A complete, in-depth treatment — from Coulomb's Law to Capacitors — crafted for JEE Advanced
The foundational chapter — from quantisation of charge to Gauss's Law and field lines of complex distributions.
Electric charge is an intrinsic property of matter. Two kinds exist — positive (proton, +e) and negative (electron, −e). Like charges repel; unlike attract.
Every observable charge \(q\) is an integer multiple of the elementary charge \(e\):
This means charge comes in discrete packets. You cannot have \(0.5e\) of charge (quarks carry fractional charge but are never found free — confinement). For JEE, always treat charge as quantised.
The total electric charge of an isolated system is conserved. Charge can neither be created nor destroyed — only transferred. In pair production: \(\gamma \to e^+ + e^-\) — net charge remains zero.
Total charge of a system = algebraic sum of individual charges. \(Q = q_1 + q_2 + q_3 + \cdots\)
Unlike mass, charge does not change with velocity (Lorentz invariant). A moving charge still has the same magnitude as at rest.
| Method | Mechanism | Charge Transfer |
|---|---|---|
| Friction | Surface contact, electron transfer | Opposite charges appear on both bodies |
| Conduction | Physical contact with charged body | Same sign charge appears on both |
| Induction | Bringing charged body near (no contact) | Opposite sign on near side, same sign on far side; net charge = 0 if isolated |
The electrostatic force between two point charges \(q_1\) and \(q_2\) separated by distance \(r\) in vacuum is:
where \(k = 9\times10^9\ \text{N m}^2\text{C}^{-2}\), \(\varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}\), and \(\hat{r}_{12}\) is the unit vector from \(q_1\) to \(q_2\).
Dielectric constant \(K = \varepsilon_r \geq 1\). For water, \(K \approx 80\), so force is reduced 80×.
The total force on a charge \(q_0\) due to a system of charges \(q_1, q_2, \dots, q_n\) is the vector sum of individual forces:
Place \(q_1\) at origin \(\vec{r}_1 = \vec{0}\), \(q_2\) at \(\vec{r}_2\). Separation vector: \(\vec{r}_{12} = \vec{r}_2 - \vec{r}_1\), \(|\vec{r}_{12}| = r\).
Unit vector: \(\hat{r}_{12} = \dfrac{\vec{r}_{12}}{r}\). Force on \(q_2\) due to \(q_1\): \(\vec{F}_{12} = \dfrac{kq_1 q_2}{r^2}\hat{r}_{12}\)
By Newton's third law: \(\vec{F}_{21} = -\vec{F}_{12}\). Both forces are equal, opposite, and along the line joining the charges — confirming central force nature.
Step 1: Force magnitude between any two charges: \(F = k\dfrac{q^2}{a^2} = 9\times10^9 \times \dfrac{(4\times10^{-6})^2}{(0.2)^2} = 9\times10^9 \times \dfrac{16\times10^{-12}}{0.04} = 3.6\ \text{N}\)
Step 2: \(F_{13}\): \(q_1(+)\) and \(q_3(+)\) → repulsive, directed away from \(q_1\), at 60° to base.
\(F_{23}\): \(q_2(-)\) and \(q_3(+)\) → attractive, directed toward \(q_2\), also at 60° to base but mirrored.
Step 3: By symmetry, x-components cancel. Both y-components add (both point away from \(q_1q_2\) midpoint): \[F_{net} = 2F\cos30° = 2 \times 3.6 \times \frac{\sqrt{3}}{2} = 3.6\sqrt{3} \approx 6.24\ \text{N}\]
Direction: Along the perpendicular bisector of \(q_1q_2\), pointing away from \(q_1q_2\).
The electric field \(\vec{E}\) at a point is the force per unit positive test charge placed at that point:
Due to a point charge \(Q\) at distance \(r\):
Dipole: charges \(+q\) at \((+a,0)\) and \(-q\) at \((-a,0)\). Dipole moment: \(\vec{p} = 2qa\hat{x}\).
At axial point \(P\) at distance \(r\) from center (\(r \gg a\)): \[E_{+q} = \frac{kq}{(r-a)^2},\ E_{-q} = \frac{kq}{(r+a)^2}\] Both along \(+x\) (net) since \(E_{+q} > E_{-q}\).
\[E_{axial} = kq\left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right] = kq\cdot\frac{4ar}{(r^2-a^2)^2}\]
For \(r \gg a\): \((r^2-a^2)^2 \approx r^4\), so: \[\boxed{E_{axial} = \frac{2kp}{r^3} = \frac{p}{2\pi\varepsilon_0 r^3}}\] Direction: along \(\vec{p}\) (from −q to +q).
