Electric Charges and Fields — Coulomb's Law, Field Lines, Gauss's Law

The Starting Point of Electrostatics

This chapter is where Class 12 physics begins, and it sets the stage for everything in electrostatics. Charges create electric fields, fields exert forces, and Gauss’s law gives us a powerful shortcut for symmetric charge distributions. For CBSE boards, this chapter is worth 8-10 marks. In JEE Main, 1-2 questions appear every session.

graph TD
    A[Charge Configuration] --> B{Symmetry?}
    B -->|Spherical| C[Gauss's Law with sphere]
    B -->|Cylindrical| D[Gauss's Law with cylinder]
    B -->|Planar| E[Gauss's Law with pillbox]
    B -->|No symmetry| F[Direct Coulomb + superposition]
    C --> G[E = kQ/r² outside]
    D --> H[E = λ/2πε₀r]
    E --> I[E = σ/2ε₀]
    F --> J[Vector sum of fields]

Key Terms & Definitions

Electric Charge — A fundamental property of matter. Two types: positive and negative. Like charges repel, unlike charges attract. SI unit: coulomb (C).

Quantization of Charge — Charge always comes in integer multiples of the electron charge $e = 1.6 \times 10^{-19}$ C. So $q = ne$ .

Conservation of Charge — Total charge in an isolated system is constant. Charge can transfer between objects but cannot be created or destroyed.

Electric Field — The force per unit positive test charge at a point: $\vec{E} = \vec{F}/q_0$ . It’s a vector quantity with units N/C or V/m.

Electric Field Lines — Imaginary lines showing the direction of force on a positive charge. They start on positive charges and end on negative charges. Never cross each other.

Essential Formulas

F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r^2} = k \cdot \frac{q_1 q_2}{r^2}

where $k = 9 \times 10^9$ N m $^2$ /C $^2$ and $\varepsilon_0 = 8.854 \times 10^{-12}$ C $^2$ /N m $^2$ .

In a medium with dielectric constant $\kappa$ : $F = \frac{kq_1 q_2}{\kappa r^2}$ (force reduces by factor $\kappa$ ).

Configuration	Field Expression
Point charge	$E = \frac{kq}{r^2}$
Infinite line charge ( $\lambda$ )	$E = \frac{\lambda}{2\pi\varepsilon_0 r}$
Infinite plane ( $\sigma$ )	$E = \frac{\sigma}{2\varepsilon_0}$
Uniformly charged sphere (outside)	$E = \frac{kQ}{r^2}$
Uniformly charged sphere (inside)	$E = \frac{kQr}{R^3}$

\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_0}

The total electric flux through any closed surface equals the net charge enclosed divided by $\varepsilon_0$ . Powerful when symmetry exists.

Dipole moment: $\vec{p} = q \cdot \vec{d}$ (from negative to positive charge).

Field on axial line (far from dipole): $E = \frac{2kp}{r^3}$

Field on equatorial line: $E = \frac{kp}{r^3}$

Torque on dipole in field: $\vec{\tau} = \vec{p} \times \vec{E}$

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE)

Two charges $+3\mu$ C and $-3\mu$ C are 20 cm apart. Find the force between them.

F = k\frac{q_1 q_2}{r^2} = 9 \times 10^9 \times \frac{3 \times 10^{-6} \times 3 \times 10^{-6}}{(0.2)^2} = \frac{81 \times 10^{-3}}{0.04} = \mathbf{2.025 \text{ N}}

Example 2 (Medium — JEE Main)

Find the electric field at the centre of a square of side $a$ with charges $+q, +q, -q, -q$ at the corners (alternating).

By symmetry, fields from the two $+q$ charges point in the same direction (along a diagonal), and fields from the two $-q$ charges also point in the same direction (along the same diagonal). All four contribute equally.

Distance from corner to centre $= \frac{a}{\sqrt{2}}$ . Each field $= \frac{kq}{a^2/2} = \frac{2kq}{a^2}$ .

