Chapter 1: Electric Charges and Fields

Electrostatics is the study of charges at rest. Despite its name, the concepts here — Coulomb's law, the electric field, flux and Gauss's law — are the foundation of every later chapter in Class XII (potential, capacitance, current, electromagnetism, even atomic structure). This chapter develops two parallel descriptions of the same physics:

1. Force-based: Coulomb's law tells us how two point charges push or pull each other.
2. Field-based: A charge sets up an electric field $\vec{E}$ everywhere in space; another charge placed in that field experiences a force $\vec{F} = q\vec{E}$.

The field picture is more powerful because it allows us to use Gauss's law, which converts hard integration problems (continuous distributions) into easy symmetry arguments.

Concept Map

• 1.1 Electric charge — origin, quantization, conservation, additivity
• 1.2 Coulomb's law — vector form, role of medium
• 1.3 Superposition of forces
• 1.4 Electric field — definition and units
• 1.5 Electric field of a point charge and system of charges
• 1.6 Electric field lines — properties and uses
• 1.7 Electric dipole — moment, axial field, equatorial field, general point
• 1.8 Dipole in a uniform external field — torque and potential energy
• 1.9 Continuous charge distributions — $\lambda, \sigma, \rho$
• 1.10 Electric flux $\Phi_E$
• 1.11 Gauss's law — statement and qualitative proof
• 1.12 Applications of Gauss's law — wire, sheet, parallel sheets, shell, solid sphere

1.1 Electric Charge

Definition

Electric charge is an intrinsic property of elementary particles (electrons, protons, quarks) responsible for electric and magnetic phenomena. By convention, the proton carries a positive charge $+e$ and the electron a negative charge $-e$, where

The SI unit is the coulomb (C). One coulomb is the charge transferred by a current of one ampere in one second.

Fundamental Properties of Charge

1. Additivity. Total charge of a system is the algebraic sum of individual charges: $$Q_{\text{total}} = q_1 + q_2 + \cdots + q_n.$$
2. Conservation. The net charge of an isolated system is constant. Charges can be transferred but never created or destroyed (e.g., in $\beta$-decay a neutron $\to$ proton $+ e^- + \bar{\nu}_e$; total charge $0 = (+1) + (-1) + 0$).
3. Quantization. Any observable charge is an integer multiple of the elementary charge: $$q = n e, \quad n \in \mathbb{Z}.$$ (Quarks carry $\pm e/3, \pm 2e/3$ but are never observed free.)
4. Invariance. Charge does not depend on the speed of the reference frame (unlike mass, which has a relativistic correction).

Methods of Charging

• Friction. Rubbing two neutral bodies transfers electrons (the body that loses electrons becomes $+$, the receiver becomes $-$). Example: glass rubbed with silk $\to$ glass becomes $+$.
• Conduction (contact). A charged body touched to a neutral conductor shares charge until both reach the same potential.
• Induction. Bringing a charged body near (without contact) a conductor redistributes the conductor's free electrons. If the far end is earthed momentarily and then the source removed, the conductor retains a charge opposite to the inducing body.

Worked Example

A body has a charge of $-1.0 \text{ nC}$. How many excess electrons does it carry?

Pitfalls

• Quantization is observed only on the microscopic scale; for $1\text{ C}$ the number of electrons is $\sim 6\times 10^{18}$, so charge appears continuous.
• Induced charges on a body are equal and opposite; the body remains overall neutral unless earthed.
• Mass changes when a body is charged (it gains or loses electrons), but the change is utterly negligible ($\sim 10^{-21}\text{ kg per nC}$).

1.2 Coulomb's Law

Definition

The electrostatic force between two point charges $q_1, q_2$ separated by a distance $r$ in vacuum is directed along the line joining them, with magnitude

Here $\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$ is the permittivity of free space, and

The force is attractive for unlike charges and repulsive for like charges.

Derivation — Vector Form

Let $\vec{r}_{12}$ be the position vector from charge $q_1$ to charge $q_2$. The unit vector is $\hat{r}_{12} = \vec{r}_{12}/r$.

