Properties of Electric Charge
- Scalar quantity; SI unit = Coulomb (C); CGS unit = e.s.u. (electrostatic unit) = franklin
- 1 C = 3×10⁹ e.s.u.; Largest unit = Faraday = 96500 C; Smallest unit = e.s.u.
- Conserved: total charge of an isolated system remains constant
- Quantized: Q = ne where n = integer and e = 1.6×10⁻¹⁹ C
- Invariant: charge does NOT depend on speed of the charged body
- Charge is transferable; follows simple scalar addition
- Mass can exist without charge; charge CANNOT exist without mass
- Positive charge = deficiency of electrons; Negative charge = excess of electrons
Charge on Conductors vs Insulators
- Hollow conducting sphere → excess charge remains on the surface
- Solid conducting sphere → excess charge remains on the surface
- Hollow insulating sphere → charge remains on the surface
- Solid insulating sphere → charge distributed throughout the volume
Methods of Charging a Body
- Friction: only for insulators; equal and opposite charges created on both bodies
- Conduction: sharing of charge until potentials equalize; new charge = R₁/(R₁+R₂) × total charge
- Induction: conductor — induced charge = inducing charge; Dielectric — induced charge < inducing charge
Properties
- Acts between two point charges (or spherically symmetric distributions)
- Always attractive for unlike charges; generally repulsive for like charges
- Central force (along line joining charges), conservative, long range
- Follows inverse square law; depends on medium
- Action-reaction pair (|F₁₂| = |F₂₁|)
- Does NOT depend on presence of other charges (superposition)
F = kq₁q₂/r² where k = 1/(4πε₀) = 9×10⁹ N·m²/C²
ε₀ = 8.85×10⁻¹² C²/(N·m²) (permittivity of free space)
In medium: Fmed = Fair/εr = Fair/K (K = dielectric constant)
ε = ε₀εr = ε₀K
Specific Charge (q/m)
- Electron: q/m = e/mₑ (maximum — electron has smallest mass)
- Proton: q/m = e/mₚ
- Deuteron: q/m = e/(2mₚ) (same charge as proton but double mass)
- α-particle: q/m = 2e/(4mₚ) = e/(2mₚ) (same as deuteron)
Position of Zero Force Between Charges
Finding Position of Third Charge
Case 1 — Like charges (Q and nQ): Third charge placed between them at distance x from smaller charge Q:
x = d/(√n + 1) from Q (closer to smaller charge Q)
Case 2 — Unlike charges (Q and −nQ): Third charge placed OUTSIDE, beyond smaller charge:
x = d/(√n − 1) from Q (beyond smaller charge Q)
For system equilibrium (all 3 charges): third charge must be negative if outer charges are positive and positive if outer charges are negative.
Q: Charges +Q and +4Q placed d apart. Find position of 3rd charge for zero force.
n = 4; x = d/(√4+1) = d/3 from Q → place at d/3 from smaller charge Q
Ans: x = d/3 from Q (between the charges) ✓
Definition
- E = electrostatic force experienced by unit positive charge = F/q
- Vector; direction = direction of force on positive test charge
- SI unit = N/C = V/m; Dimension [MLT⁻³A⁻¹]
- On +ve charge: force along E; on −ve charge: force opposite to E
| Source | Region | Electric Field |
|---|
| Point charge Q | Distance r | E = kQ/r² |
| Ring (charge Q, radius R) | Axis at distance x | E = kQx/(R²+x²)3/2 |
| Ring — maximum | x = R/√2 | Emax = 2kQ/(3√3 R²) |
| Infinite line charge (λ) | Distance r | E = 2kλ/r = λ/(2πε₀r) |
| Infinite non-conducting sheet (σ) | Any side | E = σ/(2ε₀) |
| Infinite conducting sheet (σ) | Outside | E = σ/ε₀ |
| Hollow/Solid conducting sphere (Q) | Outside (r > R) | E = kQ/r² |
| Same sphere | Surface (r = R) | E = kQ/R² |
| Same sphere | Inside (r < R) | E = 0 |
| Solid non-conducting sphere (Q) | Inside (r < R) | E = kQr/R³ |
| Circular arc (charge q, radius R, angle θ) | Centre | E = kq sin(θ/2) / (Rθ/2 × R) → E = 2kλsin(θ/2)/R |
| Half ring (λ) | Centre | E = 2kλ/R |
Properties of Electric Field Lines
- Imaginary lines; tangent at any point gives direction of E at that point
- Number of field lines per unit area ∝ magnitude of E (field line density)
- Start from +ve charge, end at −ve charge (or go to infinity)
- Never intersect each other
- Always perpendicular to surface of conductor
- E = 0 inside conductor → no field lines inside conductor
- Conservative field — never form closed loops
- Sharp turns not possible — always continuous and smooth
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Gauss's Law and Electric Flux
Electric Flux (Φ)
- Φ = counting of electric field lines passing through a surface
- Scalar; unit = N·m²/C = V·m
- For uniform field: Φ = E·A·cosθ (θ = angle between E and area vector)
- For variable field: Φ = ∮E·dA
- Outgoing flux = +ve; Incoming flux = −ve
- For a closed surface in uniform field: Φnet = 0
Gauss's Law: Φ = ∮E·dA = Qenclosed/ε₀
True for any closed surface of any shape. E on LHS is due to ALL charges; Qenc on RHS is only enclosed charge.
