Electrostatics — Coulomb's Law, Electric Field & Potential for Class 12

Electrostatics: The Force That Holds the Universe Together

Electrostatics is the study of charges at rest — and it’s one of those chapters where the physics is genuinely beautiful once you see the pattern. The entire chapter flows from one idea: like charges repel, unlike charges attract, and we can quantify exactly how much.

For Class 12 boards, this is Chapters 1 and 2 of NCERT (Electric Charges & Fields + Electrostatic Potential and Capacitance). For JEE, electrostatics carries roughly 2-3 questions per paper in JEE Main and features heavily in JEE Advanced numericals. The good news — the concepts here build cleanly on each other, so mastering Coulomb’s Law makes everything else easier.

Let’s build your mental model from the ground up, then work through the problems that actually show up in exams.

Key Terms & Definitions

Electric Charge — A fundamental property of matter. Measured in Coulombs (C). An electron carries $-1.6 \times 10^{-19}$ C, a proton carries $+1.6 \times 10^{-19}$ C. Charge is always conserved — you can’t create or destroy it, only transfer it.

Coulomb’s Law — The force between two point charges $q_1$ and $q_2$ separated by distance $r$ :

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

where $k = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$ and $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$

Electric Field ( $\vec{E}$ ) — Force per unit positive test charge at a point. Unit: N/C or V/m.

\vec{E} = \frac{\vec{F}}{q_0}

Electric Potential ( $V$ ) — Work done per unit positive charge to bring it from infinity to that point. Unit: Volt (V = J/C). Potential is a scalar — this makes it much easier to work with than field in many problems.

Capacitance ( $C$ ) — Charge stored per unit potential difference. Unit: Farad (F).

C = \frac{Q}{V}

Dielectric — An insulating material that reduces the electric field inside a capacitor. When you insert a dielectric with constant $K$ , capacitance becomes $KC$ .

Methods & Concepts

Superposition Principle

This is the engine of every electrostatics numerical. Forces (and fields) from multiple charges simply add as vectors — each charge acts independently of the others.

For $n$ charges acting on a test charge, the net force is:

\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \cdots + \vec{F}_n

Why this matters: When charges are placed symmetrically (equilateral triangle, square corners), components often cancel. Always draw the geometry first — it saves calculation time in JEE Main’s 3-minute-per-question pace.

Electric Field Lines

Field lines tell you the direction of $\vec{E}$ at any point. Three rules you must know cold:

Lines start on positive charges, end on negative charges (or go to infinity)
Lines never cross (field has a unique direction at each point)
The density of lines gives field strength — crowded lines = stronger field

CBSE board exams regularly ask you to draw field line diagrams for two point charges (+q, +q) and (+q, -q). For two equal positive charges, lines repel and there’s a neutral point between them. For a dipole (+q, -q), lines curve from + to −. Losing marks here is avoidable — practise these diagrams.

Electric Flux and Gauss’s Law

Electric flux $\Phi_E$ through a surface is:

\Phi_E = \oint \vec{E} \cdot d\vec{A}

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}

The total flux through any closed surface equals the enclosed charge divided by $\varepsilon_0$ .

Gauss’s Law is most powerful when we choose a Gaussian surface that matches the symmetry of the charge distribution. Three standard applications:

Infinite line charge (linear charge density $\lambda$ ): Use a coaxial cylinder of radius $r$ and length $L$ .

E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \implies E = \frac{\lambda}{2\pi\varepsilon_0 r}

Infinite plane sheet (surface charge density $\sigma$ ): Use a pillbox Gaussian surface.

E = \frac{\sigma}{2\varepsilon_0}

Uniformly charged sphere (charge $Q$ , radius $R$ ):

Outside ( $r > R$ ): $E = \frac{Q}{4\pi\varepsilon_0 r^2}$ (behaves like a point charge)
Inside ( $r < R$ ): $E = \frac{Qr}{4\pi\varepsilon_0 R^3}$ (varies linearly with $r$ )

For conductors, $E = 0$ inside and $E = \sigma/\varepsilon_0$ just outside the surface. The factor of 2 difference from a plane sheet trips up students — remember, for a conductor both faces of the pillbox contribute to $\sigma$ , so you get twice the field from an equivalent insulating sheet.

Electric Potential and Potential Energy

Potential due to a point charge $q$ at distance $r$ :

V = \frac{kq}{r} = \frac{q}{4\pi\varepsilon_0 r}

For a system of charges, potential is scalar — just add algebraically (no vector headaches).