Step 1. Force on $q_2$ due to $q_1$:

Step 2. By Newton's third law,

Step 3. The sign of the product $q_1 q_2$ encodes the nature of the force automatically:

• $q_1 q_2 > 0 \Rightarrow \vec{F}_{21}$ points along $\hat{r}_{12}$ (repulsion).
• $q_1 q_2 < 0 \Rightarrow$ opposite to $\hat{r}_{12}$ (attraction).

Comparison with Gravitation

For an electron–proton pair:

Force in a Medium

If a dielectric (relative permittivity $\varepsilon_r = K$) fills the space between the charges,

Equivalently, the effective distance in vacuum that gives the same force is $r\sqrt{K}$.

Worked Example

Two point charges $q_1 = +3\,\mu\text{C}$ and $q_2 = -5\,\mu\text{C}$ are placed $0.20\text{ m}$ apart in air. Find the force.

Pitfalls

• Coulomb's law applies strictly to point charges (or spherical charges, by shell theorem) — never to arbitrary shapes without integration.
• Always plug in magnitudes of charges in the scalar form and decide attractive/repulsive separately, or carry signs only in the vector form.
• The presence of a medium reduces the force; this is not screening but polarization of the medium.

1.3 Forces Between Multiple Charges — Superposition

Definition

The electrostatic force is linear: the force on a charge $q_0$ due to a collection $q_1, q_2, \dots, q_n$ equals the vector sum of the individual two-body Coulomb forces:

The presence of $q_3$ does not alter the force that $q_1$ exerts on $q_2$.

Derivation — Worked Vector Example

Three charges at the corners of an equilateral triangle of side $a$: $q_1 = q_2 = +q$ at the base, $q_3 = +q$ at the top. Find the net force on $q_3$.

Step 1. Each base charge exerts a force of magnitude

on $q_3$, directed away from the base charge.

Step 2. Symmetry: horizontal components cancel; vertical components add.

Step 3. Each force makes $60°$ with the vertical, so vertical component is $F_0 \cos 60° = F_0/2$. But the angle the line from each base charge to $q_3$ makes with the horizontal is $60°$, so the upward component is $F_0 \sin 60° = F_0\sqrt{3}/2$.

Step 4. Net force on $q_3$:

directed vertically (away from the base).

Worked Example

Charges $+2\,\mu\text{C}$ and $-2\,\mu\text{C}$ are at $(0,0)$ and $(0.3\,\text{m},0)$. Find the force on a third charge $+1\,\mu\text{C}$ at $(0.3\,\text{m}, 0.4\,\text{m})$.

Distance from $+2\,\mu\text{C}$ to test charge: $\sqrt{0.3^2+0.4^2}=0.5\text{ m}$. Distance from $-2\,\mu\text{C}$ to test charge: $0.4\text{ m}$.

Force from $+2\,\mu\text{C}$: $F_1 = \dfrac{(9\times10^9)(2\times10^{-6})(1\times10^{-6})}{(0.5)^2} = 0.072\text{ N}$, repulsive, along $(0.3,0.4)/0.5 = (0.6, 0.8)$. So $\vec{F}_1 = (0.0432, 0.0576)\text{ N}$.

Force from $-2\,\mu\text{C}$: $F_2 = \dfrac{(9\times10^9)(2\times10^{-6})(1\times10^{-6})}{(0.4)^2} = 0.1125\text{ N}$, attractive, so toward $(0.3,0)$, i.e. direction $(0,-1)$. So $\vec{F}_2 = (0, -0.1125)\text{ N}$.

Net: $\vec{F} = (0.0432, -0.0549)\text{ N}$, magnitude $0.0698\text{ N}$.

Pitfalls

• Forces are vectors — always resolve into components.
• Each two-body force is unaffected by other charges (no screening between point charges in vacuum).
• Distances are between the pair, never from the centroid or some other convenient point.