Flux Through Common Configurations
- Charge Q at centre of cube: Φcube = Q/ε₀; Each face: Φface = Q/(6ε₀)
- Charge Q at corner of cube: Φcube = Q/(8ε₀); 8 cubes share the corner charge
- Charge Q at centre of one face: Φcube = Q/(2ε₀); 2 cubes share it
- Charge Q at mid-point of an edge: Φcube = Q/(4ε₀); 4 cubes share it
- Charge Q in plane surface: Φ through plane = 0 (symmetric equal flux on both sides)
- Infinite plane with charge: Φ = σA/ε₀ (for Gaussian pillbox)
Important Notes on Gauss's Law
- Always valid; but may not always be useful to calculate E
- Useful for symmetric charge distributions (spherical, cylindrical, planar)
- E inside closed conductor = 0; E just outside conductor = σ/ε₀ (normal to surface)
- Gauss law valid regardless of location of enclosed charge
Q: A charge Q is placed at centre of a hemisphere of radius R. Find flux through flat circular base.
Total flux from Q = Q/ε₀ (through full sphere). By symmetry, flat base subtends half the solid angle as closed sphere → Φbase = Q/(2ε₀)
Ans: Φ = Q/(2ε₀) ✓
Definition
- Two equal and opposite charges (+q, −q) separated by small distance 2a
- Dipole moment P = q × 2a; vector — directed from −q to +q
- Unit = C·m; Dimension [A·T·L]
| Location | Electric Field | Potential |
|---|
| Axial point (distance r, r>>a) | Eaxial = 2kP/r³ (along P) | V = kP/r² |
| Equatorial point (distance r, r>>a) | Eeq = kP/r³ (antiparallel to P) | V = 0 |
| At angle θ from axis | E = kP√(1+3cos²θ)/r³ | V = kP cosθ/r² |
| Angle α between E and r | tan α = tanθ/2 | — |
Torque on dipole in uniform field: τ = P×E = PE sinθ (tends to align with field)
PE of dipole in uniform field: U = −PE cosθ (stable at θ = 0°, unstable at θ = 180°)
F on dipole in non-uniform field: F = P·dE/dr (along increasing E)
Force on dipole in uniform field = 0; in non-uniform field ≠ 0
Definition
- Work done in bringing unit positive charge from infinity to that point without changing KE
- Scalar; unit = Volt (V) = J/C = N·m/C; also = Wb/s (Weber per second)
- Potential decreases in the direction of electric field
- ΔV does NOT depend on reference; V depends on reference
- E = −dV/dr = −(slope of V-r graph)
| Source | Electric Potential V |
|---|
| Point charge Q | V = kQ/r |
| Ring (Q, radius R), on axis at x | V = kQ/√(R²+x²) |
| Ring — at centre (x=0) | V = kQ/R |
| Half-ring / Circular arc (Q, radius R) | V = kQ/R (same as at centre of ring) |
| Hollow/Solid conducting sphere (Q) | Inside: V = kQ/R (constant); Outside: V = kQ/r |
| Solid non-conducting sphere (Q) | Inside: V = kQ(3R²−r²)/(2R³); Surface: kQ/R; Centre: 3kQ/(2R) |
| Dipole (small, r>>a) | V = kP cosθ/r²; On equatorial: V = 0 |
| Infinite line charge (λ) | V₁−V₂ = 2kλ log(r₂/r₁) (no absolute V) |
| Infinite plane (σ) | V₁−V₂ = σ(r₁−r₂)/(2ε₀) (no absolute V) |
Key Assertion-Reason Type Facts
- When Wcf = −ve → PE increases ✓
- When Wcf = +ve → PE decreases ✓
- +ve charge can create −ve potential at a point → Yes (if reference is appropriately chosen) ✓
- V = constant → E must be zero ✓
- E = 0 at a point → V may be non-zero ✓
- V = 0 at a point → E may be non-zero ✓
Equipotential Surfaces
- Surface where V = constant at all points
- E is always perpendicular to equipotential surface
- Work done along an equipotential surface = 0
- No work done to move charge along equipotential
- For point charge → spherical; for line charge → cylindrical; for sheet → planar
- Conductor surface is always equipotential; conductor body creates equipotential region
Potential Energy of System of Charges
- For 2 charges: U = kq₁q₂/r
- Total number of PE terms for N charges = N(N−1)/2
- For N charges at corners of equilateral triangle (side a): U = −kq²/a × (N(N−1)/2) if all same charge magnitude
- Self energy of hollow/solid conducting sphere: U = kQ²/(2R)
- Self energy of solid non-conducting sphere: U = 3kQ²/(5R)
Q: Find potential at centre of equilateral triangle (side a) with charges +q, +q, +q at corners.
