Unit 9: Electrostatics & Capacitance

Electrostatics is the single largest topic in JEE — typically 10–14 % of Main and even more in Advanced. Two students with identical mechanics scores can be separated by tens of marks if one has internalised Gauss's law and the other hasn't. Expect:

• JEE-Main: One question on Coulomb / superposition, one on Gauss (sphere, sheet or shell), one on potential/dipole, one on a capacitor circuit (series–parallel with dielectric slab).
• JEE-Advanced: Multi-step problems combining field, force, work, and dielectric energy — often with image charges or a moving conductor.

Three skeletons drive the entire chapter:

1. $E = kq/r^2$ (point charge), with $k = 1/(4\pi\varepsilon_0) \approx 9\times 10^9$ N·m²/C².
2. $\oint \vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$ (Gauss).
3. $E = -\nabla V$, $V = kq/r$.

Master these, and 90 % of problems collapse to applying the right symmetry.

Concept Map

ELECTROSTATICS
│
├── Coulomb's Law:  F = kq₁q₂/r²
│
├── Electric Field
│     ├── Point:  E = kq/r²
│     ├── Dipole: axial E = 2kp/r³, equat. E = kp/r³
│     ├── Continuous: ring, disk, wire, sheet
│     └── Gauss → sphere, cylinder, sheet
│
├── Potential
│     ├── Point: V = kq/r
│     ├── Dipole: V = kp cosθ / r²
│     └── E = -∇V (radial: -dV/dr)
│
├── Energy
│     ├── Two charges:  U = kq₁q₂/r
│     ├── Dipole in field:  U = -p·E
│     └── Self-energy of charge distributions
│
├── Conductors
│     ├── E = 0 inside, σ/ε₀ just outside
│     ├── Surface is equipotential
│     └── Image charges
│
└── Dielectrics
       ├── Polarisation P, χ_e, κ = 1 + χ_e
       ├── E_in = E_0/κ inside dielectric
       └── Capacitor with slab

CAPACITORS
│
├── Parallel plate: C = ε₀A/d
├── Spherical: C = 4πε₀ R_a R_b/(R_b−R_a)
├── Cylindrical: C = 2πε₀L/ln(R_b/R_a)
│
├── Combinations
│     ├── Series: 1/C_eq = Σ 1/C
│     └── Parallel: C_eq = Σ C
│
├── Energy
│     ├── U = ½CV² = Q²/(2C)
│     └── Density u = ½ε₀E²
│
└── RC Charging:  q(t) = q₀(1 − e^(-t/RC))

Topic 1: Coulomb's Law and the Electric Field

Sub-topic A: Coulomb's Law (Vector Form)

The force on charge $q_1$ at $\vec r_1$ due to $q_2$ at $\vec r_2$:

where $\vec r_{12} = \vec r_1 - \vec r_2$ (from $q_2$ to $q_1$). Like charges repel, unlike attract.

In a medium of relative permittivity $\kappa$ (a.k.a. dielectric constant), the force is reduced by factor $\kappa$:

Sub-topic B: Principle of Superposition

The net force on a charge due to an assembly is the vector sum of the individual Coulomb forces. Coulomb's law is linear — no three-body terms.

Sub-topic C: Electric Field of a Point Charge

Force on a test charge: $\vec F = q\vec E$.

Sub-topic D: Electric Field of an Electric Dipole

A dipole = two equal-and-opposite charges $\pm q$ separated by distance $2a$. Dipole moment $\vec p = q(2\vec a)$, pointing from $-q$ to $+q$, magnitude $p = 2qa$.

Axial point (distance $r$ along the axis, $r \gg a$):

For $r \gg a$: $E_{\text{axial}} \approx 2kp/r^3$, along $\vec p$.

Equatorial point (perpendicular bisector):

Both fields have the same magnitude $kq/(r^2+a^2)$, and their horizontal components add while vertical cancel:

For $r \gg a$: $E_{\text{eq}} \approx kp/r^3$, antiparallel to $\vec p$.

General point at angular position $\theta$ from axis, $r \gg a$:

makes angle $\alpha$ with $\vec r$ where $\tan\alpha = \tfrac{1}{2}\tan\theta$.

Sub-topic E: Continuous Charge Distributions

Replace charge $q$ by:

• linear: $dq = \lambda\,d\ell$
• surface: $dq = \sigma\,dA$
• volume: $dq = \rho\,dV$

and integrate. Standard cases:

Ring of charge $Q$, radius $R$, on its axis at distance $x$:

Maximum at $x = R/\sqrt 2$.

