.2)².
- = 9 × 10⁹ × 20 × 10⁻¹² / 0.04 = 0.18/0.04.
- F = 4.5 N. The charges are unlike, so the force is attractive.
Answer: 4.5 N, attractive
A metal sphere is given a charge of +8 nC. How many electrons have been removed from it?
Show answer & solution
Concept: quantisation, q = ne.
- n = q/e = 8 × 10⁻⁹ / 1.6 × 10⁻¹⁹.
- n = 5 × 10¹⁰ electrons.
Answer: 5 × 10¹⁰ electrons
Find the electric field 20 cm from a point charge of 4 nC. What force would a 2 μC charge feel at that point?
Show answer & solution
Concept: E = kq/r², F = q₀E.
- E = 9 × 10⁹ × 4 × 10⁻⁹ / (0.2)² = 36/0.04 = 900 N C⁻¹.
- F = 2 × 10⁻⁶ × 900 = 1.8 × 10⁻³ N.
Answer: 900 N C⁻¹; 1.8 × 10⁻³ N
A dipole consists of ±2 nC charges 3 mm apart. It is placed in a uniform field of 5 × 10⁴ N C⁻¹ at 30° to the field. Find the dipole moment and the torque on it.
Show answer & solution
Concept: p = q × (separation); τ = pE sinθ.
- p = 2 × 10⁻⁹ × 3 × 10⁻³ = 6 × 10⁻¹² C m.
- τ = 6 × 10⁻¹² × 5 × 10⁴ × sin 30° = 3 × 10⁻⁷ × 0.5.
- τ = 1.5 × 10⁻⁷ N m.
Answer: 6 × 10⁻¹² C m; 1.5 × 10⁻⁷ N m
A closed surface encloses charges of +5 μC and −2 μC. A third charge of +1 μC lies outside it. Find the net electric flux through the surface (ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²).
Show answer & solution
Concept: Gauss's law uses only the enclosed charge.
- qin = 5 − 2 = 3 μC (the outside charge is ignored).
- Φ = 3 × 10⁻⁶ / 8.854 × 10⁻¹².
- Φ ≈ 3.39 × 10⁵ N m² C⁻¹.
Answer: about 3.39 × 10⁵ N m² C⁻¹
Advanced level
Charges +q and +9q are fixed 40 cm apart. Where on the line joining them is the net electric field zero?
Show answer & solution
Concept: for like charges the null point lies between them, nearer the smaller charge.
- Let the point be x from +q: kq/x² = k(9q)/(40 − x)².
- Take square roots: 1/x = 3/(40 − x) ⇒ 40 − x = 3x.
- x = 10 cm from +q (30 cm from +9q).
Answer: 10 cm from the +q charge
Two identical small conducting spheres carry +6 μC and −2 μC and attract each other with force F. They are touched together and put back at the same separation. Find the new force in terms of F.
Show answer & solution
Concept: identical spheres share the total charge equally.
- Before: F ∝ 6 × 2 = 12 (attractive).
- Total charge = 6 − 2 = +4 μC ⇒ each sphere gets +2 μC.
- After: F′ ∝ 2 × 2 = 4 ⇒ F′ = F/3.
- Both are now positive, so the force is repulsive.
Answer: F/3, now repulsive
A short dipole has moment 4 × 10⁻⁹ C m. Find the field 20 cm from its centre (a) on its axis, (b) on its equatorial line.
Show answer & solution
Concept: axial E = 2kp/r³, equatorial E = kp/r³.
- r³ = (0.2)³ = 8 × 10⁻³ m³.
- Axial: E = 2 × 9 × 10⁹ × 4 × 10⁻⁹ / 8 × 10⁻³ = 72/0.008 = 9000 N C⁻¹, along p.
- Equatorial: E = 9000/2 = 4500 N C⁻¹, opposite to p.
Answer: 9000 N C⁻¹ (axial); 4500 N C⁻¹ (equatorial)
A very long straight wire carries a uniform charge of 3 μC per metre. Find the electric field 18 cm from it. What happens to the field if the distance is doubled?
Show answer & solution
Concept: by Gauss's law, E = λ/(2πε₀r) = 2kλ/r.
- E = 2 × 9 × 10⁹ × 3 × 10⁻⁶ / 0.18.
- = 54 × 10³ / 0.18 = 3 × 10⁵ N C⁻¹.
- E ∝ 1/r, so doubling r halves E to 1.5 × 10⁵ N C⁻¹.
