Concept Core
Charge, field, potential, and capacitors — from Coulomb to stored energy.
Coulomb's Law
Force between two point charges q₁ and q₂ separated by distance r:
F = kq₁q₂/r² = q₁q₂/(4πε₀r²)
k = 9×10⁹ N·m²/C², ε₀ = 8.85×10⁻¹² C²/N·m²
Force is along line joining charges. Like charges repel, unlike attract. Obeys Newton's 3rd law.
Electric Field
Electric field E⃗ = force per unit positive test charge:
E = F/q = kQ/r² (due to point charge Q at distance r)
Field lines: leave + charges, enter − charges, never cross. Denser lines = stronger field.
For a uniform field: work done = qEd (d along field direction)
Electric Potential
Potential V = work done per unit charge to bring a positive test charge from infinity:
V = kQ/r (due to point charge Q)
Work done = q(V₂ − V₁) = qΔV
Potential is a scalar. Equipotential surfaces: no work done moving charge along them. E⃗ = −dV/dr (field is −ve gradient of potential).
Gauss's Law
Total electric flux through a closed surface = enclosed charge/ε₀:
Φ = ∮E⃗·dA⃗ = Q_enc/ε₀
Applications: inside a conductor, E = 0. On surface of spherical charge distribution, E = kQ/r². Field inside a hollow sphere = 0.
Capacitors
A capacitor stores charge: Q = CV. Capacitance C depends on geometry.
Parallel plate: C = ε₀A/d
With dielectric: C = Kε₀A/d (K = dielectric constant)
Energy stored: U = ½CV² = Q²/2C = QV/2
Capacitors in Series & Parallel
Series: 1/C_eff = 1/C₁ + 1/C₂ + ... (same charge, voltage divides)
Parallel: C_eff = C₁ + C₂ + ... (same voltage, charge adds)
Analogy: capacitors in series/parallel are OPPOSITE to resistors — parallel capacitors add directly.
Formula Vault
Electrostatics formulas — Coulomb, field, potential, and capacitors.
Coulomb's Law
F = kq₁q₂/r²
k = 9×10⁹ N·m²/C²
Electric Field
E = kQ/r²
Direction away from + charge
Electric Potential
V = kQ/r
Scalar; V = 0 at infinity
Work Done
W = q(V₁ − V₂) = qΔV
Against field: +ve work
E from V
E = −dV/dr
Field points from high to low V
Gauss's Law
Φ = Q_enc/ε₀
ε₀ = 8.85×10⁻¹² C²/Nm²
Capacitance
Q = CV
C in farads (F)
Parallel Plate Cap.
C = Kε₀A/d
K=1 vacuum; K>1 with dielectric
Energy in Capacitor
U = ½CV² = Q²/2C
Three equivalent forms
Series Capacitors
1/C = 1/C₁ + 1/C₂
Same charge; voltage divides
Parallel Capacitors
C = C₁ + C₂
Same voltage; charge adds
Electric Flux
Φ = EA cosθ
θ = angle between E and area normal
Worked Examples
5 problems — Coulomb's law, potential, capacitors, Gauss, and energy.
EasyForce between two charges: q₁=2μC, q₂=3μC, r=0.3m▾
Calculate the force between charges 2 μC and 3 μC separated by 30 cm.
1
F = kq₁q₂/r² = (9×10⁹)(2×10⁻⁶)(3×10⁻⁶)/(0.3)²
2
= 9×10⁹ × 6×10⁻¹² / 0.09 = 54×10⁻³ / 0.09 = 0.6 N
✓ F = 0.6 N (repulsive, same sign)
EasyThree 2μF capacitors in parallel — find equivalent capacitance▾
Find the equivalent capacitance of three 2 μF capacitors connected in parallel.
1
Parallel: C_eff = C₁ + C₂ + C₃ = 2 + 2 + 2 = 6 μF
✓ C_eff = 6 μF
MediumEnergy stored in a 4μF capacitor charged to 100V▾
Find the energy stored in a 4 μF capacitor charged to 100 V.
1
U = ½CV² = ½ × 4×10⁻⁶ × (100)²
2
= ½ × 4×10⁻⁶ × 10000 = ½ × 0.04 = 0.02 J = 20 mJ
✓ Energy = 20 mJ
EAPCET LevelFind E field inside and outside a uniformly charged sphere▾
A sphere of radius R has total charge Q uniformly distributed. Find E at rR (outside).
1
Outside (r > R): Apply Gauss's law with spherical surface of radius r. Q_enc = Q.
