Electric Charges
and Fields
From the nature of charge to Gauss's law — every definition, formula, and derivation you need for boards and JEE, in one place.
Electric Charge
Electric charge is an intrinsic property of matter responsible for electric and magnetic effects. It occurs in two kinds — positive and negative — with like charges repelling and unlike charges attracting.
- Additivity — charges add up algebraically like scalars, respecting sign.
- Conservation — total charge of an isolated system stays constant; charge is only transferred, never created or destroyed.
- Quantization — charge exists only as integral multiples of the elementary charge $e$.
where $e = 1.6 \times 10^{-19}\ \text{C}$
- A very common 1-mark question: state the basic properties of electric charge — learn all three definitions word-for-word.
- Don't confuse quantization with conservation — examiners test this distinction directly.
- Quantization means a charge of $0.5e$ or $1.7e$ is impossible — only exact integral multiples of $e$ exist. Frequently tested as an assertion-reason question.
Coulomb's Law
⭐ High WeightageCoulomb's law gives the electrostatic force between two stationary point charges — the foundation of all electrostatics.
$\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{Nm}^2/\text{C}^2$ in vacuum/air
Force on charge 1 due to charge 2, directed along the unit vector from 2 to 1
In a medium of relative permittivity (dielectric constant) $K = \varepsilon/\varepsilon_0$, the force reduces by a factor of $K$:
$$F_{\text{medium}} = \frac{F_{\text{vacuum}}}{K}$$- Frequently combined with equilibrium-of-three-charges problems — assign signs carefully before writing force equations.
- Watch for dielectric-medium variants where $K$ is buried inside the problem statement (e.g. "force reduces to 1/4th in a medium" $\Rightarrow K=4$).
Principle of Superposition
When multiple charges act on a given charge, the net force is the vector sum of the forces due to each charge individually — as if the others weren't present.
- Superposition problems usually reduce to resolving vectors along x-y axes. For symmetric arrangements (equilateral triangle, square), exploit symmetry to cancel components rather than computing every force explicitly.
Electric Field
Rather than charges acting on each other directly at a distance, a charge creates an electric field in the space around it; any other charge placed in that field experiences a force.
SI unit: N/C or V/m
- Always mention "the test charge should be vanishingly small" while defining electric field — examiners specifically look for this qualifier in the 2-mark definition.
Electric Field Lines
Field lines visually represent the electric field — imaginary curves whose tangent at any point gives the field direction there.
- Start on positive charges (or infinity) and end on negative charges (or infinity).
- Never cross each other — an intersection would mean two field directions at one point.
- Denser lines indicate a stronger field.
- Continuous curves with no breaks in a charge-free region.
- Field lines can never form closed loops in electrostatics — this only holds for electrostatic fields, not the induced fields seen later in electromagnetic induction.
Electric Dipole
⭐ High WeightageA pair of equal and opposite charges separated by a small distance forms an electric dipole — a configuration central to molecular physics and a JEE favourite.
Direction: from negative to positive charge. Unit: C·m
A dipole in a uniform field feels zero net force (forces on $+q$ and $-q$ cancel) but a net torque that aligns it with the field:
- Standard 3-mark derivation: derive $\tau = pE\sin\theta$ for a dipole in a uniform field. Draw the diagram with perpendicular distance $2a\sin\theta$ clearly marked — diagram marks matter here.
- For $r \gg a$: $E_{\text{axial}} = 2\times E_{\text{equatorial}}$ at the same distance. Axial field is along $\vec{p}$; equatorial field is opposite to $\vec{p}$ — a frequent direct-answer question.
A dipole of moment $p = 4\times10^{-9}\ \text{C·m}$ sits in a uniform field $E = 5\times10^{4}\ \text{N/C}$ at $30°$ to the field. Find the torque.