Point \(P\) on perpendicular bisector at distance \(r\). Distance from each charge: \(d = \sqrt{r^2+a^2}\).
Fields due to +q and −q have equal magnitude: \(E' = \dfrac{kq}{r^2+a^2}\). Perpendicular components cancel; horizontal components add: \[E_{eq} = 2E'\cos\theta = \frac{2kq}{r^2+a^2}\cdot\frac{a}{\sqrt{r^2+a^2}} = \frac{kp}{(r^2+a^2)^{3/2}}\]
For \(r \gg a\): \[\boxed{E_{eq} = \frac{kp}{r^3}}\] Direction: antiparallel to \(\vec{p}\) (from +q to −q).
Equilibrium: \(\theta=0\) (stable), \(\theta=\pi\) (unstable).
| Distribution | Element | Field Element |
|---|---|---|
| Linear (λ, C/m) | \(dq = \lambda\, dl\) | \(d\vec{E} = \dfrac{k\lambda\,dl}{r^2}\hat{r}\) |
| Surface (σ, C/m²) | \(dq = \sigma\, dA\) | \(d\vec{E} = \dfrac{k\sigma\,dA}{r^2}\hat{r}\) |
| Volume (ρ, C/m³) | \(dq = \rho\, dV\) | \(d\vec{E} = \dfrac{k\rho\,dV}{r^2}\hat{r}\) |
Line charge density \(\lambda\), point P at perpendicular distance \(r\). Consider element \(dx\) at distance \(x\) from foot of perpendicular. Distance from element to P: \(\ell = \sqrt{r^2+x^2}\).
By symmetry, axial components cancel. Perpendicular (radial) components add: \[dE_\perp = \frac{k\lambda\,dx}{r^2+x^2}\cdot\frac{r}{\sqrt{r^2+x^2}} = \frac{k\lambda r\,dx}{(r^2+x^2)^{3/2}}\]
Integrate from \(-\infty\) to \(+\infty\): using \(\int_{-\infty}^{\infty}\dfrac{dx}{(r^2+x^2)^{3/2}} = \dfrac{2}{r^2}\): \[\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r}}\]
Flux through a closed surface counts net field lines emerging. Outward flux is positive, inward is negative.
The total electric flux through any closed surface (Gaussian surface) is equal to the net charge enclosed divided by \(\varepsilon_0\):
Outside (r > R): Spherical Gaussian surface of radius r. By symmetry, \(E\) is radial and uniform on surface. \(Q_{enc} = \frac{4}{3}\pi R^3\rho = Q\) \[E\cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \implies \boxed{E = \frac{Q}{4\pi\varepsilon_0 r^2}} \quad (r > R)\] Behaves like a point charge.
Inside (r < R): \(Q_{enc} = \rho \cdot \frac{4}{3}\pi r^3 = Q\cdot\frac{r^3}{R^3}\) \[E\cdot 4\pi r^2 = \frac{Q r^3}{\varepsilon_0 R^3} \implies \boxed{E = \frac{Qr}{4\pi\varepsilon_0 R^3}} \quad (r < R)\] \(E \propto r\) inside — linear increase!
Maximum field at surface \(r = R\): \(E_{max} = \dfrac{Q}{4\pi\varepsilon_0 R^2} = \dfrac{\sigma}{\varepsilon_0}\) (where σ = surface charge density).
Choose Gaussian surface: a cylinder (pillbox) with cross-section area A, with flat faces on either side of the sheet. The curved surface has \(\vec{E} \perp d\vec{A}\), so no contribution.
\(Q_{enc} = \sigma A\). Flux through both flat faces = \(2EA\). \[2EA = \frac{\sigma A}{\varepsilon_0} \implies \boxed{E = \frac{\sigma}{2\varepsilon_0}}\] Field is uniform and perpendicular to sheet, independent of distance!
Field inside a spherical shell is zero — this is the basis of electrostatic shielding.
Total charge Q: \(Q = \int_0^R \rho\cdot 4\pi r^2 dr = \int_0^R \frac{\rho_0 r}{R}\cdot 4\pi r^2 dr = \frac{4\pi\rho_0}{R}\cdot\frac{R^4}{4} = \pi\rho_0 R^3\)
For r > R: \(Q_{enc} = Q = \pi\rho_0 R^3\) \[E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{\pi\rho_0 R^3}{4\pi\varepsilon_0 r^2} = \frac{\rho_0 R^3}{4\varepsilon_0 r^2}\]
For r < R: \(Q_{enc} = \int_0^r \frac{\rho_0 r'}{R}\cdot 4\pi r'^2 dr' = \frac{4\pi\rho_0}{R}\cdot\frac{r^4}{4} = \frac{\pi\rho_0 r^4}{R}\) \[E\cdot 4\pi r^2 = \frac{\pi\rho_0 r^4}{\varepsilon_0 R} \implies \boxed{E = \frac{\rho_0 r^2}{4\varepsilon_0 R}}\]
Note: \(E \propto r^2\) inside (instead of \(r\) for uniform charge).
