Electric Charges and Fields

The chapter builds in four movements. First, charge itself: what it is, how it arises from rubbing, and its three defining properties — quantisation, additivity, conservation — plus the distinction between conductors and insulators. Second, Coulomb's law and the electric field: the force law between point charges, the superposition principle for many charges, the definition of the field as force per unit test charge, and the visual language of field lines. Third, the electric dipole: its field on the axis and on the equatorial plane, its dipole moment, and the torque and force it experiences in external fields — the model behind polar molecules and induced polarisation. Fourth, electric flux and Gauss's law: the geometric idea of flux, the statement of Gauss's law, and its payoff in three symmetric applications — the infinite line charge, the infinite plane sheet, and the uniformly charged spherical shell — plus continuous charge densities that make these calculations rigorous.

What electric charge is, and its three basic properties

Rubbing certain materials together — glass with silk, plastic with fur — leaves both objects able to attract light bits of paper. Careful experiments going back centuries show there are exactly two kinds of this "electrification," historically named positive and negative by Benjamin Franklin: like charges repel, unlike charges attract. At the atomic level this is simply the transfer of electrons — a body that loses electrons becomes positively charged, one that gains electrons becomes negatively charged. No charge is created in the process; it only moves from one body to another. Three properties make charge mathematically tractable. Quantisation means every free charge is an integer multiple of the elementary charge $e$:

where: $q$ is the charge on a body, $n$ is a (signed) integer, and $e$ is the magnitude of charge on a single electron or proton. Physically, charge comes in indivisible lumps — you cannot have half an electron's worth of charge. At the macroscopic scale, where charges of a few microcoulombs involve some $10^{13}$ elementary charges, this graininess is invisible and charge behaves as if continuous, just as a fine dotted line looks solid from a distance.

Additivity means charges combine like ordinary signed numbers: the total charge of a system of $n$ point charges is simply $q_1 + q_2 + \cdots + q_n$, with proper signs. Conservation means the total charge of an isolated system is constant in time — charge can be transferred from one part of a system to another, or new charged particles can be created in pairs of opposite sign (as in a neutron decaying to a proton and an electron), but the algebraic sum never changes. Together these three properties are why charge, unlike mass, is treated as a signed scalar in every calculation that follows. Materials divide broadly into conductors, which have charges (typically electrons) free to move throughout the material and redistribute charge instantly over the whole surface, and insulators, where charge stays put wherever it is deposited. Metals, the human body and the earth are conductors; glass, plastic and dry wood are insulators. This is why a metal rod held in bare hands cannot be charged by friction — any charge you produce leaks away through your body to the ground — while the same rod held by an insulating handle charges readily.

Coulomb's law: the force between two point charges

Charles Augustin de Coulomb, using a torsion balance sensitive enough to measure tiny forces, established experimentally that the force between two point charges varies as the inverse square of the distance between them and is proportional to the product of their magnitudes. When the charged bodies are small enough compared to their separation to be treated as points, the magnitude of the force is:

where: $q_1, q_2$ are the magnitudes of the two point charges, $r$ is the distance between them, and $\varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space, with $\dfrac{1}{4\pi\varepsilon_0} \approx 9 \times 10^{9}\ \text{N m}^2\text{C}^{-2}$. The physical reading is direct: double either charge and the force doubles; double the separation and the force falls to a quarter. Because force is a vector, the full statement needs direction too. Writing $\hat{r}_{21}$ as the unit vector pointing from charge 1 to charge 2, the force on charge 2 due to charge 1 is:

where: the sign of the product $q_1 q_2$ automatically fixes the direction — if both charges share a sign, $\vec{F}_{21}$ points along $\hat{r}_{21}$ (repulsion, pushing charge 2 away from charge 1); if they are opposite signs, $\vec{F}_{21}$ points opposite to $\hat{r}_{21}$ (attraction). This single vector equation handles both cases without ever needing separate formulas, and it satisfies Newton's third law exactly: $\vec{F}_{12} = -\vec{F}_{21}$.