1.4 Electric Field

Definition

The electric field $\vec{E}$ at a point is the force experienced per unit positive test charge placed at that point, in the limit that the test charge is vanishingly small (so it does not disturb the source distribution):

Units. $\text{N C}^{-1}$ or equivalently $\text{V m}^{-1}$.

Why the Limit?

A finite test charge would polarize nearby conductors or redistribute source charges, changing the very field we want to measure. The limit removes this back-reaction.

Source vs Test Charge

• The source charges create the field (whether or not a test charge is present).
• The test charge probes the field; it experiences $\vec{F} = q_0 \vec{E}$.

Worked Example

An electron is placed in a uniform field $\vec{E} = 2 \times 10^4 \,\hat{i}\text{ N C}^{-1}$. Find its acceleration.

Pitfalls

• $\vec{E}$ is defined even where no test charge sits — it's a property of the source distribution.
• Direction: always taken from $+$ source outward or toward $-$ source.
• For a continuous distribution, $\vec{E}$ is finite even on the surface (provided no $\delta$-singular point), unlike point-charge $\vec{E}$ which diverges at the source.

1.5 Electric Field due to Point Charge and System of Charges

Definition / Derivation — Point Charge

For a point charge $q$ at the origin, the field at $\vec{r}$ is

Derivation. Place a positive test charge $q_0$ at $\vec{r}$. Coulomb's law gives

The field points radially outward for $q>0$ and inward for $q< 0$.

System of Charges (Superposition)

For charges $q_i$ at positions $\vec{r}_i$, the field at point $\vec{r}$ is

Worked Example

Two charges $+q$ at $x = -a$ and $+q$ at $x = +a$. Find $\vec{E}$ at the point $(0, y)$.

By symmetry, $x$-components cancel. Each contributes $E_0 = \dfrac{kq}{a^2+y^2}$, making angle $\theta$ with the $y$-axis where $\cos\theta = y/\sqrt{a^2+y^2}$.

For $y\gg a$: $E_y \approx 2kq/y^2$ (acts like a charge $2q$). For $y\ll a$: $E_y \approx 2kqy/a^3 \to 0$ as $y\to 0$ (point on perpendicular bisector at center is zero).

Pitfalls

• The field of a point charge diverges at its own location — a point charge is a mathematical idealization.
• For a symmetric distribution, exploit symmetry first; do not blindly integrate.

1.6 Electric Field Lines

Definition

An electric field line is a curve drawn so that its tangent at every point gives the direction of $\vec{E}$ at that point.

Properties

1. Field lines start on $+$ charges and end on $-$ charges (or go to infinity).
2. Two field lines never cross — at a crossing $\vec{E}$ would have two directions.
3. The density of lines (lines per unit perpendicular area) is proportional to $\vert \vec{E}\vert$.
4. Field lines are continuous — no breaks in free space (except at point charges).
5. Field lines do not form closed loops in electrostatics (a consequence of $\oint \vec{E}\cdot d\vec{l}=0$).
6. Lines are normal to conducting surfaces (in equilibrium).
7. The number of lines emanating from a charge is proportional to the charge.

Standard Patterns

• Single $+$ point charge: radially outward, isotropic.
• Single $-$ point charge: radially inward.
• Two equal $+$ charges: lines curve away; a neutral point (where $\vec{E}=0$) lies midway.
• Equal and opposite charges (dipole): lines run from $+$ to $-$, denser near the charges, looping outside.
• Uniform field: parallel, equally spaced lines.

Worked Example

Sketch the field of a dipole consisting of $+q$ at $(-a,0)$ and $-q$ at $(+a,0)$. Identify a point where $\vec{E}$ is parallel to the $y$-axis on the line $x = 0$.

On the perpendicular bisector ($x=0$), by symmetry the field is purely along $-x$ (from $+$ to $-$). On the axis ($y=0$), the field is along $-x$ between the charges and along the same direction outside on each side (computed by adding two outward/inward radial contributions).