From the energy landscape of charges to the engineering marvel of capacitors — energy storage, dielectrics, and everything in between.
The electric potential \(V\) at a point is the work done per unit positive charge in bringing a test charge from infinity to that point, without acceleration:
Work done in moving \(dq_0\) from \(r\) to \(r+dr\) against field: \(dW = -F\,dr = -\dfrac{kQq_0}{r^2}dr\)
Total work from \(\infty\) to \(r\): \(W = -\int_\infty^r \frac{kQq_0}{r'^2}dr' = kQq_0\left[\frac{1}{r'}\right]_\infty^r = \frac{kQq_0}{r}\)
\[V = \frac{W}{q_0} = \boxed{\frac{kQ}{r} = \frac{Q}{4\pi\varepsilon_0 r}}\]
Unlike electric field, potential is a scalar — superposition is algebraic, not vector addition. This makes problems much easier!
Point P at distance r from center, at angle θ with dipole axis. Distances from ±q: \(r_+ \approx r - a\cos\theta\), \(r_- \approx r + a\cos\theta\) (for r ≫ a).
\[V = k\left(\frac{q}{r_+} + \frac{-q}{r_-}\right) = kq\left(\frac{1}{r-a\cos\theta} - \frac{1}{r+a\cos\theta}\right)\]
\[= kq\cdot\frac{2a\cos\theta}{r^2-a^2\cos^2\theta} \approx \frac{k\cdot 2qa\cos\theta}{r^2} = \boxed{\frac{kp\cos\theta}{r^2}}\]
Check: Axial (θ=0): \(V = kp/r^2\). Equatorial (θ=90°): \(V = 0\). ✓
The electric field points in the direction of decreasing potential. Equipotential surfaces are perpendicular to field lines.
(a) Potential at center: Distance from each corner to center = \(\dfrac{a\sqrt{2}}{2} = \dfrac{\sqrt{2}}{2}\ \text{m}\)
\(V_{center} = 4 \times \dfrac{kq}{a/\sqrt{2}} = 4 \times \dfrac{9\times10^9 \times 10^{-6}}{\sqrt{2}/2} = 4 \times 9000\sqrt{2} \approx 50,912\ \text{V}\)
Exact: \(V = \dfrac{4kq\sqrt{2}}{a} = \dfrac{4 \times 9\times10^9 \times 10^{-6} \times \sqrt{2}}{1} = 36\sqrt{2}\times10^3 \approx 50.9\ \text{kV}\)
(b) Total PE: Count pairs: 4 sides (length a) + 2 diagonals (length a√2). \[U = 4\cdot\frac{kq^2}{a} + 2\cdot\frac{kq^2}{a\sqrt{2}} = \frac{kq^2}{a}\left(4 + \sqrt{2}\right)\] \[= \frac{9\times10^9 \times (10^{-6})^2}{1}(4+\sqrt{2}) = 9\times10^{-3}(4+\sqrt{2}) \approx 48.7\ \text{mJ}\]
\(E_x = -\dfrac{\partial V}{\partial x} = -6xy\) → at (1,2,-1): \(E_x = -12\ \text{V/m}\)
\(E_y = -\dfrac{\partial V}{\partial y} = -(3x^2 + 8y - 2z)\) → at (1,2,-1): \(E_y = -(3+16+2) = -21\ \text{V/m}\)
\(E_z = -\dfrac{\partial V}{\partial z} = -(-2y) = 2y\) → at (1,2,-1): \(E_z = 4\ \text{V/m}\)
\[\vec{E} = -12\hat{x} - 21\hat{y} + 4\hat{z}\ \text{V/m}\] \[|\vec{E}| = \sqrt{144+441+16} = \sqrt{601} \approx 24.5\ \text{V/m}\]
1. \(\vec{E} = 0\) inside a conductor in static equilibrium.
2. Any excess charge resides entirely on the surface.
3. \(\vec{E}\) just outside a conductor surface: \(E = \sigma/\varepsilon_0\), perpendicular to surface.