The superposition principle

Coulomb's law describes only two charges at a time. Real problems usually involve many. Experiment confirms that the force on any one charge due to a collection of others is simply the vector sum of the individual Coulomb forces, each calculated as though the other charges were not present:

where: $\vec{F}_1$ is the net force on charge $q_1$, and each term in the sum is the ordinary Coulomb force from one other charge, added by the parallelogram law. This is not a trivial statement — it asserts both that pairwise forces are unaffected by the presence of a third charge, and that no extra "three-body" force terms appear. Superposition is the reason electrostatics is solvable at all: any charge distribution, however complicated, reduces to adding up simple two-charge interactions.

The electric field: force reimagined as a property of space

If a charge $q$ sitting at point $P$ is removed, is there "nothing" left at $P$? Physicists answer no: the source charge $Q$ sets up an electric field everywhere around it, and any test charge placed at $P$ merely reveals a force already latent in the field there. The field due to a point charge $Q$, at a position a distance $r$ away, is defined operationally as the force per unit charge in the limit of a vanishingly small test charge $q$:

where: $\vec{E}$ has SI unit newton per coulomb (equivalently volt per metre), $\hat{r}$ points from the source charge outward to the field point, and the limit $q \to 0$ is taken so that the test charge's own field does not disturb the source. The physical reading: the electric field is a property of the source charge $Q$ alone — it exists at every point in space whether or not a test charge happens to be there, and once known, the force on any charge $q$ placed anywhere is recovered instantly from $\vec{F} = q\vec{E}$. For a system of many source charges, superposition applies again: the net field at any point is the vector sum of the fields due to each source charge individually, $\vec{E} = \vec{E}_1 + \vec{E}_2 + \cdots + \vec{E}_n$. A positive source charge produces a field pointing radially outward; a negative source charge, radially inward.

Electric field lines: picturing a vector field

Drawing an arrow for $\vec{E}$ at every point of space is unworkable, so Faraday introduced field lines: continuous curves whose tangent at every point gives the direction of $\vec{E}$ there, with their crowding showing the field's strength — dense where the field is strong, sparse where it is weak. Four properties make field lines a reliable diagnostic tool, and each follows from the physics, not from artistic convention. Field lines start on positive charges and end on negative charges (or run to infinity for an isolated single charge); they cannot break in a charge-free region; two field lines can never cross (a crossing would mean two directions for $\vec{E}$ at one point, which is meaningless); and electrostatic field lines never form closed loops, because the electrostatic field is conservative — a property proved formally in the next chapter. The figure below shows the essential library of patterns you must be able to sketch on demand: an isolated positive charge, an isolated negative charge, and a dipole.

Electric flux: counting field lines through a surface

To make "density of field lines" precise, define an area element as a vector $\Delta\vec{S} = \Delta S\,\hat{n}$, where $\hat{n}$ is the unit normal to the small flat patch (by convention, the outward normal for a closed surface). The electric flux through that patch is:

where: $\theta$ is the angle between $\vec{E}$ and the normal $\hat{n}$. The physical reading mirrors fluid flow: just as the volume of liquid crossing a tilted hoop depends on $\cos\theta$ between the flow and the hoop's normal, the number of field lines crossing a tilted patch scales the same way — maximal when the patch faces the field square on ($\theta = 0$), zero when the patch is edge-on to the field ($\theta = 90^\circ$). Summing $\Delta\phi$ over every element of an entire surface $S$ gives the total flux $\phi = \sum \vec{E}\cdot\Delta\vec{S}$, exact in the limit of infinitesimally small elements. Flux has SI unit $\text{N m}^2\text{C}^{-1}$, and it is this quantity — not the field itself — that Gauss's law connects directly to enclosed charge.

