Electrostatics

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.

F = kq₁q₂/r² = (9×10⁹)(2×10⁻⁶)(3×10⁻⁶)/(0.3)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.

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.

U = ½CV² = ½ × 4×10⁻⁶ × (100)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).

Outside (r > R): Apply Gauss's law with spherical surface of radius r. Q_enc = Q.

E × 4πr² = Q/ε₀ → E = Q/(4πε₀r²) = kQ/r² (same as point charge)

Inside (r < R): For uniform volume distribution, Q_enc = Q(r/R)³

E × 4πr² = Q(r/R)³/ε₀ → E = kQr/R³ (increases linearly with r)

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?

Isolated capacitor: Charge Q is constant (no battery to replenish).

Inserting dielectric: C increases to KC (where K>1).

Energy U = Q²/2C. Since C increased and Q is constant: U = Q²/(2KC) = U_original/K.

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)

Coulomb's law & E field

Electric potential

Gauss's law applications

Dielectric in capacitors

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).

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

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