4. The entire conductor is an equipotential body.
5. Charge density is highest at sharp points/corners (lightning rod principle).
The interior of a hollow conductor is completely shielded from external electric fields. Even if external field changes, interior remains field-free. This is the principle behind Faraday cages.
Choose a small Gaussian pillbox straddling the surface. Area element A. Inside conductor: \(E_{in} = 0\). Outside: \(E_{out}\) normal to surface.
Flux: \(\Phi = E_{out}\cdot A + 0 = \dfrac{\sigma A}{\varepsilon_0}\) \[\boxed{E_{out} = \frac{\sigma}{\varepsilon_0}}\] Note: This is \(2\times\) the field of an isolated plane sheet (\(\sigma/2\varepsilon_0\)) because the conductor maintains \(E=0\) inside by redistribution.
Capacitance is the ability of a conductor/system to store charge per unit potential:
Two plates of area A, separated by d. Surface charge density σ = Q/A. From Gauss's law for opposite plates: \(E = \sigma/\varepsilon_0\) between plates, 0 outside.
Potential difference: \(V = E\cdot d = \dfrac{\sigma d}{\varepsilon_0} = \dfrac{Qd}{\varepsilon_0 A}\)
\[\boxed{C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}}\] Increasing A or decreasing d increases capacitance.
Inner sphere: charge +Q. Field between spheres (a < r < b): \(E = kQ/r^2\).
\[V = -\int_b^a E\,dr = kQ\int_a^b \frac{dr}{r^2} = kQ\left(\frac{1}{a} - \frac{1}{b}\right)\]
\[\boxed{C = \frac{Q}{V} = \frac{4\pi\varepsilon_0 ab}{b-a}}\] If b → ∞: \(C = 4\pi\varepsilon_0 a\) (isolated sphere).
(a) \(\dfrac{1}{C_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3+2+1}{6} = 1\) → \(C_{eq} = 1\ \mu\text{F}\)
(b) In series, same charge: \(Q = C_{eq} \cdot V = 1\times10^{-6} \times 12 = 12\ \mu\text{C}\)
(c) \(V_2 = Q/C_2 = 12\times10^{-6}/(3\times10^{-6}) = 4\ \text{V}\)
Check: \(V_1 = 6\ \text{V}\), \(V_2 = 4\ \text{V}\), \(V_3 = 2\ \text{V}\) → Sum = 12 V ✓
When charge q is already on capacitor, potential = q/C. Work to add dq more: \(dW = \dfrac{q}{C}dq\)
Total work from 0 to Q: \(U = \int_0^Q \frac{q}{C}dq = \frac{Q^2}{2C}\)
Using \(Q = CV\): \[\boxed{U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{QV}{2}}\]
This result is universal — valid for any electric field, not just inside a capacitor.
Inserting a dielectric (dielectric constant K) between plates:
The dielectric is polarised — bound surface charges appear, creating an opposing field \(E_{induced}\). The net field is reduced: \(E_{net} = E_0/K\). So for the same charge Q, the voltage is lower, meaning capacitance is higher.
| Condition | Q | V | E | C | U |
|---|---|---|---|---|---|
| Isolated (disconnected) Q = const | Same | V₀/K | E₀/K | KC₀ | U₀/K (decreases) |
| Connected to battery V = const | KQ₀ | Same | Same | KC₀ | KU₀ (increases) |
Think of it as two capacitors in series: air gap (d−t) and dielectric (t).
\[\frac{1}{C} = \frac{d-t}{\varepsilon_0 A} + \frac{t}{K\varepsilon_0 A} = \frac{1}{\varepsilon_0 A}\left(d - t + \frac{t}{K}\right)\]
\[\boxed{C = \frac{\varepsilon_0 A}{d - t + t/K} = \frac{\varepsilon_0 A}{d - t(1-1/K)}}\] If K→∞ (conductor slab): \(C = \dfrac{\varepsilon_0 A}{d-t}\) — effectively reduces gap!
Initial: \(Q = 4\times10^{-6}\times100 = 400\ \mu\text{C}\), \(U_i = \frac{1}{2}\times4\times10^{-6}\times100^2 = 20\ \text{mJ}\)
After insertion (Q = const): \(C' = KC = 4\times4 = 16\ \mu\text{F}\)
\(V' = Q/C' = 400/16 = 25\ \text{V}\) (voltage drops to V₀/K ✓)
\(U_f = \frac{Q^2}{2C'} = \frac{(400\times10^{-6})^2}{2\times16\times10^{-6}} = \frac{16\times10^{-8}}{32\times10^{-6}} = 5\ \text{mJ}\)
Energy difference: \(\Delta U = 20 - 5 = 15\ \text{mJ}\)
Explanation: The "lost" energy goes into polarising the dielectric and appears as heat or work done by the electric force in pulling the dielectric slab in.