The electric dipole: field, moment and behaviour in external fields

An electric dipole is a pair of equal and opposite point charges, $+q$ and $-q$, separated by a distance $2a$. Its dipole moment is defined as the vector:

where: $p = q(2a)$ is the magnitude (SI unit coulomb-metre) and the direction runs from the negative charge to the positive charge. Although the dipole's total charge is exactly zero, its field is not zero, because the two charges sit at slightly different distances from any field point, so their $1/r^2$ contributions do not cancel perfectly. The derivation section below shows that, far from the dipole ($r \gg a$), the field on the dipole axis has magnitude $E_{\text{axis}} = \dfrac{2p}{4\pi\varepsilon_0 r^3}$ (along $\hat{p}$) and the field on the equatorial plane (perpendicular to the axis, through the centre) has magnitude $E_{\text{eq}} = \dfrac{p}{4\pi\varepsilon_0 r^3}$ (anti-parallel to $\hat{p}$) — half the axial value, and pointing the opposite way. The defining feature of any dipole field is the $1/r^3$ fall-off, steeper than the $1/r^2$ of a single point charge, because a dipole is, in effect, two nearly-cancelling monopoles. Placed in a uniform external field $\vec{E}$, a dipole feels zero net force (the forces $q\vec{E}$ and $-q\vec{E}$ on its two ends cancel exactly), but because those equal and opposite forces act at different points, they produce a net torque:

where: $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. The torque is zero when the dipole is aligned with the field ($\theta = 0$, the stable configuration) or anti-aligned ($\theta = 180^\circ$, unstable), and maximal at $\theta = 90^\circ$. In a non-uniform field, the two ends experience unequal forces, so besides a torque the dipole also feels a net force — pulled toward the region of stronger field if $\vec{p}$ is parallel to $\vec{E}$, and toward weaker field if anti-parallel. This is precisely how a charged comb, though it induces no net charge on a scrap of paper, still attracts it: the comb's non-uniform field induces a small dipole moment in the paper, and that induced dipole is then pulled toward the stronger field near the comb.

Continuous charge distributions

Real charged objects — a rod, a plate, a sphere — carry so many elementary charges that treating them individually is hopeless; instead we smooth the charge out into a continuous density. Linear charge density $\lambda = \Delta Q/\Delta l$ (unit $\text{C m}^{-1}$) applies to charge spread along a length, such as a wire. Surface charge density $\sigma = \Delta Q/\Delta S$ (unit $\text{C m}^{-2}$) applies to charge spread over an area, such as a conductor's surface. Volume charge density $\rho = \Delta Q/\Delta V$ (unit $\text{C m}^{-3}$) applies to charge spread through a volume. In each case, $\Delta l$, $\Delta S$ or $\Delta V$ is chosen large enough to contain many elementary charges (so the density is smooth) yet small enough to be effectively a point on the macroscopic scale — exactly the same idea used to define the density of an ordinary fluid while ignoring its molecular graininess. With these densities, the field due to a continuous distribution is found by replacing the discrete sum of Coulomb's law with an integral over the charge-carrying region, using the same superposition idea as for point charges.

Gauss's law: flux through any closed surface

Consider a point charge $q$ at the centre of an imaginary sphere of radius $r$. Every point on the sphere is at the same distance from $q$, and $\vec{E}$ there is radial — parallel to the outward normal everywhere — so the flux through the whole sphere works out (shown in the derivation below) to be exactly $q/\varepsilon_0$, independent of $r$. This is not a coincidence of the sphere's shape; it turns out to be true for any closed surface enclosing charge $q$, of any shape whatsoever. This general result is Gauss's law:

where: the closed-surface (Gaussian) integral on the left is the total electric flux, and $q_{\text{enc}}$ is only the charge enclosed inside the chosen closed surface — charges outside contribute nothing to the flux total, even though they do contribute to $\vec{E}$ at individual points on the surface. The physical reading is subtle and important: the field $\vec{E}$ appearing on the left is the total field due to all charges, inside and outside; but when you sum $\vec{E}\cdot d\vec{S}$ over the entire closed surface, only the enclosed charge survives in the final total — the contributions of every outside charge cancel out over the closed surface as a whole. A Gaussian surface may never be drawn so that it passes through a discrete point charge (the field there is undefined), though it may freely pass through a continuous charge distribution. Gauss's law is always true, but it is only a fast calculation tool when the charge distribution has enough symmetry (spherical, cylindrical, or planar) to guess the direction and constancy of $\vec{E}$ on some cleverly chosen surface before doing any algebra. The next section, and the derivations below, use exactly that strategy for the chapter's three standard results — the infinitely long charged wire, the infinite charged sheet, and the uniformly charged spherical shell.

