Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. F = kq₁q₂/r², where k = 9×10⁹ N·m²/C².

What is the difference between electric potential and electric potential energy?

Electric potential (V) at a point is the work done per unit positive charge to bring a test charge from infinity to that point: V = W/q. Electric potential energy (U) is the work done to bring a charge q from infinity to that point: U = qV. Potential is a property of the field; potential energy depends on the charge placed.

What happens to capacitance when a dielectric is inserted?

When a dielectric of dielectric constant K is inserted between the plates of a capacitor, the capacitance increases by a factor K: C' = KC. This is because the dielectric reduces the electric field inside (by creating an opposing internal field), which lowers the voltage for the same charge, thus increasing C = Q/V.

How do you find the equivalent capacitance of capacitors in series?

For capacitors in series, the reciprocals of individual capacitances add: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃. The equivalent capacitance is always less than the smallest individual capacitance.

Tamil Nadu Class 12 Physics — Electrostatics Notes, Formulae & Important Questions

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Board Exam Weight: Electrostatics carries 7–10 marks in Tamil Nadu Class 12 Physics board exams (Volume 1, Chapter 1). NEET typically has 2–3 questions from this chapter each year.

1. Electric Charge and Properties

Electric charge is a fundamental property of matter. Two types exist: positive and negative.

Properties of charge:

Charge is quantised: q = ne, where n is an integer and e = 1.6×10⁻¹⁹ C
Charge is conserved: total charge in an isolated system remains constant
Like charges repel; unlike charges attract
Charge is additive: total charge = algebraic sum of all charges

Coulomb's Law

The force between two point charges q₁ and q₂ separated by distance r in vacuum:

F = k q₁q₂ / r²

where k = 1/(4πε₀) = 9×10⁹ N·m²/C² and ε₀ = 8.85×10⁻¹² C²/N·m² (permittivity of free space).

Force is along the line joining the two charges
Attractive if charges are opposite; repulsive if same sign
In a medium of dielectric constant K: F = kq₁q₂ / (Kr²)

2. Electric Field

The electric field E at a point is the force per unit positive test charge placed at that point:

E = F / q₀   (N/C or V/m)

Field due to a point charge q at distance r:

E = kq / r²

Direction: Away from positive charge; toward negative charge.

Electric Field Lines (Properties)

Start on positive charges, end on negative charges
Never intersect each other
Tangent at any point gives the field direction
Closer lines = stronger field
Number of lines proportional to charge magnitude

Electric Field — Special Cases

Configuration	Formula
Point charge	E = kq/r²
Infinite line charge (linear charge density λ)	E = λ/(2πε₀r)
Infinite plane sheet (surface charge density σ)	E = σ/(2ε₀)
Between parallel plates	E = σ/ε₀

3. Electric Potential

Electric potential V at a point is the work done by an external agent to bring a unit positive charge from infinity to that point (without acceleration):

V = W / q₀ = kq / r   (Volts)

Relation between E and V:

E = -dV/dr

The negative sign means the field points in the direction of decreasing potential (from high V to low V).

Equipotential surfaces: Surfaces on which V is constant. The electric field is always perpendicular to equipotential surfaces.

Potential Due to a System of Charges

Since potential is a scalar, the total potential at a point due to multiple charges is the algebraic sum:

V = k(q₁/r₁ + q₂/r₂ + q₃/r₃ + ...)

Work Done Moving a Charge

W = q(V_A - V_B) = q × ΔV

4. Gauss's Law

Electric flux Φ through a surface:

Φ = E · A · cosθ   (N·m²/C)

Gauss's Law: The total electric flux through any closed surface (Gaussian surface) equals the net charge enclosed divided by ε₀:

∮ E · dA = Q_enclosed / ε₀

Applications of Gauss's Law

1. Infinite plane sheet of charge (σ = surface charge density):

E = σ / (2ε₀)    [on each side]

2. Uniformly charged sphere (outside, r > R):

E = kQ / r²    [same as point charge]

3. Uniformly charged sphere (inside, r < R):

E = kQr / R³   [increases linearly with r]

Answer

At centre (r = 0): E = 0

5. Capacitors and Capacitance

A capacitor stores electric charge and energy. Capacitance C is defined as:

C = Q / V   (Farads)

Parallel Plate Capacitor

For plates of area A separated by distance d in vacuum:

C = ε₀A / d

With a dielectric of constant K:

C = Kε₀A / d = KC₀

Energy Stored in a Capacitor

U = ½CV² = Q²/(2C) = QV/2

Capacitors in Series

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃

Same charge on each capacitor
Voltages add: V = V₁ + V₂ + V₃
C_eq < smallest C

Capacitors in Parallel

C_eq = C₁ + C₂ + C₃

Same voltage across each capacitor
Charges add: Q = Q₁ + Q₂ + Q₃
C_eq > largest C

6. Dielectrics

A dielectric is an insulating material placed between capacitor plates. The dielectric constant K (also called relative permittivity ε_r) is always K ≥ 1.

Effect of inserting dielectric	Result
Capacitance	Increases: C' = KC
Electric field	Decreases: E' = E/K
Voltage (charge fixed)	Decreases: V' = V/K
Energy stored (charge fixed)	Decreases: U' = U/K
Energy stored (voltage fixed)	Increases: U' = KU

7. Worked Numericals

Numerical 1 — Coulomb's Law

Problem: Two charges +3 μC and −4 μC are placed 30 cm apart in air. Find the force between them.