Pitfalls

• Field lines do not represent the trajectory of a charged particle (that would require force, but a particle has inertia and may not move along the field line).
• For a non-uniform field, lines diverge or converge — but the spacing tells you the magnitude only qualitatively.

1.7 Electric Dipole

Definition

An electric dipole consists of two equal and opposite point charges $\pm q$ separated by a small distance $2a$. The dipole moment is

directed from $-q$ to $+q$. SI unit: $\text{C m}$.

Derivation — Axial Field

Let the dipole lie on the $x$-axis with $-q$ at $-a$ and $+q$ at $+a$. Find $\vec{E}$ at a point $P$ on the axis, distance $r$ from the center, with $r > a$.

Step 1. Distance from $+q$ to $P$: $r-a$. Distance from $-q$ to $P$: $r+a$.

Step 2. Field from $+q$ at $P$ (outward, along $+\hat{x}$):

Field from $-q$ at $P$ (inward, along $-\hat{x}$):

Step 3. Net axial field along $+\hat{x}$:

Step 4. Simplify:

Step 5.

The field on the axis is parallel to $\vec{p}$.

Derivation — Equatorial Field

Point $P$ on the perpendicular bisector at distance $r$ from the center.

Step 1. Distance from each charge to $P$: $\sqrt{r^2+a^2}$.

Step 2. Magnitudes equal: $E_+ = E_- = \dfrac{kq}{r^2+a^2}$.

Step 3. Components perpendicular to the dipole axis cancel; components along the axis (anti-parallel to $\vec{p}$) add. Geometry: each field makes angle $\theta$ with the axis, where $\cos\theta = a/\sqrt{r^2+a^2}$.

Step 4.

anti-parallel to $\vec{p}$:

Note: $\vert \vec{E}_{\text{axial}}\vert = 2\vert \vec{E}_{\text{eq}}\vert$ for the same $r\gg a$.

Field at a General Point (angle $\theta$ from axis)

For $r\gg a$, with $\theta$ measured from $\vec{p}$:

• Radial component: $E_r = \dfrac{2kp\cos\theta}{r^3}$.
• Tangential component: $E_\theta = \dfrac{kp\sin\theta}{r^3}$.
• Magnitude: $E = \dfrac{kp}{r^3}\sqrt{1+3\cos^2\theta}$.

Worked Example

A dipole has $q=2\text{ nC}$, $2a = 1\text{ cm}$. Find $E$ at $20\text{ cm}$ on the axial line.

$p = q(2a) = 2\times 10^{-9}\times 10^{-2} = 2\times 10^{-11}\text{ C m}$. $r\gg a$, so $E \approx \dfrac{2kp}{r^3} = \dfrac{2(9\times 10^9)(2\times 10^{-11})}{(0.2)^3} = \dfrac{0.36}{0.008} = 45\text{ N C}^{-1}$.

Pitfalls

• $\vec{p}$ points from $-$ to $+$, not the other way.
• The axial field is along $\vec{p}$; the equatorial field is opposite to $\vec{p}$. Many students get the equatorial sign wrong.
• The $r\gg a$ "ideal dipole" formulas are approximations — for finite dipoles, use the exact expressions.

1.8 Dipole in a Uniform External Field

Definition

A dipole $\vec{p}$ in a uniform field $\vec{E}$ experiences no net force (the forces on $+q$ and $-q$ are equal and opposite) but a net torque.

Derivation — Torque

Place the dipole so $\vec{p}$ makes angle $\theta$ with $\vec{E}$. The force on $+q$ is $+q\vec{E}$ and on $-q$ is $-q\vec{E}$; these form a couple.

Step 1. Perpendicular distance between the lines of action of the two forces: $2a\sin\theta$.

Step 2. Magnitude of torque:

Step 3. Vector form (torque tends to align $\vec{p}$ with $\vec{E}$):

Derivation — Potential Energy

The work done by the field when the dipole rotates from $\theta_0$ to $\theta$ is

Define $U(\theta_0 = \pi/2)=0$ (perpendicular orientation is the reference):

• $U_{\min} = -pE$ at $\theta=0$ (stable equilibrium, $\vec{p}\parallel\vec{E}$).
• $U_{\max} = +pE$ at $\theta=\pi$ (unstable equilibrium, anti-parallel).