All four components add along the diagonal: $E = 4 \times \frac{2kq}{a^2} \times \frac{1}{\sqrt{2}} \cdot \sqrt{2} = \frac{4\sqrt{2}kq}{a^2}$

Example 3 (Hard — JEE Advanced)

A uniformly charged ring of radius $R$ and charge $Q$ . Find the electric field on the axis at distance $x$ from the centre.

Each element $dq$ contributes a field $dE = \frac{k \cdot dq}{(R^2 + x^2)}$ . By symmetry, perpendicular components cancel. The axial component:

E = \frac{kQx}{(R^2 + x^2)^{3/2}}

Maximum at $x = R/\sqrt{2}$ (differentiate and set to zero).

Exam-Specific Tips

CBSE Board: Expect derivations — Gauss’s law for a uniformly charged sphere and for an infinite plane sheet are frequently asked (5-mark questions). Also practise dipole field derivations and torque on a dipole.

JEE Main: Superposition problems (multiple point charges) and Gauss’s law applications are common. The charged ring field formula appears directly in questions.

Common Mistakes to Avoid

Mistake 1 — Forgetting the vector nature of electric field. When adding fields from multiple charges, you must do vector addition, not scalar addition. Draw the field vectors first.

Mistake 2 — Using Gauss’s law without symmetry. Gauss’s law is always true, but it’s useful for calculation only when symmetry lets you pull $E$ out of the integral.

Mistake 3 — Confusing flux with field. Flux is $\Phi = \vec{E} \cdot \vec{A}$ , a scalar. Field is $\vec{E}$ , a vector. Zero flux doesn’t mean zero field.

Mistake 4 — Wrong direction of dipole moment. The dipole moment vector $\vec{p}$ points from the negative charge to the positive charge. Many students reverse this.

Mistake 5 — Using the $1/r^3$ dipole formula too close. The far-field dipole approximation ( $E \propto 1/r^3$ ) is valid only when $r \gg d$ (distance between charges). At points comparable to $d$ , use exact Coulomb calculation.

Practice Questions

Q1. Find the electric field at a point 0.3 m from a charge of $5\mu$ C.

$E = kq/r^2 = 9 \times 10^9 \times 5 \times 10^{-6} / 0.09 = 5 \times 10^5$ N/C.

Q2. Two equal and opposite charges of 2 nC are 1 mm apart. Find the dipole moment.

$p = qd = 2 \times 10^{-9} \times 10^{-3} = 2 \times 10^{-12}$ C m.

Q3. An infinite plane has surface charge density $\sigma = 10^{-6}$ C/m $^2$ . Find the electric field.

$E = \sigma/(2\varepsilon_0) = 10^{-6}/(2 \times 8.854 \times 10^{-12}) \approx 5.65 \times 10^4$ N/C.

Q4. A charge $Q$ is at the centre of a cube. What is the flux through one face?

Total flux $= Q/\varepsilon_0$ . By symmetry, each of 6 faces gets $Q/(6\varepsilon_0)$ .

Q5. Find the torque on a dipole of moment $5 \times 10^{-6}$ C m in a field of $10^6$ N/C at $30°$ .

$\tau = pE\sin\theta = 5 \times 10^{-6} \times 10^6 \times 0.5 = 2.5$ N m.

Q6. Inside a uniformly charged solid sphere of radius $R$ and total charge $Q$ , find $E$ at distance $r$ from centre.

$E = kQr/R^3$ . The field increases linearly from centre to surface.

Q7. Three charges $+q$ , $+q$ , $+q$ are at vertices of an equilateral triangle of side $a$ . Find the field at the centroid.

By symmetry, the three field vectors at the centroid make $120°$ angles and have equal magnitudes. Their vector sum is zero. $E = 0$ .

Q8. A proton and an electron are released from rest in a uniform field. Which has greater acceleration?

Both experience the same force ( $F = qE$ ). Electron has much smaller mass, so $a_e = eE/m_e \gg a_p = eE/m_p$ . Electron accelerates much faster.

FAQs

Why do electric field lines never cross?

At any point, the field has a unique direction. If two lines crossed, the field would have two directions at that point — impossible.

What is the difference between electric field and electric potential?