Potential difference between two points A and B:

V_A - V_B = -\int_B^A \vec{E} \cdot d\vec{l}

Electrostatic potential energy of a system of two charges:

U = k\frac{q_1 q_2}{r_{12}}

For three charges, sum all three pairs: $U = k\left(\frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}}\right)$

Capacitors

A parallel plate capacitor with plate area $A$ , separation $d$ :

C_0 = \frac{\varepsilon_0 A}{d} \quad \text{(air/vacuum)}

C = \frac{K\varepsilon_0 A}{d} \quad \text{(with dielectric constant } K\text{)}

U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{QV}{2}

Series combination: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$ (same charge, different voltages)

Parallel combination: $C_{eq} = C_1 + C_2 + \cdots$ (same voltage, different charges)

Solved Examples

Example 1 — Easy (CBSE Level)

Two charges of +4 μC and −4 μC are placed 20 cm apart. Find the electric field at the midpoint of the line joining them.

The midpoint is 10 cm = 0.1 m from each charge.

Field due to +4 μC at midpoint (pointing away from +, i.e., towards −charge side):

E_1 = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.1)^2} = \frac{36000}{0.01} = 3.6 \times 10^6 \text{ N/C}

Field due to −4 μC at midpoint (pointing towards −, i.e., in the same direction as $E_1$ ):

E_2 = 3.6 \times 10^6 \text{ N/C}

Both fields point in the same direction (from + to −), so:

E_{net} = E_1 + E_2 = 7.2 \times 10^6 \text{ N/C}

Example 2 — Medium (JEE Main Level)

A charge $Q$ is uniformly distributed over a thin ring of radius $R$ . Find the electric field at a point on the axis at distance $x$ from the centre.

Each small element $dq$ of the ring is at distance $\sqrt{x^2 + R^2}$ from the axial point.

The field component perpendicular to the axis cancels by symmetry (opposite elements cancel). Only the axial component survives:

dE_{axial} = \frac{k\,dq}{x^2 + R^2} \cdot \frac{x}{\sqrt{x^2+R^2}}

Integrating over the full ring ( $\int dq = Q$ ):

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

Check: At $x \to \infty$ , this reduces to $\frac{kQ}{x^2}$ — the ring looks like a point charge. Always verify limiting cases.

Example 3 — Hard (JEE Advanced Level)

A capacitor of capacitance $C$ is charged to potential $V_0$ and disconnected from the battery. A dielectric slab of constant $K$ is then inserted. Find the change in energy stored.

Before insertion: $Q = CV_0$ , $U_i = \frac{1}{2}CV_0^2$

After insertion, charge is conserved (battery disconnected): $Q$ remains $CV_0$ .

New capacitance: $C' = KC$

New potential: $V' = Q/C' = CV_0/KC = V_0/K$

New energy: $U_f = \frac{Q^2}{2C'} = \frac{C^2V_0^2}{2KC} = \frac{CV_0^2}{2K}$

\Delta U = U_f - U_i = \frac{CV_0^2}{2K} - \frac{CV_0^2}{2} = \frac{CV_0^2}{2}\left(\frac{1}{K} - 1\right)

Since $K > 1$ , $\Delta U < 0$ — energy decreases. The missing energy goes into the mechanical work of pulling the dielectric in (the slab is attracted into the capacitor).

This problem appeared in JEE Advanced 2019. The key trap is whether the battery is connected or disconnected. Battery connected → voltage constant, charge changes. Battery disconnected → charge constant, voltage changes. Identify this first before writing any formula.

Exam-Specific Tips

For CBSE Class 12 Boards

3-mark questions typically ask you to derive Gauss’s Law applications (infinite wire, plane sheet, sphere). Write these derivations with diagrams — CBSE awards step marks.
5-mark questions combine potential + field + capacitor energy in one problem. Show each step clearly.
Van de Graaff generator and earth-shielding of capacitors are direct 2-mark questions — memorise the one-paragraph explanations from NCERT.

For JEE Main

Electrostatics appears in 2-3 questions per paper (roughly 8-12 marks). High-yield topics based on PYQ analysis:

Capacitor combinations with dielectric (appeared JEE Main Jan 2024, Apr 2023, Feb 2022)
Electric field on axis of dipole vs equatorial line
Charge distribution on conductors + energy stored
Gauss’s Law applications — field inside/outside sphere

Practice calculating $\frac{1}{4\pi\varepsilon_0}$ as $9 \times 10^9$ quickly and keep $\varepsilon_0 = 8.85 \times 10^{-12}$ handy for formula-based questions.