Inner sphere surface (r = R₁): Charge Q is on inner sphere. \(\sigma_1 = Q/(4\pi R_1^2)\)
Inner surface of shell (r = R₂): By Gauss's law, field inside conductor = 0, so Gaussian surface inside the shell material encloses zero net charge. Hence charge −Q is induced on inner surface. \(\sigma_2 = -Q/(4\pi R_2^2)\)
Outer surface of shell (r = R₃): Shell was uncharged, so inner surface has −Q, outer must have +Q. \(\sigma_3 = +Q/(4\pi R_3^2)\)
Check: Field inside shell material: charges on inner and outer surface exactly cancel the contribution from inner sphere. ✓
Key principle: When connected, both spheres reach the same potential. Charge redistributes.
Total charge: \(Q = Q_1 + Q_2 = 6 - 2 = 4\ \text{nC}\)
At same potential: \(\dfrac{kq_1'}{r_1} = \dfrac{kq_2'}{r_2}\) → \(\dfrac{q_1'}{q_2'} = \dfrac{r_1}{r_2} = \dfrac{3}{12} = \dfrac{1}{4}\)
Also: \(q_1' + q_2' = 4\ \text{nC}\)
Solving: \(q_1' = 0.8\ \text{nC}\), \(q_2' = 3.2\ \text{nC}\)
Common potential: \(V = kq_1'/r_1 = 9\times10^9 \times 0.8\times10^{-9}/0.03 = 240\ \text{V}\)
Work done by field: \(W = q(\vec{E}\cdot\vec{d}) = q(E_x\cdot\Delta x + E_y\cdot\Delta y)\)
\(W = 2\times10^{-6}(5000\times3 + 3000\times4) = 2\times10^{-6}(15000+12000)\)
\(W = 2\times10^{-6}\times27000 = 54\ \text{mJ}\)
Note: Electrostatic field is conservative — path doesn't matter, only start and end points.
Step 1: By symmetry, perpendicular components cancel. Only axial component survives.
Step 2: Distance from element to P: \(\ell = \sqrt{R^2+x^2}\). Component along axis: \(\cos\theta = x/\ell\).
\[dE_{axial} = \frac{k\,dq}{\ell^2}\cos\theta = \frac{k\,dq}{(R^2+x^2)}\cdot\frac{x}{\sqrt{R^2+x^2}}\]
\[\boxed{E = \frac{kQx}{(R^2+x^2)^{3/2}}}\]
Maximum: Set \(dE/dx = 0\): \[\frac{d}{dx}\left[\frac{x}{(R^2+x^2)^{3/2}}\right] = 0 \implies (R^2+x^2)^{3/2} = x\cdot\frac{3}{2}(R^2+x^2)^{1/2}\cdot 2x\] \[R^2+x^2 = 3x^2 \implies \boxed{x = \frac{R}{\sqrt{2}}}\]
\(E_{max} = \dfrac{kQ\cdot R/\sqrt{2}}{(3R^2/2)^{3/2}} = \dfrac{kQ}{R^2}\cdot\dfrac{1}{3\sqrt{3}/2\sqrt{2}} = \dfrac{2kQ}{3\sqrt{3}R^2}\)
Step 1: Treat disc as concentric rings. Ring of radius r, width dr: \(dq = \sigma\cdot 2\pi r\,dr\).
Step 2: Field from this ring at P: \(dE = \dfrac{k\cdot dq\cdot x}{(r^2+x^2)^{3/2}}\)
\[E = \int_0^R \frac{kx\cdot\sigma\cdot 2\pi r\,dr}{(r^2+x^2)^{3/2}}\]
Let \(u = r^2+x^2\), \(du = 2r\,dr\): \[E = k\sigma\pi x\int_{x^2}^{R^2+x^2}\frac{du}{u^{3/2}} = k\sigma\pi x\cdot\left[-\frac{2}{\sqrt{u}}\right]_{x^2}^{R^2+x^2}\]
\[\boxed{E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{x}{\sqrt{R^2+x^2}}\right)}\]
Limit R → ∞: \(E \to \dfrac{\sigma}{2\varepsilon_0}\) — infinite plane sheet result ✓
Limit x → 0: \(E \to \dfrac{\sigma}{2\varepsilon_0}\) — field just above disc surface ✓