Worked Example

A dipole $p = 5\times 10^{-9}\text{ C m}$ makes $30°$ with $\vec{E} = 10^4\text{ N C}^{-1}$. $\tau = pE\sin 30° = (5\times 10^{-9})(10^4)(0.5) = 2.5\times 10^{-5}\text{ N m}$. $U = -pE\cos 30° = -(5\times 10^{-9})(10^4)(\sqrt{3}/2) \approx -4.33\times 10^{-5}\text{ J}$.

Pitfalls

• In a non-uniform field a dipole also experiences a net force $\vec{F} = (\vec{p}\cdot\nabla)\vec{E}$ — beyond Class XII syllabus but useful to remember.
• The reference for $U$ is conventional; do not mix two conventions.

1.9 Continuous Charge Distributions

Definition

When charges are spread continuously, point-charge Coulomb's law is replaced by integrals using densities:

The field at $\vec{r}$ is

Worked Derivation — Field on the axis of a uniformly charged ring

Ring of radius $R$, total charge $Q$, point $P$ on axis at distance $z$ from center.

Step 1. Take an element $dq$ on the ring. Distance to $P$: $\sqrt{R^2+z^2}$.

Step 2. Magnitude of $d\vec{E}$ from $dq$: $\dfrac{k\,dq}{R^2+z^2}$.

Step 3. Components perpendicular to the axis cancel by symmetry. Axial component: $dE_z = \dfrac{k\,dq}{R^2+z^2}\cdot\dfrac{z}{\sqrt{R^2+z^2}}$.

Step 4. Integrate around the ring ($\int dq = Q$):

Step 5. Limits:

• $z\to 0$: $E_z \to 0$ (center of ring, by symmetry).
• $z\gg R$: $E_z \to kQ/z^2$ (acts like point charge).
• Maximum at $z = R/\sqrt{2}$.

Pitfalls

• Always exploit symmetry first; the "perpendicular components cancel" argument saves enormous algebra.
• $dq$ depends on the element's geometry — be precise about whether to integrate over arc length, area, or volume.

1.10 Electric Flux

Definition

The electric flux through a small area element $d\vec{A}$ is

where $\theta$ is the angle between $\vec{E}$ and the outward normal $\hat{n}$ to the area. For a finite surface $S$:

For a closed surface,

Units. $\text{N m}^2\text{ C}^{-1}$ or $\text{V m}$.

Worked Example

A uniform field $\vec{E} = 200\,\hat{i}\text{ N C}^{-1}$ passes through a square of side $0.5\text{ m}$ whose normal makes $60°$ with $\hat{i}$.

$\Phi_E = E A \cos\theta = 200 \times 0.25 \times \cos 60° = 25 \text{ V m}$.

Pitfalls

• Flux is a scalar (signed); direction lives in the choice of $\hat{n}$.
• For a closed surface, $\hat{n}$ is the outward normal by convention.

1.11 Gauss's Law

Statement

The total electric flux through any closed surface $S$ (a "Gaussian surface") equals $1/\varepsilon_0$ times the net charge enclosed:

Qualitative Proof via Solid Angle

Step 1. For a point charge $q$ at the center of a sphere of radius $r$:

Step 2. The flux is independent of $r$ — the $1/r^2$ falloff of $E$ exactly cancels the $r^2$ growth of area.

Step 3. For an arbitrary closed surface enclosing $q$: divide it into elementary patches. Each subtends a small solid angle $d\Omega$ at $q$. The flux through the patch is $q\,d\Omega/(4\pi\varepsilon_0)$. Integrating $\int d\Omega = 4\pi$ over a closed surface gives $\Phi = q/\varepsilon_0$, regardless of shape or position of $q$ inside.