Board level - force and field between two point charges

Two small charged spheres carry charges $q_1 = 2\times10^{-7}\ \text{C}$ and $q_2 = 3\times10^{-7}\ \text{C}$ and are placed $30\ \text{cm}$ apart in air. Find (a) the electrostatic force between them, and (b) the electric field at the location of $q_2$ due to $q_1$ alone.

Step 1. Convert to SI and apply Coulomb's law, Eq. (2): $r = 0.30\ \text{m}$, so $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2} = (9\times10^9)\times\dfrac{(2\times10^{-7})(3\times10^{-7})}{(0.30)^2}$.

Step 2. Evaluate the numerator and denominator separately: numerator $= (9\times10^9)(6\times10^{-14}) = 5.4\times10^{-4}$; denominator $= 0.09$. So $F = 5.4\times10^{-4}/0.09 = 6.0\times10^{-3}\ \text{N}$. Since both charges are positive, the force is repulsive, pushing the spheres apart along the line joining them.

Step 3. Find the field at $q_2$'s location due to $q_1$ using Eq. (5): $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r^2} = (9\times10^9)\times\dfrac{2\times10^{-7}}{0.09} = 2.0\times10^{4}\ \text{N/C}$, directed away from $q_1$ (since $q_1$ is positive). As a check, $F = q_2 E = (3\times10^{-7})(2.0\times10^4) = 6.0\times10^{-3}\ \text{N}$, matching Step 2 exactly, since $\vec{F} = q\vec{E}$ is just Coulomb's law rewritten.

Answer: $F = 6.0\times10^{-3}\ \text{N}$ (repulsive), and $E = 2.0\times10^{4}\ \text{N/C}$ at $q_2$'s location, directed away from $q_1$.

Competitive level (JEE / NEET) - net force at the centre of a charged square

Four point charges $q_A = 2\ \mu\text{C}$, $q_B = -5\ \mu\text{C}$, $q_C = 2\ \mu\text{C}$ and $q_D = -5\ \mu\text{C}$ sit at the corners $A, B, C, D$ of a square of side $10\ \text{cm}$, taken in order. Find the net force on a charge $q_0 = 1\ \mu\text{C}$ placed at the centre of the square.

Step 1. Exploit symmetry before computing anything. Opposite corners $A$ and $C$ both carry $+2\ \mu\text{C}$, equidistant from the centre; the forces they exert on $q_0$ are equal in magnitude and exactly opposite in direction (each pushes $q_0$ away from itself, along the diagonal, and the two diagonals' halves point oppositely through the centre), so these two forces cancel completely.

Step 2. The same argument applies to $B$ and $D$, which both carry $-5\ \mu\text{C}$: each pulls $q_0$ toward itself along its diagonal, equal in magnitude, opposite in direction, so these two forces also cancel completely.

Step 3. Since every force cancels in pairs by the square's symmetry, the net force on $q_0$ is zero — no distance or magnitude calculation is even needed once the symmetry is recognised. This is a deliberate trap: a student who launches into computing four separate Coulomb forces and adding components will still reach zero, but will burn several minutes doing so.

Answer: the net force on the $1\ \mu\text{C}$ charge at the centre is exactly $\vec{0}$, by the pairwise cancellation of forces from charges at opposite corners. Timing note: under a minute if the symmetry is spotted immediately. The general lesson for JEE/NEET is to always scan a charge configuration for symmetry (equal charges at equal distances, in opposing pairs) before setting up any vector-component calculation — many "four charge" or "polygon" problems are designed to collapse this way, and grinding through full vector algebra when the answer is zero by symmetry is the single biggest time sink in this topic.