2
E × 4πr² = Q/ε₀ → E = Q/(4πε₀r²) = kQ/r² (same as point charge)
3
Inside (r < R): For uniform volume distribution, Q_enc = Q(r/R)³
4
E × 4πr² = Q(r/R)³/ε₀ → E = kQr/R³ (increases linearly with r)
5
At r = 0: E = 0. At r = R: both formulas give kQ/R² (continuous).
✓ Outside: E = kQ/r²; Inside: E = kQr/R³ (linear)
Trap QuestionA dielectric slab is inserted in a charged, isolated capacitor — does energy increase or decrease?▾
A capacitor is charged to Q and disconnected from the battery. A dielectric (K>1) is inserted. What happens to energy?
1
Isolated capacitor: Charge Q is constant (no battery to replenish).
2
Inserting dielectric: C increases to KC (where K>1).
3
Energy U = Q²/2C. Since C increased and Q is constant: U = Q²/(2KC) = U_original/K.
4
Energy decreases by factor K. The dielectric is pulled in by electrostatic force — mechanical work converts to reduced electrical energy.
✓ Energy decreases — dielectric insertion reduces U by factor K when charge is constant
Mistake DNA
4 electrostatics errors from EAPCET distractor analysis.
🔋
Capacitors in Series vs Parallel: Applying Resistor Rules
Capacitors in series combine like resistors in parallel (1/C total). Students reverse the rule.
❌ Wrong
Series capacitors:
C = C₁ + C₂ ✗
(that's parallel!)
✓ Correct
Series: 1/C=1/C₁+1/C₂ ✓
Parallel: C=C₁+C₂ ✓
Opposite to resistors
Memory trick: capacitors and resistors are OPPOSITES. Parallel capacitors add (like series resistors). Series capacitors use 1/C (like parallel resistors).
📍
Electric Field Inside a Conductor is Zero — Always
Students forget that E = 0 everywhere inside a conductor (not just at the centre). All excess charge resides on the surface.
❌ Wrong
Inside conductor:
E ≠ 0 (there must be
some field) ✗
✓ Correct
E = 0 everywhere inside ✓
All charge on surface ✓
Equipotential body ✓
Electrostatic shielding: no electric field can penetrate a conductor. The interior is shielded. This is why Faraday cages work.
⚡
Work Done Against Electric Field: Sign Error
W = q(V_A − V_B) = −q(V_B − V_A). Moving positive charge from low to high potential requires work done against the field.
❌ Wrong
Moving +q from
V=10V to V=20V:
W = q(20−10) ✗
(wrong sign; this is
work by field)
✓ Correct
Work BY external force:
W = q(V_final−V_initial)
= q(20−10) = +10q J ✓
Field does −work
Clarify: W_external = q(V_f − V_i). W_field = −q(V_f − V_i) = q(V_i − V_f). When positive charge moves to higher potential, W_external > 0 (you push against the field).
🔮
Potential is Zero Inside a Hollow Sphere — False!
Inside a hollow sphere with charge Q on surface, the field E=0 but potential V ≠ 0. V is uniform inside, equal to kQ/R.
❌ Wrong
Inside hollow sphere:
V = 0 ✗
(only E = 0 inside)
✓ Correct
E = 0 inside ✓
But V = kQ/R (constant) ✓
V ≠ 0 unless Q = 0
E = −dV/dr. If E = 0 inside, then dV/dr = 0, meaning V is constant (not zero). V inside = V at surface = kQ/R.
Chapter Intelligence
Electrostatics is the foundation of all electromagnetism. Every chapter from here builds on it.
EAPCET Weightage (2019–2024)
Capacitors (C, series/parallel, energy)~9 Coulomb's law & E field~7 Gauss's law applications~5 Dielectric in capacitors~3 High-Yield PYQ Patterns
Equivalent capacitance of combinationsEnergy stored in capacitorElectric field due to point chargeWork done moving charge between potentialsField inside/outside charged sphereDielectric effect on capacitanceForce between charges Coulomb
Exam Strategy
- Capacitor combinations: identify series (shared wire between just those two) vs parallel (both connected to same two nodes). Apply formulas accordingly.
- Energy stored: three equivalent forms — use ½CV² when V is given, Q²/2C when Q is given, QV/2 as a check.
- Gauss's law: choose a Gaussian surface matching the symmetry (sphere for point/spherical charge, cylinder for infinite line/surface).
- Remember: inside a conductor E=0, V=constant (not zero). Inside a hollow sphere E=0, V=kQ/R (also not zero).
- This chapter leads directly to Current Electricity (potential difference drives current) and then Magnetism (moving charges create fields).