Step 4. Charges outside the surface contribute zero net flux (lines that enter must exit). Superposition extends the result to any charge distribution.

Key Properties

• Gauss's law is always true, but it's useful only when symmetry lets us pull $E$ out of the integral.
• The total flux depends only on the enclosed charge, not on its location inside or on charges outside.
• The field $\vec{E}$ on the Gaussian surface, however, depends on all charges.

Pitfalls

• A common error: thinking that if $q_{\text{enc}}=0$ then $\vec{E}=0$ everywhere on the surface. The integral vanishes; the field generally does not.
• Choose Gaussian surfaces respecting symmetry: sphere for point/spherical symmetry, cylinder for line symmetry, pillbox for planar symmetry.

1.12 Applications of Gauss's Law

(a) Infinite Straight Charged Wire

A wire with linear density $\lambda$. Use a coaxial cylinder of radius $r$ and length $\ell$.

Step 1. By symmetry, $\vec{E}$ is radial and depends only on $r$.

Step 2. Flux through curved surface: $E \cdot 2\pi r\ell$. Flux through end caps: zero ($\vec{E}\perp\hat{n}$).

Step 3. Enclosed charge: $\lambda\ell$.

Step 4. Gauss's law: $E \cdot 2\pi r\ell = \lambda\ell/\varepsilon_0$, so

(b) Infinite Plane Sheet of Charge

Surface density $\sigma$. Use a Gaussian "pillbox" cylinder with caps of area $A$, perpendicular to the sheet.

Step 1. Symmetry: $\vec{E}$ is normal to the sheet, equal magnitudes on each side.

Step 2. Flux: two caps contribute $EA$ each; lateral side contributes zero.

Step 3. Enclosed charge: $\sigma A$.

Step 4. Gauss's law: $2EA = \sigma A/\varepsilon_0$, so

(c) Two Parallel Sheets

Sheets with densities $+\sigma$ and $-\sigma$.

• Outside both sheets: contributions cancel; $E = 0$.
• Between the sheets: contributions add; $E = \sigma/\varepsilon_0$, directed from $+$ to $-$.

For two sheets with general densities $\sigma_1, \sigma_2$:

• Outside (beyond $\sigma_2$): $E = (\sigma_1+\sigma_2)/(2\varepsilon_0)$.
• Between: $E = (\sigma_1-\sigma_2)/(2\varepsilon_0)$.
• Outside (beyond $\sigma_1$): $E = -(\sigma_1+\sigma_2)/(2\varepsilon_0)$ (direction toward the sheets).

(d) Uniformly Charged Spherical Shell

Total charge $Q$, radius $R$.

Outside ($r>R$). Gaussian sphere of radius $r$. By symmetry $E$ radial. $E\cdot 4\pi r^2 = Q/\varepsilon_0$:

The shell looks like a point charge from outside (shell theorem).

Inside ($r<R$). Gaussian sphere encloses no charge, so

On the surface ($r=R^+$): $E = \sigma/\varepsilon_0$ where $\sigma = Q/(4\pi R^2)$.

(e) Uniformly Charged Solid Sphere

Total charge $Q$, radius $R$, uniform volume density $\rho = Q/(\tfrac{4}{3}\pi R^3)$.

Outside ($r>R$). Same as shell: $E = kQ/r^2$.

Inside ($r<R$). Enclosed charge $q_{\text{enc}} = \rho\cdot\tfrac{4}{3}\pi r^3 = Q(r/R)^3$.

Field grows linearly with $r$ inside, peaks at $E = kQ/R^2$ on the surface, then falls as $1/r^2$ outside.

Worked Example

A solid sphere of radius $10\text{ cm}$ carries uniform charge $5\,\mu\text{C}$. Find $E$ at $r = 5\text{ cm}$ and at $r = 20\text{ cm}$.

At $r = 5\text{ cm}$ (inside): $E = kQr/R^3 = (9\times 10^9)(5\times 10^{-6})(0.05)/(0.1)^3 = 2.25\times 10^{6}\text{ N C}^{-1}$.