Board level - field of a short dipole on axis and on the equator

Two charges $+10\ \mu\text{C}$ and $-10\ \mu\text{C}$ are $5.0\ \text{mm}$ apart, forming a dipole. Find the electric field magnitude at a point $15\ \text{cm}$ from the centre, (a) on the dipole axis, and (b) on the equatorial line, using the far-field dipole formulas.

Step 1. Compute the dipole moment from Eq. (7): $p = q(2a) = (10\times10^{-6}\ \text{C})(5.0\times10^{-3}\ \text{m}) = 5.0\times10^{-8}\ \text{C m}$. Check that the far-field approximation is valid: $r/a = 0.15/0.0025 = 60 \gg 1$, so the far-field formulas apply comfortably.

Step 2. Axial field from the derivation above, $E_{\text{axis}} = \dfrac{2p}{4\pi\varepsilon_0 r^3} = (9\times10^9)\times\dfrac{2(5.0\times10^{-8})}{(0.15)^3}$. The numerator is $(9\times10^9)(1.0\times10^{-7}) = 9.0\times10^{2}$, and $(0.15)^3 = 3.375\times10^{-3}$, so $E_{\text{axis}} = 9.0\times10^{2}/3.375\times10^{-3} = 2.67\times10^{5}\ \text{N/C}$, directed along $\hat{p}$ (from the $-q$ end toward the $+q$ end).

Step 3. Equatorial field, $E_{\text{eq}} = \dfrac{p}{4\pi\varepsilon_0 r^3} = \dfrac{1}{2}E_{\text{axis}} = 1.33\times10^{5}\ \text{N/C}$, directed anti-parallel to $\hat{p}$.

Answer: $E_{\text{axis}} \approx 2.7\times10^{5}\ \text{N/C}$ along the dipole moment, and $E_{\text{eq}} \approx 1.3\times10^{5}\ \text{N/C}$ opposite to the dipole moment — exactly half the axial value, as Eqs. (7) and the derivation predict.

Competitive level (JEE / NEET) - Gauss's law inside and outside a uniformly charged sphere

A solid, non-conducting sphere of radius $R = 10\ \text{cm}$ carries a total charge $Q = 3\ \mu\text{C}$ spread uniformly through its volume. Find the electric field magnitude (a) at $r = 5\ \text{cm}$ from the centre (inside the sphere), and (b) at $r = 20\ \text{cm}$ from the centre (outside the sphere).

Step 1. Outside the sphere ($r = 0.20\ \text{m} \gt R$): a uniformly charged solid sphere behaves, for exterior points, exactly like a point charge $Q$ at the centre (built up as nested shells, each obeying Eq. (12) outside itself). So $E_{\text{out}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2} = (9\times10^9)\times\dfrac{3\times10^{-6}}{(0.20)^2} = (9\times10^9)\times\dfrac{3\times10^{-6}}{0.04} = 6.75\times10^{5}\ \text{N/C}$, directed radially outward.

Step 2. Inside the sphere ($r = 0.05\ \text{m} \lt R$): unlike a thin shell, a solid sphere's interior Gaussian surface does enclose some charge — specifically, the charge contained within radius $r$, not the full $Q$. Since the charge is uniform, the enclosed charge scales with enclosed volume: $q_{\text{enc}} = Q\times\dfrac{(4/3)\pi r^3}{(4/3)\pi R^3} = Q\left(\dfrac{r}{R}\right)^3$.

Step 3. Apply Gauss's law with this reduced enclosed charge on a concentric Gaussian sphere of radius $r$: $E_{\text{in}}\times 4\pi r^2 = \dfrac{q_{\text{enc}}}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0}\left(\dfrac{r}{R}\right)^3$, so $E_{\text{in}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q r}{R^3}$ — notice the field grows linearly with $r$ inside a uniformly charged solid sphere, unlike the $1/r^2$ falloff outside.

Step 4. Substitute numbers: $E_{\text{in}} = (9\times10^9)\times\dfrac{(3\times10^{-6})(0.05)}{(0.10)^3} = (9\times10^9)\times\dfrac{1.5\times10^{-7}}{1.0\times10^{-3}} = 1.35\times10^{6}\ \text{N/C}$, directed radially outward.