Vcentre = 3 × (kq/r) where r = a/√3; V = 3kq√3/a = 3√3 kq/a
Ans: V = 3√3 kq/a ✓
Definition and Basic Formula
- Capacitor = electrical device to store electrostatic energy by storing charge
- Combination of two conductors with equal and opposite charges
- Capacitance C = Q/V (ratio of charge to potential difference)
- Scalar; SI unit = Farad (F) = C/V; Dimension [M⁻¹L⁻²T⁴A²]
- Battery = source of energy (not charge); maintains constant potential difference
| Type | Capacitance | Notes |
|---|
| Isolated sphere (radius R) | C = 4πε₀R = R/k | Earth capacitance ≈ 711 μF |
| Parallel plate (area A, separation d) | C₀ = ε₀A/d | Air between plates |
| Parallel plate with dielectric K | C = Kε₀A/d = KC₀ | K inserted |
| n identical small spheres merged | Rnew = n1/3R; Cnew = n1/3C | Volume conserved |
Series: 1/Ceq = 1/C₁ + 1/C₂ + ... (same charge Q; V splits)
Parallel: Ceq = C₁ + C₂ + ... (same V; charge splits)
Law of potential drop (series): V₁/V₂ = C₂/C₁ (inverse ratio)
Law of charge distribution (parallel): Q₁/Q₂ = C₁/C₂ (direct ratio)
Energy Stored in Capacitor
- U = Q²/(2C) = ½CV² = QV/2
- Energy density (u) = ½ε₀E² = σ²/(2ε₀)
- Pressure on plates: P = σ²/(2ε₀) = u (energy density)
- Work done by battery WB = QV = CV²
- Energy stored = ½CV²; Energy lost as heat = ½CV²
- Energy by battery = Energy stored + Energy lost → CV² = ½CV² + ½CV²
Effect of Dielectric Insertion
| Quantity | Battery Connected (V = const) | Battery Removed (Q = const) |
|---|
| Capacitance | C₀ → KC₀ (increases K×) | C₀ → KC₀ (increases K×) |
| Potential V | V₀ → V₀ (same) | V₀ → V₀/K (decreases) |
| Charge Q | Q₀ → KQ₀ (increases K×) | Q₀ → Q₀ (same) |
| Electric field E | E₀ → E₀ (same) | E₀ → E₀/K (decreases) |
| Energy U | U₀ → KU₀ (increases) | U₀ → U₀/K (decreases) |
| Force on plates | F₀ → KF₀ (increases) | F₀ → F₀/K (decreases) |
Wheatstone Bridge Condition for Capacitors
- If C₁/C₂ = C₃/C₄ → bridge balanced → no charge on middle capacitor → remove it
- After removing middle capacitor: two branches in series, then parallel
Q: Two capacitors C₁ = 4μF and C₂ = 6μF connected in series across 100 V. Find charge and voltage across each.
Ceq = (4×6)/(4+6) = 2.4 μF; Q = CeqV = 2.4×10⁻⁶×100 = 240 μC (same on both); V₁ = Q/C₁ = 240/4 = 60 V; V₂ = Q/C₂ = 240/6 = 40 V
Ans: Q = 240 μC; V₁ = 60 V, V₂ = 40 V ✓
Q: Capacitor C = 10 μF charged to V = 100 V. Battery disconnected. Dielectric K = 5 inserted. Find new V, E, and U.
Q = const = 1000 μC; Cnew = 5×10 = 50 μF; Vnew = Q/Cnew = 1000/50 = 20 V; Enew = E₀/K (decreases); Unew = Q²/(2Cnew) = 10⁶/(2×50×10⁻⁶) = 10⁻²J vs original ½×10×10⁻⁶×10⁴ = 5×10⁻²J → decreases to 1/K
Ans: V = 20 V; E decreases; U decreases to U₀/5 ✓
Q: A charged body is suspended between parallel plates. Find the angle of suspension.
Electric force = qE = qσ/ε₀; gravity = mg; tan θ = qE/mg = qσ/(mgε₀); θ = tan⁻¹(qσ/mgε₀)
Ans: θ = tan⁻¹(qσ/mgε₀) ✓ (note: does NOT depend on individual charges of plates)