📚 StudyHub

⚛️ Physics  ·  Class 12  ·  NEET & JEE

Electric Charges and Fields

Coulomb's law, electric field, potential, capacitors, Gauss's law.

Practice Electric Charges and Fields Quiz — 100% Free →
Reading time~8 min
Revision time~3 min
Last updated2026-07-19
1 Read the chapter ~8 min

🎯 Key Points

  • F=kq₁q₂/r², k=1/4πε₀=9×10⁹ N·m²/C²; E=F/q=kQ/r²; V=kQ/r; E=−dV/dr
  • Gauss's Law: ΦE=Q_enclosed/ε₀ — gives quick E for symmetric charge distributions (sphere, line, sheet) without integration
  • Capacitor: C=Q/V=ε₀A/d; Series: 1/C_eq=Σ(1/Cᵢ) (like resistors in parallel-pattern math); Parallel: C_eq=ΣCᵢ
  • Energy stored U=½CV²=½QV=Q²/2C; energy density u=½ε₀E²
  • Dipole: axial field E=2kp/r³ is TWICE the equatorial field E=kp/r³ at the same distance; torque τ=pEsinθ
  • Field inside a conductor in electrostatic equilibrium is always zero; all charge resides on the outer surface
Field Lines: Single Charge vs Electric Dipole+isolated +charge: lines radiate outward, all directions+dipole: lines curve from + to − (denser between them)

Field lines always point from positive to negative charge, are denser where the field is stronger, and never cross; an isolated positive charge has lines radiating symmetrically outward, while a dipole's lines curve from the positive to the negative charge.

Coulomb's Law

  • F = kq₁q₂/r² where k = 1/4πε₀ = 9 × 10⁹ N·m²/C²
  • Electric field: E = F/q = kQ/r² (field due to point charge)

Electric Field

  • Superposition: total field = vector sum of individual fields
  • Field lines originate from + charge and terminate at - charge
  • Field inside conductor = 0; field at surface is perpendicular

Gauss's Law

  • ΦE = Q_enclosed/ε₀ (total flux through closed surface)
  • For uniformly charged sphere outside: E = kQ/r²
  • For infinite line charge: E = λ/2πε₀r
  • For infinite sheet: E = σ/2ε₀

Electric Potential

  • V = kQ/r (due to point charge)
  • E = -dV/dr (field is negative gradient of potential)
  • Work done: W = q·ΔV
  • Potential energy: U = kq₁q₂/r

Capacitors

  • C = Q/V = ε₀A/d (parallel plate)
  • Series: 1/C = Σ(1/Cᵢ); Parallel: C = ΣCᵢ
  • Energy stored: U = ½CV² = ½QV = Q²/2C
  • Dielectric: C' = KC (K = dielectric constant)

Electric Dipole

  • Dipole moment: p = q × 2a (directed from negative to positive charge)
  • Field on axial line: E = 2kp/r³ (for r >> a)
  • Field on equatorial line: E = kp/r³ (for r >> a), direction opposite to dipole moment
  • Torque on dipole in uniform field: τ = pE·sin(θ) = p × E
  • Potential energy of dipole: U = -p·E·cos(θ); minimum (most stable) when dipole aligns with field

Electrostatic Properties of Conductors

  • Electric field inside a conductor in electrostatic equilibrium is always zero
  • Charge resides only on the outer surface of a charged conductor
  • Field just outside a charged conductor surface: E = σ/ε₀ (perpendicular to surface)
  • Electrostatic shielding: field inside a cavity within a conductor is zero, used in Faraday cages

Combination of Capacitors

  • Series combination: same charge Q on each capacitor; 1/C_eq = 1/C₁ + 1/C₂ + ...; effective capacitance is less than the smallest individual capacitance
  • Parallel combination: same voltage V across each; C_eq = C₁ + C₂ + ...; effective capacitance is greater than the largest individual value
  • Energy density in a parallel plate capacitor: u = ½ε₀E²