Answer: $E_{\text{in}} \approx 1.35\times10^{6}\ \text{N/C}$ at $r = 5\ \text{cm}$ (inside, growing linearly with $r$), and $E_{\text{out}} \approx 6.75\times10^{5}\ \text{N/C}$ at $r = 20\ \text{cm}$ (outside, falling as $1/r^2$). Timing note: about two minutes. The one conceptual fork every JEE/NEET aspirant must make instantly is "am I inside or outside the charge?" — outside, always use the full charge $Q$ as if it were a point charge at the centre; inside a uniformly charged solid sphere, only the fractional enclosed charge $(r/R)^3$ counts, giving a field that rises linearly to a maximum exactly at the surface $r = R$, then falls as $1/r^2$ beyond it. Sketching that $E$-versus-$r$ graph (a rising straight line inside, an inverse-square curve outside, meeting at $r = R$) is a favourite examiner check on this exact idea.

Confusing the scalar magnitude form of Coulomb's law with its vector form leads to sign errors — fix the direction from the physical rule (like charges repel, unlike attract) rather than trusting a raw algebraic sign blindly, especially with several charges involved. A related error is forgetting that $\vec{F}=q\vec{E}$ gives a force in the direction of $\vec{E}$ for a positive $q$, but opposite for a negative $q$. Students often assume the electric field "belongs" to the test charge, when it is entirely a property of the source charges; the limit $q \to 0$ is a bookkeeping device, not a physical requirement — a field exists at a point whether or not a charge sits there to measure it. Another frequent slip is applying the far-field dipole formulas ($1/r^3$) at points close to the dipole, where $r$ is not much greater than $2a$; the exact expressions from the dipole derivation must be used instead. In Gauss's law problems, the most common mistake is thinking the field $\vec{E}$ is caused only by the enclosed charge — it is not; $\vec{E}$ at any point is due to every charge in the universe, and only the net flux integral, summed over the whole closed surface, reduces to depending on the enclosed charge alone. A second mistake is applying Gauss's law to an asymmetric distribution — the law is always true, but only solvable by hand when a Gaussian surface can be chosen on which $\vec{E}$ is constant (or zero) everywhere on it. Finally, students misremember scope — Eq. (10) for a wire and Eq. (11) for a sheet are only valid far from the ends or edges of a finite real object, exact only for genuinely infinite idealisations.

• State and apply the three basic properties of charge — quantisation ($q=ne$), additivity, and conservation — and explain why quantisation is invisible at macroscopic scales.
• Write Coulomb's law in both scalar and vector form, and correctly assign the direction of the force from the signs of the two charges.
• Apply the superposition principle to find the net force or net field due to three or more point charges, including recognising when symmetry makes the answer zero without detailed calculation.
• Define the electric field as the limiting force per unit test charge, and explain why the field is independent of the test charge used to measure it.
• Sketch the field-line patterns of an isolated positive charge, an isolated negative charge, and a dipole, and state all four defining properties of field lines.
• Define electric flux, compute it for simple flat and closed surfaces, and state Gauss's law precisely, including exactly what "enclosed charge" means.
• Derive, or reconstruct on demand, the fields of an infinite line charge, an infinite plane sheet, and a uniformly charged spherical shell (inside and outside), using a Gaussian surface matched to the symmetry.
• Compute the dipole moment of a charge pair, and state the axial and equatorial field formulas along with the torque $\vec\tau = \vec p \times \vec E$ in a uniform field.
• Distinguish linear, surface and volume charge density, and know which applies to a wire, a sheet, and a bulk-charged solid respectively.
• Explain, with a specific example, why a symmetric closed surface with zero net enclosed charge can still have a non-zero electric field at points on that surface.

Ask the tutor: if you would like a worked walk-through of any Gauss's-law geometry you find in a numerical problem — line, sheet, shell, or solid sphere — just ask and we can set up the Gaussian surface together step by step.