Field is force per unit charge (vector). Potential is energy per unit charge (scalar). $E = -dV/dr$ .

Can Gauss’s law find the field for any charge distribution?

Gauss’s law is always valid, but it gives a useful answer only when symmetry lets you simplify the surface integral. For irregular distributions, use Coulomb’s law with superposition.

What is the significance of the dielectric constant?

A dielectric medium reduces the electrostatic force and field by a factor $\kappa$ . This is why capacitors use dielectrics — they reduce the field and allow more charge to be stored.

How does quantization of charge affect everyday life?

At macroscopic scales, the charge quantum ( $1.6 \times 10^{-19}$ C) is so tiny that charge appears continuous. Quantization matters at the atomic/molecular level.

Additional Concepts

Electric flux

Electric flux through a surface is $\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta$ . For a closed surface, Gauss’s law gives $\Phi = q_{enc}/\varepsilon_0$ .

Key insight: the flux through a closed surface depends ONLY on the charge inside, not on charges outside. External charges contribute zero net flux (field lines enter and exit equally).

Continuous charge distributions

Distribution	Linear ( $\lambda$ )	Surface ( $\sigma$ )	Volume ( $\rho$ )
Definition	Charge/length (C/m)	Charge/area (C/m $^2$ )	Charge/volume (C/m $^3$ )
Element	$dq = \lambda\,dl$	$dq = \sigma\,dA$	$dq = \rho\,dV$

Potential energy of a dipole in a field

U = -\vec{p} \cdot \vec{E} = -pE\cos\theta

Stable equilibrium: $\theta = 0°$ ( $p$ parallel to $E$ ), $U = -pE$ (minimum).

Unstable equilibrium: $\theta = 180°$ , $U = +pE$ (maximum).

Work done to rotate from $\theta_1$ to $\theta_2$ : $W = pE(\cos\theta_1 - \cos\theta_2)$ .

The torque $\tau = pE\sin\theta$ is maximum at 90° but the PE is zero there. At 0°, torque is zero but PE is minimum. These complementary facts are favourite MCQ traps.

Additional Practice Questions

Q9. A charge of 10 $\mu$ C is at the centre of a hemisphere of radius 0.1 m. Find the flux through the curved surface.

Total flux through a complete sphere = $Q/\varepsilon_0$ . A hemisphere has half the closed surface, but the flat base also has flux. By symmetry, flux through curved surface = $Q/(2\varepsilon_0) = 10 \times 10^{-6}/(2 \times 8.854 \times 10^{-12}) = 5.65 \times 10^5$ N m $^2$ /C.

Q10. Two point charges $+q$ and $-q$ are separated by distance $d$ . Find the electric field at the midpoint.

Both fields point in the same direction (from $+q$ toward $-q$ ). Each has magnitude $E = kq/(d/2)^2 = 4kq/d^2$ . Total: $E = 8kq/d^2 = \frac{2q}{\pi\varepsilon_0 d^2}$ .

Conductors and Insulators in Electrostatics

Inside a conductor, the electric field is zero in the electrostatic equilibrium. All excess charge resides on the surface. This is why a car (metal body) protects passengers during lightning — the charge flows on the outside, and the field inside is zero (Faraday cage effect).

Electrostatic shielding: A hollow conductor shields its interior from external fields. This principle is used in sensitive electronic equipment and in coaxial cables.

Electric field at the surface of a conductor: $E = \sigma/\varepsilon_0$ (perpendicular to the surface), where $\sigma$ is the local surface charge density. Charge accumulates more at sharper points (higher curvature), which is why lightning rods are pointed — they concentrate charge and ionise the surrounding air.

CBSE 2023 asked: “Show that the electric field inside a charged conductor is zero.” This is a 3-mark derivation using Gauss’s law with a Gaussian surface just inside the conductor surface.

Superposition principle

The electric field at any point due to multiple charges is the vector sum of fields due to each charge individually. This works because Coulomb’s law is linear — doubling a charge doubles the force.

\vec{E}_{total} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \cdots

For continuous distributions, the sum becomes an integral.