Disk of charge density $\sigma$, radius $R$, on its axis:

Limits: $R\to\infty$ gives infinite sheet $E = \sigma/(2\varepsilon_0)$ (independent of $x$).

Infinite straight wire of $\lambda$: $E = \lambda/(2\pi\varepsilon_0 r)$ — Gaussian cylinder derivation below.

Infinite plane sheet: $E = \sigma/(2\varepsilon_0)$ — Gaussian pillbox.

Two parallel sheets, $\pm\sigma$: field between $= \sigma/\varepsilon_0$; outside $= 0$.

Worked Problem 1

Two point charges $q$ each are placed at adjacent corners of a square of side $a$. Find the electric field at the centre.

Solution. Centre is at distance $r = a/\sqrt 2$ from each charge. Each contributes $E_0 = kq/r^2 = 2kq/a^2$, pointing away from its charge. The vector sum: the two fields are perpendicular (each pointing along the diagonal of the square). Resultant $= E_0\sqrt 2 = 2\sqrt 2 \, kq/a^2$.

Worked Problem 2 (dipole)

A dipole has $p = 2\times 10^{-29}$ C·m. Find $E$ at $r = 1$ cm along the axis.

Solution. $E = 2kp/r^3 = 2\cdot 9\times 10^9 \cdot 2\times 10^{-29}/(10^{-2})^3 = 360$ V/m.

Topic 2: Gauss's Law and Applications

Sub-topic A: Electric Flux

For a flat surface $\vec A$ in uniform $\vec E$:

For a curved surface or non-uniform field:

Sub-topic B: Gauss's Law (Statement and Proof)

Proof for a point charge enclosed in a sphere: $\vec E = kq/r^2\hat r$, $d\vec A = r^2\sin\theta\,d\theta\,d\phi\,\hat r$:

Result is independent of the surface (only depends on enclosed charge) — generalises to any closed surface and any distribution by superposition.

Sub-topic C: Applications

1. Uniformly charged spherical shell, total charge $Q$, radius $R$.

Gaussian sphere of radius $r$:

• $r > R$: $E\cdot 4\pi r^2 = Q/\varepsilon_0\Rightarrow E = kQ/r^2$ (as if charge at centre).
• $r < R$: $Q_{\text{enc}} = 0 \Rightarrow E = 0$.

2. Uniformly charged solid sphere, $\rho = 3Q/(4\pi R^3)$.

• $r > R$: same as point charge, $E = kQ/r^2$.
• $r < R$: $Q_{\text{enc}} = Q(r/R)^3\Rightarrow E = kQr/R^3$ — linear in $r$.

3. Non-uniformly charged solid sphere, $\rho(r) = \rho_0 r/R$.

$Q(r) = \int_0^r 4\pi r'^2 \rho_0 r'/R\,dr' = \pi\rho_0 r^4/R$ $E(r) = Q(r)/(4\pi\varepsilon_0 r^2) = \rho_0 r^2/(4\varepsilon_0 R)$

4. Infinite line charge $\lambda$.

Gaussian cylinder radius $r$, length $L$: $E\cdot 2\pi r L = \lambda L/\varepsilon_0\Rightarrow E = \lambda/(2\pi\varepsilon_0 r)$.

5. Infinite plane sheet, surface charge $\sigma$.

Gaussian pillbox of area $A$, half on each side: $2EA = \sigma A/\varepsilon_0\Rightarrow E = \sigma/(2\varepsilon_0)$.

6. Cylindrical symmetry (long charged cylinder of radius $R$):

• $r > R$: $E = \lambda/(2\pi\varepsilon_0 r)$ where $\lambda =$ charge per length.
• $r < R$: depends on whether charge is volume or surface distributed.

Worked Problem 3

A solid non-conducting sphere of radius $R$ has charge density $\rho(r) = \alpha r$, where $\alpha$ is a constant. Find $E$ at radial distance $r > R$ and $r < R$.

Solution. Total enclosed up to radius $r$: $Q(r) = \int_0^r 4\pi r'^2 \cdot \alpha r'\, dr' = \pi\alpha r^4$ Total $Q = \pi\alpha R^4$.

• For $r < R$: $E = Q(r)/(4\pi\varepsilon_0 r^2) = \alpha r^2/(4\varepsilon_0)$.
• For $r > R$: $E = Q/(4\pi\varepsilon_0 r^2) = \alpha R^4/(4\varepsilon_0 r^2)$.

Topic 3: Electric Potential

Sub-topic A: Definition

Work done by an external agent in moving unit positive charge from infinity to a point (against the field), against electrostatic force, with no kinetic energy gain.