For JEE Advanced

JEE Advanced tests conceptual depth, not just formula plugging. Focus areas:

Energy stored in capacitors during charge redistribution
Non-uniform charge distributions and their fields
Work-energy theorem applied to charges moving in fields
Multi-conductor systems and induced charges

Common Mistakes to Avoid

Mistake 1: Forgetting vector addition for forces/fields. Coulomb’s Law gives magnitude — the direction depends on geometry. Draw a clear diagram, resolve into components, then add. Students lose 2-3 marks per question by adding magnitudes instead of doing proper vector addition.

Mistake 2: Confusing E = σ/ε₀ (conductor) with E = σ/2ε₀ (infinite sheet). A conducting surface has $E = \sigma/\varepsilon_0$ just outside it. An isolated infinite sheet has $E = \sigma/2\varepsilon_0$ on each side. The difference: a conductor has charge only on its surface with field cancelling inside.

Mistake 3: Ignoring the battery condition in capacitor problems. Always state: is the battery connected or disconnected when the change happens? Connected → $V$ constant. Disconnected → $Q$ constant. Getting this wrong guarantees a wrong answer.

Mistake 4: Treating potential as a vector. Potential is a scalar. When finding potential due to multiple charges, simply add the values algebraically — no angle, no component resolution. Students who vector-add potentials waste time and get wrong answers.

Mistake 5: Wrong sign in potential energy. $U = kq_1q_2/r$ can be negative (unlike charges) or positive (like charges). Negative PE means the system is bound — work must be done to separate them. Don’t reflexively write $U$ as positive.

Practice Questions

Q1. Two point charges +3 μC and +3 μC are placed 12 cm apart. Find the position of the neutral point (where net electric field is zero).

By symmetry, the neutral point lies on the line joining the charges, between them. Let the neutral point be at distance $x$ from the first charge.

\frac{k \cdot 3 \times 10^{-6}}{x^2} = \frac{k \cdot 3 \times 10^{-6}}{(0.12 - x)^2}

This gives $x = 0.12 - x$ , so $x = 0.06$ m = 6 cm.

The neutral point is at the midpoint. (This is always true for two equal like charges.)

Q2. A charge of 8.85 μC is placed at the centre of a cube of side 10 cm. Find the total electric flux through the cube.

By Gauss’s Law:

\Phi = \frac{Q_{enc}}{\varepsilon_0} = \frac{8.85 \times 10^{-6}}{8.85 \times 10^{-12}} = 10^6 \text{ N m}^2 \text{ C}^{-1}

The size of the cube doesn’t matter — only the enclosed charge.

Q3. Find the work done in moving a charge of 2 μC from a point at potential 100 V to a point at potential 200 V.

W = q(V_B - V_A) = 2 \times 10^{-6} \times (200 - 100) = 2 \times 10^{-4} \text{ J} = 0.2 \text{ mJ}

Q4. Three capacitors of 2 μF, 3 μF, and 6 μF are connected in series across a 12 V battery. Find the charge on each capacitor and the voltage across the 2 μF capacitor.

$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \implies C_{eq} = 1 \text{ }\mu\text{F}$

Charge on each (series → same charge): $Q = C_{eq} \times V = 1 \times 10^{-6} \times 12 = 12 \text{ }\mu\text{C}$

Voltage across 2 μF: $V = Q/C = 12 \times 10^{-6} / 2 \times 10^{-6} = 6 \text{ V}$

Q5. An electric dipole of moment $p = 10^{-8}$ C·m is placed in a uniform field $E = 10^3$ N/C at 30° to the field. Find the torque and the potential energy.

Torque: $\tau = pE\sin\theta = 10^{-8} \times 10^3 \times \sin 30° = 10^{-5} \times 0.5 = 5 \times 10^{-6}$ N·m

Potential energy: $U = -pE\cos\theta = -10^{-8} \times 10^3 \times \cos 30° = -10^{-5} \times \frac{\sqrt{3}}{2} \approx -8.66 \times 10^{-6}$ J

Q6. A parallel plate capacitor with air between plates has $C = 8$ pF. The separation is 3 mm and area is $A$ . A dielectric slab of thickness 2 mm and $K = 6$ is inserted. Find the new capacitance.

The capacitor is now two capacitors in series: air gap of 1 mm and dielectric of 2 mm.