$\tau = pE\sin\theta = 4\times10^{-9}\times5\times10^{4}\times\sin30° = 1\times10^{-4}\ \text{N·m}$
Continuous Charge Distribution
For a large number of closely-spaced charges, it's convenient to treat the charge as continuously distributed, described using charge densities.
| Density | Definition | Unit |
|---|---|---|
| Linear ($\lambda$) | $\lambda = dq/dl$ | C/m |
| Surface ($\sigma$) | $\sigma = dq/dA$ | C/m² |
| Volume ($\rho$) | $\rho = dq/dV$ | C/m³ |
- Charge densities set up the integrals used to derive field expressions for rods, rings, and discs — practice setting up $dq$ for a ring and a disc, both appear often in JEE Advanced.
Electric Flux
Electric flux measures how much field passes through a given surface.
$\theta$ = angle between $\vec{E}$ and the outward normal. Unit: N·m²/C
- Flux is maximum when the field is perpendicular to the surface ($\theta=0°$) and zero when parallel to it ($\theta=90°$) — a common 1-mark conceptual question.
Gauss's Law & Applications
⭐ High WeightageGauss's law connects the total flux through any closed surface to the charge it encloses — turning painful integrals into elegant one-line answers for symmetric charge distributions.
Using a cylindrical Gaussian surface coaxial with the wire:
$$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$Radially outward for $\lambda>0$; falls off as $1/r$
Using a "pillbox" Gaussian surface straddling the sheet:
$$E = \frac{\sigma}{2\varepsilon_0}$$Uniform — independent of distance from the sheet
| Region | Field |
|---|---|
| Outside ($r>R$) | $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ |
| On surface ($r=R$) | $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R^2}$ |
| Inside ($r<R$) | $E = 0$ |
Outside the shell it behaves as if all charge were concentrated at the center; inside, the enclosed charge is zero so the field vanishes entirely.
- The E-vs-r graph for a charged spherical shell (rising to a peak at $R$, falling as $1/r^2$ outside, zero inside) is frequently asked — practice drawing it with correctly labelled axes.
- All three applications (wire, sheet, shell) are derived by choosing a Gaussian surface that matches the symmetry — memorize the shape of surface used for each case, not just the final formula.
A long straight wire carries linear charge density $\lambda = 2\times10^{-6}\ \text{C/m}$. Find the field at a point $5\ \text{cm}$ from the wire.
$E = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{2\times10^{-6}\times9\times10^{9}\times2}{0.05} = 7.2\times10^{5}\ \text{N/C}$, directed radially outward
Complete Comparison
Axial (end-on):
$$E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$$Equatorial (broadside-on):
$$E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$$Infinite wire: $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ (falls as $1/r$)
Infinite sheet: $E=\dfrac{\sigma}{2\varepsilon_0}$ (constant)
Spherical shell (outside): $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ (falls as $1/r^2$)
Solved Examples
Two point charges of $+2\ \mu\text{C}$ and $+3\ \mu\text{C}$ are placed $30\ \text{cm}$ apart in air. Find the force between them.
$F = \dfrac{9\times10^9\times2\times10^{-6}\times3\times10^{-6}}{(0.3)^2} = 0.6\ \text{N}$ (repulsive, both charges positive)
A spherical shell of radius $R = 10\ \text{cm}$ carries charge $q = 1\ \mu\text{C}$. Find $E$ at $r = 5\ \text{cm}$ and $r = 20\ \text{cm}$.
At $r=5\ \text{cm}$ (inside shell, $r<R$): $E = 0$
At $r=20\ \text{cm}$ (outside shell, $r>R$): $E = \dfrac{9\times10^9\times1\times10^{-6}}{(0.2)^2} = 2.25\times10^{5}\ \text{N/C}$
A charged conducting sheet has surface charge density $\sigma = 4.4\times10^{-6}\ \text{C/m}^2$. Find the electric field near the sheet.
$E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{4.4\times10^{-6}}{2\times8.85\times10^{-12}} \approx 2.49\times10^{5}\ \text{N/C}$
- Forgetting that force is a vector — always assign a sign convention before adding forces from multiple charges.
- Using $\sigma/\varepsilon_0$ instead of $\sigma/2\varepsilon_0$ for a single charged sheet (the factor of 2 is only dropped for a charged conductor's surface, a distinct case).
- Assuming field inside a spherical shell is small but non-zero — it is exactly zero.