For point charge with $V(\infty) = 0$:

For a discrete distribution: $V = \sum_i kq_i/r_i$ (scalar — easier than $E$ sum).

Sub-topic B: Potential of Common Distributions

Dipole, general angular point ($r\gg a$):

Ring (axis):

Disk (axis):

Spherical shell, $Q$, $R$:

• Outside ($r > R$): $V = kQ/r$.
• Inside ($r \le R$): $V = kQ/R$ — constant!

Solid sphere, uniform $\rho$:

• Outside: $V = kQ/r$.
• Inside: $V(r) = kQ(3R^2 - r^2)/(2R^3)$ — parabolic.

Sub-topic C: $\vec E = -\nabla V$

For one-dimensional / radial cases:

In an equipotential surface (constant $V$), $\vec E$ is perpendicular to the surface.

Sub-topic D: Energy of a Charge Configuration

$U = \tfrac{1}{2}\sum_i q_i V_i$ where $V_i$ is the potential at the location of $q_i$ due to all other charges.

For two charges: $U = kq_1q_2/r$.

For a dipole in external $\vec E$: $U = -\vec p\cdot\vec E$. Torque $\vec\tau = \vec p\times\vec E$. Stable equilibrium when $\vec p \parallel \vec E$, unstable when antiparallel.

Worked Problem 4

Three charges $+q, +q, -q$ are at the vertices of an equilateral triangle of side $a$. Find the work to assemble them.

Solution. Three pair-interactions: $U = \frac{k q\cdot q}{a} + \frac{kq\cdot(-q)}{a}+\frac{kq\cdot(-q)}{a} = -\frac{kq^2}{a}$

Work done by external agent equals $U$ since they all started at infinity.

Worked Problem 5

A charge $+Q$ is placed at the centre of a spherical shell of inner radius $R_1$, outer radius $R_2$, carrying total charge $-2Q$. Find $V$ at the centre.

Solution. Three contributions:

• From point charge $+Q$ at distance 0: requires special care — for the centre, use $V_{\text{point}}$ at the inner radius. Wait, the centre is the location of $+Q$ itself, so we sum the potentials from other distributions at the centre.
• Inner surface induced charge $-Q$ at radius $R_1$: $V = -kQ/R_1$.
• Outer surface charge $-Q$ at radius $R_2$: $V = -kQ/R_2$.

So $V_{\text{centre}}$ due to the shell at the centre = $-kQ/R_1 - kQ/R_2$.

(Adding the diverging self-potential of the point charge is unphysical for the test question. The asked quantity is $V$ at the centre due to other charges, i.e. due to the shell.)

Topic 4: Conductors in Electrostatic Equilibrium

Sub-topic A: Properties

1. $\vec E = 0$ inside.
2. The entire conductor (volume + surface) is at the same potential.
3. Excess charge resides on the outer surface.
4. Just outside the surface: $\vec E = \sigma/\varepsilon_0$, perpendicular to the surface (Gaussian pillbox).
5. Field discontinuity across a surface charge: $\Delta E_\perp = \sigma/\varepsilon_0$.

Sub-topic B: Cavity Problems

If a conductor has a cavity, with charge $q$ inside the cavity:

• The conductor's inner surface acquires charge $-q$ (by Gauss).
• An equal $+q$ appears on the outer surface (by charge conservation).
• Field inside the conducting bulk is zero.
• Field outside the conductor is identical to that of a point charge $+q$ at the centre of the outer surface, regardless of the cavity's position!

Sub-topic C: Method of Images

To find the field of a point charge near an infinite grounded conducting plane: replace the conductor by an image charge $-q$ at the mirror position. The actual force on the real charge is $F = -kq^2/(2d)^2$ (attractive).

For a charge $q$ at distance $d$ from a grounded sphere of radius $R$: image charge $-qR/d$ at $R^2/d$ from centre.

Worked Problem 6

A point charge $+q$ is placed at the centre of a spherical conducting shell of inner radius $a$, outer radius $b$, uncharged. Find $E$ in all three regions.

Solution.

• For $r < a$: $E = kq/r^2$ (point-charge field).
• For $a < r < b$: $E = 0$ (inside the conductor).
• For $r > b$: $E = kq/r^2$ (induced $-q$ on inner surface, $+q$ on outer surface).

Topic 5: Dielectrics

Sub-topic A: Polarisation

In a dielectric, an external field induces dipoles. The polarisation $\vec P$ (dipole moment per unit volume) satisfies $\vec P = \varepsilon_0 \chi_e \vec E$, where $\chi_e$ is electric susceptibility.