Basic Properties of Electric Charge

  • Quantisation: charge exists only in integer multiples of the elementary charge e = 1.6×10⁻¹⁹ C, i.e. q = ±ne (n = 1, 2, 3, …); fractional charge is never observed in free particles
  • Conservation: the total charge of an isolated system is constant; charge can be transferred but never created or destroyed
  • Additivity: total charge is the algebraic (signed) sum of individual charges, since charge is a scalar
  • Like charges repel, unlike charges attract; a charged body attracts a neutral body by induction

Coulomb's Law in Vector Form and Superposition

  • Vector form: F₁₂ = (kq₁q₂/r²) r̂₁₂, where r̂₁₂ is the unit vector pointing from charge 1 to charge 2; the sign of q₁q₂ automatically gives repulsion (+) or attraction (−)
  • Principle of superposition: the net force on any charge is the vector sum of the forces due to every other charge, each computed independently as if the others were absent: F_net = F₁ + F₂ + F₃ + …
  • Coulomb forces obey the inverse-square law and act along the line joining the two point charges (a central force)
  • In a medium of dielectric constant K, the force is reduced by a factor K: F_medium = F_vacuum/K

Properties of Electric Field Lines

  • Field lines start on positive charges (or infinity) and end on negative charges (or infinity); they are continuous curves with no breaks
  • The tangent to a field line at any point gives the direction of E there
  • Lines are denser where the field is stronger; the number of lines per unit area (⊥ to lines) is proportional to E
  • Two field lines never cross — if they did, the field would have two directions at that point
  • Field lines do not form closed loops (electrostatic field is conservative) and are always perpendicular to the surface of a conductor
Four panels of electric field line patterns: a point dipole, a pair of equal and opposite point charges labelled plus and minus, a polarized disc, and a parallel plate capacitor, all with field lines directed from positive to negative

Electric field lines start on positive charge and end on negative charge, never cross, and crowd where the field is strong. Image: Geek3, CC BY-SA 4.0, via Wikimedia Commons.

Continuous Charge Distributions

  • For large-scale charged bodies, charge is treated as continuous, described by a density: linear λ = dq/dl (C/m), surface σ = dq/dA (C/m²), or volume ρ = dq/dV (C/m³)
  • The field is found by integrating the contribution of each element dq: E = k ∫ (dq/r²) r̂ over the whole distribution
  • Total charge: q = ∫λ dl = ∫σ dA = ∫ρ dV depending on the geometry

Applications of Gauss's Law

By choosing a Gaussian surface matching the symmetry of the charge, E comes out of the flux integral and is found without integration:

Charge distributionField magnitude EDistance behaviour
Infinite line charge (density λ)λ/2πε₀r∝ 1/r
Infinite plane sheet (density σ)σ/2ε₀uniform (independent of r)
Two oppositely charged sheets (between)σ/ε₀uniform
Charged spherical shell (outside, r > R)kQ/r²∝ 1/r² (acts as point charge)
Charged spherical shell (inside, r < R)0zero everywhere inside
  • Field of a conducting sheet (charge on both faces) is σ/ε₀ just outside — twice that of a single thin charged sheet
  • For a solid uniformly charged sphere, the internal field grows linearly: E = kQr/R³ for r < R

🚀 JEE Advanced Edge

Capacitor with dielectric partially filling the gap: When a dielectric slab of thickness t (less than the full plate separation d) is inserted, treat it as the original capacitor in series with an air-gap capacitor of thickness (d−t): C = ε₀A / (d−t+t/K).

Charge redistribution when capacitors are connected: When a charged capacitor is connected to an uncharged one, charge redistributes until both reach the same potential (conservation of charge, not voltage); energy is LOST in this process (dissipated as heat/radiation during the transient), even though charge is conserved — a classic "energy not conserved but charge is" JEE trap.