At $r = 20\text{ cm}$ (outside): $E = kQ/r^2 = (9\times 10^9)(5\times 10^{-6})/(0.2)^2 = 1.125\times 10^{6}\text{ N C}^{-1}$.

Pitfalls

• Field of an infinite sheet is $\sigma/(2\varepsilon_0)$ — without a conducting backing.
• Field just outside a conductor's surface is $\sigma/\varepsilon_0$ (not $\sigma/(2\varepsilon_0)$); the doubling reflects the boundary condition that $\vec{E}=0$ inside the conductor.
• Inside a uniformly charged solid sphere, $\vec{E}\neq 0$; only inside a shell is $\vec{E}=0$.

Solved Problems

Problem 1 (Easy)

Two charges $+4\,\mu\text{C}$ and $-4\,\mu\text{C}$ are $5\text{ cm}$ apart. Find the dipole moment and the field at a point $50\text{ cm}$ from the center on the axial line.

Solution. $p = q(2a) = (4\times 10^{-6})(0.05) = 2\times 10^{-7}\text{ C m}$. Since $r=0.5\text{ m}\gg a$,

Problem 2 (Easy)

A point charge $+10\,\mu\text{C}$ is enclosed by a Gaussian cube. What is the flux through one face?

Solution. Total flux $\Phi = q/\varepsilon_0 = (10\times 10^{-6})/(8.85\times 10^{-12}) = 1.13\times 10^6\text{ V m}$. By symmetry (charge at center), each face has $\Phi/6 = 1.88\times 10^5\text{ V m}$.

If the charge were at a corner, the flux through each of the three faces meeting at that corner would be zero (the field is parallel to those faces); the remaining flux $q/(8\varepsilon_0)$ would split among 3 faces (corner is shared by 8 cubes).

Problem 3 (Medium)

A thin non-conducting rod of length $L$ has uniform linear density $\lambda$. Find $\vec{E}$ at a point on the perpendicular bisector at distance $d$.

Solution. Place rod along $x$-axis from $-L/2$ to $L/2$, field point at $(0,d)$.

Element $dq = \lambda\,dx$ at position $x$. Distance to field point: $\sqrt{x^2+d^2}$. By symmetry $E_x = 0$ (cancellation). The $y$-component is

Integrate:

For $L\to\infty$: $E_y \to 2k\lambda/d = \lambda/(2\pi\varepsilon_0 d)$, recovering the infinite wire.

Problem 4 (Medium)

A charge $+Q$ is placed at the center of a hollow conducting spherical shell of inner radius $a$ and outer radius $b$. Find $E$ at $r<a$, $a<r<b$, and $r>b$. Also find induced charges on inner and outer surfaces.

Solution. For $r<a$: only $+Q$ inside Gaussian sphere; $E = kQ/r^2$.

For $a<r<b$ (inside conductor): $E = 0$ (electrostatic equilibrium). Gauss's law applied here ($\Phi = 0$) gives enclosed charge $= 0$. Since $+Q$ is at the center, the induced charge on the inner surface must be $-Q$.

By charge conservation on the (initially uncharged) shell, the outer surface carries $+Q$.

For $r>b$: enclosed charge $= +Q$ (point) $-Q$ (inner) $+Q$ (outer) $= +Q$. So $E = kQ/r^2$.

Problem 5 (Medium)

A dipole $\vec{p}$ is placed in a non-uniform field along $x$: $\vec{E}(x) = E_0\hat{i}(1+\alpha x)$. The dipole has its $\vec{p}$ along $\hat{i}$. Find the net force.

Solution. The $+q$ end at $x+a$ feels $qE_0(1+\alpha(x+a))$. The $-q$ end at $x-a$ feels $-qE_0(1+\alpha(x-a))$. Net force:

So $F = p\,(dE/dx)$, consistent with $F = p\,\partial_x E$ for $\vec{p}\parallel\vec{E}$.