Dielectric constant (relative permittivity):

Inside a dielectric, the field is reduced:

Sub-topic B: Capacitor with Dielectric Slab

A parallel-plate capacitor of plate area $A$, separation $d$, has $C_0 = \varepsilon_0 A/d$.

Full filling with dielectric $\kappa$: $C = \kappa C_0$.

Slab of thickness $t$ in gap (rest air):

Derivation: net potential difference $V = E_{\text{air}}(d-t) + E_{\text{dielectric}}\cdot t = E_0 (d-t) + (E_0/\kappa)t = E_0(d - t + t/\kappa)$. Since $E_0 = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, get $V$ in terms of $Q$ and read off $C$.

Two dielectrics side-by-side: capacitors in parallel:

Two dielectrics stacked: capacitors in series:

Sub-topic C: Energy with Dielectric

A capacitor disconnected from battery, charge $Q$ fixed: inserting dielectric reduces energy (the system does work pulling in the dielectric):

If kept connected to battery, $V$ fixed, $C$ increases by $\kappa$, energy $\tfrac{1}{2}CV^2$ increases by $\kappa$ — battery does additional work.

Topic 6: Capacitors

Sub-topic A: Definition and Standard Geometries

A capacitor stores charge $\pm Q$ on two conductors at potential difference $V$: $C = Q/V$. Unit: Farad (1 F = 1 C/V; in practice $\mu$F, nF, pF).

Parallel-plate (area $A$, separation $d$, vacuum):

Spherical capacitor (radii $a < b$, vacuum, charge $\pm Q$):

Cylindrical capacitor (inner radius $a$, outer $b$, length $L$):

Field between: $E = \lambda/(2\pi\varepsilon_0 r)$.

Isolated sphere of radius $R$: $C = 4\pi\varepsilon_0 R = R/k$.

Sub-topic B: Series and Parallel

Series: same $Q$ on each. $V_{\text{tot}} = V_1 + V_2$:

Parallel: same $V$. $Q_{\text{tot}} = \sum Q_i$:

Sub-topic C: Energy Stored

Work done charging from $0$ to $Q$:

Energy density in the field:

This is a profound result — energy is stored in the field, not on the plates.

Sub-topic D: RC Charging

Capacitor $C$, resistor $R$, battery $\varepsilon$. KVL:

Time constant $\tau = RC$. After $\tau$, $q$ reaches $(1 - 1/e)\approx 63\%$ of max.

During discharge (battery removed):

Worked Problem 7

Three capacitors $1$, $2$, $3 \mu$F in series across a $9$ V battery. Find $Q$ on each and $V$ across each.

Solution. $1/C_{eq} = 1 + 0.5 + 0.333 = 1.833$ ⇒ $C_{eq} \approx 0.545 \mu$F. $Q = C_{eq} V = 0.545 \times 9 = 4.91 \mu$C, same on each.

$V_1 = Q/C_1 = 4.91$ V, $V_2 = 2.45$ V, $V_3 = 1.64$ V. Check: $4.91+2.45+1.64\approx 9$ V. ✓

Worked Problem 8 (JEE-Advanced)

A parallel-plate capacitor $C_0$ is connected to a battery of EMF $V_0$. With the battery still connected, a dielectric slab of dielectric constant $\kappa = 3$ is inserted to fill the gap. Find the new charge on the plates, work done by the battery, change in energy, and work done by the external agent.

Solution. Initial: $Q_0 = C_0 V_0$, $U_0 = \tfrac{1}{2}C_0 V_0^2$. With dielectric (battery still on): $C = 3C_0$, $V = V_0$, so $Q = 3C_0 V_0 = 3Q_0$. New energy: $U = \tfrac{1}{2}(3C_0)V_0^2 = 3U_0$.

Battery delivers $\Delta Q = 2Q_0$ at $V_0$: $W_{\text{batt}} = 2Q_0 V_0 = 4 U_0$. Change in capacitor energy: $\Delta U = +2U_0$.

By energy conservation: $W_{\text{batt}} = \Delta U + W_{\text{ext-out}}$, where $W_{\text{ext-out}}$ is work done on the agent (since the slab is pulled in). So $W_{\text{ext-out}} = 4U_0 - 2U_0 = 2U_0$, i.e. the agent must apply force to hold the slab back (or the slab moves in spontaneously, releasing $2U_0$).

Worked Problem 9 (parallel combinations)

Two capacitors of $4\mu$F and $6\mu$F are charged to $100$ V and $50$ V respectively. They are connected in parallel with positive plates together. Find the common potential and charge redistribution.