Worked problem: A parallel plate capacitor of capacitance 2μF is charged to 100V, then connected to an uncharged 3μF capacitor. Find the common potential. Approach: Charge conservation: Q_initial = C₁V₁ = 2×100 = 200μC. Common V = Q_total/(C₁+C₂) = 200/(2+3) = 40V.

2 Revise ~3 min before the exam

📐 Formula Sheet

  • Coulomb's law: F = kq₁q₂/r², k = 1/4πε₀ = 9 × 10⁹ N·m²/C²
  • Electric field: E = F/q₀  |  point charge: E = kq/r²
  • Dipole moment: p = q × 2a (points from −q to +q)
  • Dipole field — axial: E = 2kp/r³  |  equatorial: E = kp/r³
  • Torque on a dipole: τ = p × E = pE·sinθ  |  PE: U = −pE·cosθ
  • Gauss's law: Φ = ∮E·dA = qenclosed/ε₀
  • From Gauss: line charge E = λ/2πε₀r  |  sheet E = σ/2ε₀  |  conductor surface E = σ/ε₀
  • Inside a conductor / hollow shell: E = 0
3 Practice apply it

✍️ Worked Examples

Example 1 — Force between two charges
Q: Two charges of +2 μC and −3 μC are 30 cm apart in air. Find the force between them.
Step 1 — Convert: q₁ = 2 × 10⁻⁶ C, q₂ = 3 × 10⁻⁶ C, r = 0.3 m.
Step 2 — Coulomb's law: F = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/(0.3)².
Step 3 — Compute: numerator = 9 × 10⁹ × 6 × 10⁻¹² = 0.054; divide by 0.09 ⇒ F = 0.6 N.
Answer: 0.6 N, attractive (the charges have opposite signs). Trap: leaving r in centimetres makes the force 10,000× too small.

Example 2 — Field inside a hollow shell
Q: A hollow conducting sphere of radius 10 cm carries charge +5 μC. Find E at 5 cm and at 20 cm from the centre.
Step 1 — At 5 cm we are inside the shell. A Gaussian surface there encloses no charge, so E = 0.
Step 2 — At 20 cm we are outside, so the shell behaves as a point charge at the centre.
Step 3 — Compute: E = kq/r² = (9 × 10⁹)(5 × 10⁻⁶)/(0.2)² = 45,000/0.04 = 1.125 × 10⁶ N/C.
Answer: 0 inside, 1.125 × 10⁶ N/C outside. Key idea: this shielding is why a car is safe in a lightning strike.

Example 3 — Torque on a dipole
Q: A dipole of moment 4 × 10⁻⁹ C·m sits at 30° to a uniform field of 5 × 10⁴ N/C. Find the torque.
Step 1 — Use τ = pE·sinθ.
Step 2 — Substitute: τ = (4 × 10⁻⁹)(5 × 10⁴)(sin30°) = (4 × 10⁻⁹)(5 × 10⁴)(0.5).
Step 3 — Compute: τ = 1 × 10⁻⁴ N·m.
Answer: 10⁻⁴ N·m. Note: torque is maximum at 90° and zero at 0° — a dipole tends to align with the field.

Practice Electric Charges and Fields Quiz — 100% Free →

Frequently Asked Questions — Electric Charges and Fields

What are the key concepts in Electric Charges and Fields?
Coulomb's law, electric field, potential, capacitors, Gauss's law.
Is Electric Charges and Fields important for NEET & JEE?
Yes. Electric Charges and Fields is part of the Physics Class 12 NCERT syllabus and is directly tested in NEET and JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Electric Charges and Fields questions on StudyHub?
Open StudyHub and select Physics → Electric Charges and Fields. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at NEET & JEE level with full step-by-step explanations.

References

  1. NCERT Class 12 Physics Textbook — Chapter: Electric Charges and Fields
  2. CBSE Curriculum — Physics (Class 12)
  3. NTA NEET UG Official Syllabus — subject-wise topic list
  4. NTA JEE Main Official Syllabus — subject-wise topic list