Answer: 3 × 10⁵ N C⁻¹; it halves
An oil drop of mass 1.6 × 10⁻¹⁵ kg stays suspended between two horizontal plates where the upward electric field is 2 × 10⁴ N C⁻¹. Find its charge and the number of excess elementary charges (g = 10 m s⁻²).
Show answer & solution
Concept: balance of forces: qE = mg.
- q = mg/E = 1.6 × 10⁻¹⁵ × 10 / 2 × 10⁴.
- q = 8 × 10⁻¹⁹ C.
- n = q/e = 8 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 5.
- The field is upward and the force must be upward, so the drop is positively charged (it has lost 5 electrons).
Answer: 8 × 10⁻¹⁹ C; 5 elementary charges (positive)
13. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
Which of these charges is not possible on a body?
- 3.2 × 10⁻¹⁹ C
- 4.8 × 10⁻¹⁹ C
- 2.4 × 10⁻¹⁹ C
- 6.4 × 10⁻¹⁹ C
Show solution
Charge must be a whole-number multiple of 1.6 × 10⁻¹⁹ C. The values 3.2, 4.8 and 6.4 are 2e, 3e and 4e. But 2.4 × 10⁻¹⁹ C is 1.5e, which is not a whole number.
Answer: (C)
Two point charges repel with force F. If the distance between them is halved, the force becomes:
- F/2
- 2F
- F/4
- 4F
Show solution
F ∝ 1/r². Halving r multiplies the force by 4. (A) and (B) wrongly treat the force as ∝ 1/r, and (C) gets the direction of change wrong.
Answer: (D)
A charge q is placed at the centre of a cube. The electric flux through one face of the cube is:
- q/ε₀
- q/(6ε₀)
- 6q/ε₀
- zero
Show solution
The total flux is q/ε₀. By symmetry, the six faces share it equally, so each face gets q/(6ε₀). (A) is the total flux, not the flux through one face. (C) wrongly multiplies instead of dividing, and (D) would be true only if the charge were outside the cube.
Answer: (B)
An electric dipole is placed in a uniform electric field. It experiences:
- A net force and a torque
- A net force but no torque
- A torque but no net force (in general)
- Neither force nor torque
Show solution
The forces qE and −qE are equal and opposite, so their sum is zero. They act at different points, so they form a couple with torque pE sinθ. This torque is zero only if p is parallel or antiparallel to E.
Answer: (C)
The electric field due to an infinite plane sheet of charge density σ at a distance r from it:
- Varies as 1/r
- Varies as 1/r²
- Is σ/(2ε₀), independent of r
- Is σ/ε₀, independent of r
Show solution
Gauss's law gives E = σ/(2ε₀) for a sheet, with no dependence on r. The value σ/ε₀ in (D) applies just outside a conductor or between two oppositely charged sheets. 1/r is the line-charge result and 1/r² is the point-charge result.
Answer: (C)
The ratio of the electric field on the axis to that on the equatorial line of a short dipole, at equal distances, is:
- 1 : 1
- 1 : 2
- 2 : 1
- 4 : 1
Show solution
Eaxial = 2kp/r³ and Eequatorial = kp/r³, so the ratio is 2 : 1. (B) is the inverse ratio.
Answer: (C)
A hollow metal sphere of radius R carries charge Q. The electric field at a distance R/2 from its centre is:
- kQ/R²
- 4kQ/R²
- Zero
- kQ/(2R²)
Show solution
A Gaussian sphere of radius R/2 encloses no charge, because all the charge is on the surface. So the field inside is zero. (B) wrongly uses the point-charge formula at R/2, and (A) is the field just outside the surface.
Answer: (C)
Two identical conducting spheres carry charges +12 μC and −4 μC and attract with force F at separation r. After they touch and are placed back at r, the force is:
- F/3, repulsive
- F/3, attractive
- F/12, repulsive
- 4F/3, repulsive
Show solution
The total charge is +8 μC, so each sphere gets +4 μC. The original product was 12 × 4 = 48, and the new product is 4 × 4 = 16, so F′ = F/3. Both spheres are now positive, so the force is repulsive. (B) misses the change of sign, (C) uses the wrong product, and (D) inverts the ratio.
Answer: (A)
14. Quick Revision Summary
- q = ne; charge is conserved and additive.
- F = kq₁q₂/r²; in a medium F/K; use superposition for many charges.
- E = kq/r²; field lines run from + to − and never cross.
- Dipole: p = q(2a); Eaxial = 2kp/r³, Eeq = kp/r³; τ = pE sinθ.
- Gauss: Φ = qin/ε₀; line 2kλ/r, sheet σ/2ε₀, shell 0 inside.