Solution. $Q_1 = 4\cdot 100 = 400\mu$C, $Q_2 = 6\cdot 50 = 300\mu$C. Total $Q = 700\mu$C, total $C = 10\mu$F. Common $V = 70$ V. Final charges $Q_1' = 4\cdot 70 = 280\mu$C, $Q_2' = 6\cdot 70 = 420\mu$C.

Heat dissipated in connecting: $\Delta U = \tfrac{1}{2}\cdot 4\cdot 100^2 + \tfrac{1}{2}\cdot 6\cdot 50^2 - \tfrac{1}{2}\cdot 10\cdot 70^2 = 20000 + 7500 - 24500 = 3000\,\mu$J = $3$ mJ.

Problem-Solving Heuristics

1. Use symmetry first. Before integrating, ask whether there is spherical, cylindrical or planar symmetry → Gauss.
2. Potential is a scalar. Easier to compute than field — once you have $V$, use $E = -dV/dr$.
3. For a dipole in non-uniform field, force $= (\vec p\cdot\nabla)\vec E$. JEE-Advanced.
4. Hollow conductor + interior cavity: field outside doesn't care where the charge in the cavity is, only its magnitude. Field inside the cavity depends on the cavity charge only.
5. Image charge trick is allowed for grounded planes and spheres. Otherwise compute the induced surface charge density.
6. Capacitor circuits: series/parallel reduce, then $V$ across "branches" or $Q$ on each in series.
7. With battery disconnected, $Q$ is fixed. With battery connected, $V$ is fixed.
8. Inserting a dielectric: with $Q$ fixed, $V$ drops by $\kappa$, $E$ drops by $\kappa$, energy drops by $\kappa$. With $V$ fixed, $Q$ rises by $\kappa$, $E$ unchanged, energy rises by $\kappa$.
9. Energy density $\tfrac{1}{2}\varepsilon_0 E^2$ is universal — used in capacitor and free-space arguments.
10. RC time constant $\tau = RC$. After $5\tau$ the system is essentially at steady state.
11. For dipole on equatorial line the field is antiparallel to $\vec p$; on axial line it's parallel. Easy to get wrong.
12. Always check units: $kq/r$ in volts, $kq/r^2$ in V/m.

Common Traps & Mistakes

• Signs in Coulomb's law. Don't forget that two positives or two negatives repel; opposite attract.
• Conducting sphere with point charge inside cavity — induced charge appears on inner surface; field inside conductor is zero; field outside depends on net enclosed.
• Potential is continuous across a surface charge (with finite $\sigma$), but $E$ has a jump $\sigma/\varepsilon_0$.
• Inserting dielectric while battery is connected vs disconnected — energetics differ drastically.
• Field of a dipole drops as $1/r^3$, not $1/r^2$ — the next-order moment.
• Closed Gaussian surface can be drawn anywhere — only enclosed charge matters; external charges contribute zero net flux.
• Gauss with non-spherical symmetry is useless for finding $E$; if the field isn't constant on your surface, you still know the flux but not the field pointwise.
• Equipotential surfaces never cross.
• For three or more charges, energy is the sum over pairs, not the sum of products of one charge with the potential due to all others (factor of 2 trap).
• Cylindrical capacitor formula logs ratio of radii — not difference.

Quick Revision Card

• Coulomb: $F = kq_1q_2/r^2$, $k = 9\times 10^9$ Nm²/C².
• Point charge field: $E = kq/r^2$; potential: $V = kq/r$.
• Dipole: axial $E = 2kp/r^3$; equatorial $E = kp/r^3$ (antiparallel to $p$).
• Gauss: $\oint \vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$.
• Shell: outside as point charge; inside $E=0$.
• Solid uniform sphere inside: $E = kQr/R^3$.
• Sheet: $E = \sigma/(2\varepsilon_0)$. Two-sheet: $\sigma/\varepsilon_0$ between, $0$ outside.
• Conductor: $E = 0$ inside, $\sigma/\varepsilon_0$ just outside (perpendicular).
• Capacitor: $C = \varepsilon_0 A/d$ (PP); $C = 4\pi\varepsilon_0 ab/(b-a)$ (sphere).
• Series: $1/C$ sums; parallel: $C$ sums.
• Energy: $U = Q^2/(2C) = \tfrac{1}{2}CV^2$, density $u = \tfrac{1}{2}\varepsilon_0 E^2$.
• Dielectric: $\kappa = 1+\chi_e$; slab partial fill $C = \varepsilon_0 A/(d-t+t/\kappa)$.
• RC charging: $q = q_0(1 - e^{-t/RC})$, $\tau = RC$.

Formula Sheet