C_1 = \frac{\varepsilon_0 A}{1 \times 10^{-3}}, \quad C_2 = \frac{6\varepsilon_0 A}{2 \times 10^{-3}} = \frac{3\varepsilon_0 A}{10^{-3}}

From original: $8 \times 10^{-12} = \frac{\varepsilon_0 A}{3 \times 10^{-3}}$ , so $\varepsilon_0 A = 24 \times 10^{-15}$ F·m.

C_1 = \frac{24 \times 10^{-15}}{10^{-3}} = 24 \text{ pF}, \quad C_2 = \frac{3 \times 24 \times 10^{-15}}{10^{-3}} = 72 \text{ pF}

\frac{1}{C_{new}} = \frac{1}{24} + \frac{1}{72} = \frac{3+1}{72} = \frac{4}{72} \implies C_{new} = 18 \text{ pF}

Q7. Two conducting spheres of radii $r_1$ and $r_2$ are connected by a long thin wire. Sphere 1 carries charge $Q$ . Find the charge on each sphere after equilibrium.

When connected, they reach the same potential. For a sphere, $V = kQ/r$ .

So $kQ_1/r_1 = kQ_2/r_2$ , which gives $Q_1/Q_2 = r_1/r_2$ .

With $Q_1 + Q_2 = Q$ :

Q_1 = Q\frac{r_1}{r_1+r_2}, \quad Q_2 = Q\frac{r_2}{r_1+r_2}

The larger sphere takes more charge — this is why charge accumulates at sharp points (small radius).

Q8. A 100 pF capacitor is charged to 100 V, then disconnected and connected to an uncharged 50 pF capacitor. Find the final voltage and energy lost.

Initial charge: $Q = 100 \times 10^{-12} \times 100 = 10^{-8}$ C

Total capacitance when connected: $C_{total} = 100 + 50 = 150$ pF

Final voltage: $V_f = Q/C_{total} = 10^{-8}/(150 \times 10^{-12}) = 66.7$ V

Initial energy: $U_i = \frac{1}{2} \times 100 \times 10^{-12} \times 100^2 = 5 \times 10^{-7}$ J

Final energy: $U_f = \frac{Q^2}{2C_{total}} = \frac{(10^{-8})^2}{2 \times 150 \times 10^{-12}} = 3.33 \times 10^{-7}$ J

Energy lost: $\Delta U = 1.67 \times 10^{-7}$ J (dissipated as heat in the connecting wire)

FAQs

Why is the electric field zero inside a conductor?

Free electrons in a conductor redistribute themselves until the net force on every charge is zero. If there were a non-zero field inside, electrons would keep moving — contradicting the assumption of electrostatic equilibrium. Any excess charge resides entirely on the surface.

What’s the difference between electric potential and electric potential energy?

Potential ( $V$ ) is a property of the point in space — it tells you the work done per unit charge. Potential energy ( $U = qV$ ) belongs to the charge placed at that point. You can have potential in empty space; you need a charge to have potential energy.

Can electric field lines be closed loops?

No — not in electrostatics. Field lines start and end on charges. In electrodynamics (changing magnetic fields), the induced electric field does form closed loops — but that’s Chapter 6 territory.

Why does inserting a dielectric increase capacitance?

The dielectric gets polarised — molecules align with the field, creating an opposing internal field. This reduces the net field between the plates, which reduces the voltage for the same charge. Since $C = Q/V$ , lower $V$ for same $Q$ means higher $C$ .

What is the significance of equipotential surfaces?

On an equipotential surface, $V$ is constant — so no work is done moving a charge along it. The electric field is always perpendicular to equipotential surfaces (if there were a component along the surface, charges would move, contradicting the constant-potential condition). For a point charge, equipotential surfaces are concentric spheres.

How do we apply Gauss’s Law to non-symmetric charge distributions?

We don’t — at least not to find $\vec{E}$ directly. Gauss’s Law is always true, but it only lets us calculate $\vec{E}$ easily when the charge distribution has spherical, cylindrical, or planar symmetry. For arbitrary distributions, we use superposition or integration.

Why is the potential energy of a system of like charges positive?

Positive PE means you had to do positive work to assemble the configuration against the repulsive force. If you released them, they’d fly apart — converting that stored PE to kinetic energy. Negative PE (unlike charges) means the system is in a bound state.

Is Coulomb’s Law valid inside a medium?

Yes, but with modification. In a medium with permittivity $\varepsilon = K\varepsilon_0$ :

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

The force is reduced by factor $K$ . For water ( $K \approx 80$ ), the Coulomb force between ions is 80 times weaker than in vacuum — which is why ionic compounds dissolve easily in water.