Problem 6 (Hard)

A solid sphere of radius $R$ has volume charge density $\rho(r) = \rho_0 (1-r/R)$ for $r\le R$, zero outside. Find $E(r)$ for $r<R$.

Solution.

By Gauss's law:

Problem 7 (Hard)

Two infinite parallel sheets have densities $\sigma_1 = +3\,\mu\text{C/m}^2$ and $\sigma_2 = -1\,\mu\text{C/m}^2$. Find $E$ everywhere.

Solution.

Region I (left of $\sigma_1$): $E_I = (\sigma_1+\sigma_2)/(2\varepsilon_0)$ directed to the left $= (2\,\mu\text{C/m}^2)/(2\varepsilon_0) = \sigma/(2\varepsilon_0)\cdot 2$.

Numerically:

(Direction: away from $\sigma_1$, toward $\sigma_2$, i.e., to the left of $\sigma_1$ it points left; to the right of $\sigma_2$, right.)

Region II (between): $E_{II} = (\sigma_1-\sigma_2)/(2\varepsilon_0) = (3-(-1))\times 10^{-6}/(2\varepsilon_0) = 4\times 10^{-6}/(2\varepsilon_0) = 2.26\times 10^5\text{ N/C}$, directed from $+$ sheet toward $-$ sheet.

Region III (right of $\sigma_2$): same magnitude as Region I but opposite sense.

JEE/NEET Edge Cases

1. Charge on a corner of a cube. Flux through the cube is $q/(8\varepsilon_0)$ (the corner is shared by 8 cubes; total flux from $q$ is $q/\varepsilon_0$ shared equally).

2. Charge at the center of one face. Flux through the cube is $q/(2\varepsilon_0)$ (the face is shared by 2 cubes).

3. Field at the midpoint of a side of a square with charges $\pm q$ alternating at corners — set up vectors carefully, use symmetry only where it exists.

4. Equilibrium of three collinear charges. For a third charge to be in equilibrium between two unlike charges, it must be placed at a specific point determined by ratio of charges; check stability sign.

5. Hollow conductor with a charge inside off-center. The outer field is still symmetric and the same as if the total charge sat at the conductor's center (a classic counter-intuitive result).

6. Continuous vs discrete. A charged ring at its center has zero field — but a uniformly charged disk has nonzero field on axis.

7. Dipole in non-uniform field can experience a net force and a torque.

8. Two charged balls hanging from threads (Coulomb's classic): in equilibrium, $\tan\theta = F_e/mg$; if immersed in a dielectric liquid the apparent angle can change due to both buoyancy and reduced Coulomb force.

9. Maximum field on the axis of a ring occurs at $z = R/\sqrt{2}$, giving $E_{\max} = \dfrac{Q}{6\sqrt{3}\pi\varepsilon_0 R^2}$.

10. Gauss's law trap. The flux is determined by enclosed charge alone, but the field at any point on the surface depends on outside charges too. Don't conflate the two.

Quick Recap

• Charge is conserved, quantized ($q=ne$), additive, and relativistically invariant.
• Coulomb's law: $\vec{F} = \dfrac{kq_1q_2}{r^2}\hat{r}$; vector form is symmetric.
• $\vec{E} = \vec{F}/q_0$ in the limit of vanishing test charge.
• Field of point charge: $\vec{E} = kq\hat{r}/r^2$.
• Dipole axial: $E = 2kp/r^3$; equatorial: $E = -kp/r^3$; $\vert E_{\text{axial}}\vert = 2\vert E_{\text{eq}}\vert$.
• Torque on dipole: $\vec{\tau} = \vec{p}\times\vec{E}$; potential energy: $U = -\vec{p}\cdot\vec{E}$.
• Gauss's law: $\oint \vec{E}\cdot d\vec{A} = q_{\text{enc}}/\varepsilon_0$.
• Standard fields: wire $\lambda/(2\pi\varepsilon_0 r)$; sheet $\sigma/(2\varepsilon_0)$; outside shell $kQ/r^2$, inside shell $0$.