Solution

Given: q₁ = 3×10⁻⁶ C, q₂ = 4×10⁻⁶ C, r = 0.30 m, k = 9×10⁹ N·m²/C²

F = k|q₁||q₂| / r²
F = (9×10⁹ × 3×10⁻⁶ × 4×10⁻⁶) / (0.30)²
F = (9×10⁹ × 12×10⁻¹²) / 0.09
F = (108×10⁻³) / 0.09
F = 1.2 N

Nature: Attractive (opposite charges). The force pulls the two charges toward each other.

Numerical 2 — Parallel Plate Capacitor

Problem: A parallel plate capacitor has plate area 2×10⁻² m², plate separation 5 mm, and is connected to 100 V. Find (a) capacitance and (b) charge stored.

Solution

Given: A = 2×10⁻² m², d = 5×10⁻³ m, V = 100 V
ε₀ = 8.85×10⁻¹² C²/N·m²

(a) C = ε₀A/d
    C = (8.85×10⁻¹² × 2×10⁻²) / (5×10⁻³)
    C = (17.7×10⁻¹⁴) / (5×10⁻³)
    C = 3.54×10⁻¹¹ F = 35.4 pF

(b) Q = CV = 35.4×10⁻¹² × 100 = 3.54×10⁻⁹ C = 3.54 nC

Numerical 3 — Capacitors in Series

Problem: Three capacitors of 2 μF, 3 μF, and 6 μF are connected in series. Find the equivalent capacitance.

Solution

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃
1/C_eq = 1/2 + 1/3 + 1/6
1/C_eq = 3/6 + 2/6 + 1/6
1/C_eq = 6/6 = 1

C_eq = 1 μF

Note: The equivalent capacitance (1 μF) is less than the smallest capacitor (2 μF).

Numerical 4 — Energy Stored in a Capacitor

Problem: A 4 μF capacitor is charged to a potential difference of 500 V. Find the energy stored.

Solution

Given: C = 4×10⁻⁶ F, V = 500 V

U = ½CV²
U = ½ × 4×10⁻⁶ × (500)²
U = ½ × 4×10⁻⁶ × 250000
U = ½ × 1 = 0.5 J

The energy stored is 0.5 J.

8. Board Exam Model Questions

2-Mark Questions

Q1. Define electric flux and state its SI unit. Electric flux through a surface is the total number of electric field lines passing through that surface. Mathematically, Φ = E·A·cosθ. SI unit: N·m²/C (or V·m).

Q2. What is an equipotential surface? Give one property. An equipotential surface is a surface on which every point has the same electric potential. Property: No work is done in moving a charge along an equipotential surface, because ΔV = 0 → W = qΔV = 0.

Q3. Define dielectric constant. What is its value for vacuum? Dielectric constant K (or relative permittivity) of a medium is the ratio of the capacitance with the dielectric to the capacitance without it: K = C'/C. For vacuum, K = 1 (by definition).

5-Mark Questions

Q4. Derive an expression for the capacitance of a parallel plate capacitor and discuss the effect of a dielectric.

Answer

A parallel plate capacitor has two plates, each of area A, separated by distance d. The electric field between the plates is E = σ/ε₀ = Q/(ε₀A). The potential difference is V = Ed = Qd/(ε₀A). Therefore:

C = Q/V = ε₀A/d

When a dielectric of constant K is inserted, the internal field opposes the original field, reducing it to E' = E/K. The new voltage is V' = V/K, so the new capacitance is:

C' = Q/V' = Q/(V/K) = KC = Kε₀A/d

Inserting a dielectric increases capacitance by factor K.

Q5. State and explain Gauss's law. Apply it to find the field due to an infinite plane sheet of charge.

Answer

Gauss's law: The total electric flux through any closed surface = Q_enclosed/ε₀.

For an infinite plane sheet of surface charge density σ: Choose a cylindrical Gaussian surface with its axis perpendicular to the sheet and flat faces of area A on each side.

Flux through curved surface = 0 (E is parallel to it)
Flux through each flat face = EA
Total flux = 2EA = σA/ε₀

ThereforeE = σ/(2ε₀). The field is uniform and directed away from the sheet (if σ > 0).

9. NEET Relevance

Topics from this chapter that appear frequently in NEET:

Topic	NEET Frequency
Coulomb's law & superposition	Very High
Electric field due to point charges	High
Electric potential and potential energy	High
Gauss's law applications	Medium
Capacitors — series/parallel	Very High
Energy stored in capacitors	High
Effect of dielectric	Medium

NEET Tip: In numericals, watch the sign of charges carefully. For potential, use algebraic sum; for force/field magnitude, use absolute values then determine direction separately.

10. Key Formulae Summary

Quantity	Formula
Coulomb's force	F = kq₁q₂/r²
Electric field	E = kq/r²
Electric potential	V = kq/r
Gauss's law	∮E·dA = Q/ε₀
Capacitance (parallel plate)	C = ε₀A/d
Capacitance with dielectric	C = Kε₀A/d
Energy in capacitor	U = ½CV²
Series capacitors	1/C = 1/C₁ + 1/C₂
Parallel capacitors	C = C₁ + C₂

Constants: k = 9×10⁹ N·m²/C², ε₀ = 8.85×10⁻¹² C²/N·m², e = 1.6×10⁻¹⁹ C

Frequently Asked Questions

The SI unit of electric charge is the Coulomb (C). One Coulomb is defined as the charge transported by a current of one ampere in one second.