Electric Charges and Fields: Full Notes

1.1 Electric Charge: Basic Properties

Rub a glass rod with silk and it picks up the ability to attract small bits of paper. Rub two different materials together and you find there are exactly two kinds of this property, conventionally called positive and negative: bodies with the same kind repel, bodies with opposite kinds attract. Nothing about the labels is special, only the rule that like repels and unlike attracts.

Charge is also additive: if a system contains several charges $q_1, q_2, \ldots, q_n$ , its total charge is simply the algebraic (scalar, signed) sum $q_1+q_2+\cdots+q_n$ , counting negative charges as negative numbers, regardless of where the individual charges sit or how they interact.

1.1.1 Quantisation of Charge: q = ne

Charge does not come in arbitrarily small amounts. Every charge observed in nature is an integer multiple of a smallest indivisible unit, the elementary charge $e$ , carried (in magnitude) by the electron and the proton:

Quantisation of charge

\textcolor{#e08a1e}{q} = ne,\qquad n = 0,\pm1,\pm2,\ldots

with $e \approx 1.6\times10^{-19}\text{ C}$ . A charge of $1.5e$ simply does not occur. The reason this is not obvious in daily life is scale: rubbing a balloon on your hair transfers on the order of $10^{-8}$ to $10^{-7}$ C, which already corresponds to $10^{11}$ – $10^{12}$ electrons, so individual steps of $e$ are far too fine-grained to notice, exactly as the grain of sand on a beach is invisible from a distance even though the beach is not a continuous solid.

1.1.2 Conservation, Conductors, and Induction

Charge is also conserved: the total charge of an isolated system cannot change. Rubbing two objects together does not create charge, it only transfers electrons from one surface to the other, so whatever positive charge one object gains, the other loses an equal amount of negative charge (or gains an equal positive charge) — the sum before and after is identical.

Materials split broadly into conductors (metals, the human body, the earth), which contain charges free to move through the bulk of the material, and insulators (glass, rubber, most plastics), whose charges are bound to individual atoms and cannot migrate. This is why a charged rod can pick up paper and also light up a conductor by induction: bring a charged rod near, but not touching, an isolated conductor, and its field pushes the conductor’s free electrons away from (or toward) the near face, piling up an induced opposite charge there and an induced like charge on the far face — with zero net charge transferred to the conductor, since nothing ever touched it.

1.2 Coulomb's Law

For two point charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum, the force between them is directly proportional to the product of the charges and inversely proportional to the square of the separation, and it acts along the line joining them:

F = k\dfrac{q_1q_2}{r^2}

where the constant $k = \dfrac{1}{4\pi\varepsilon_0} \approx 8.99\times10^9\text{ N m}^2\text{C}^{-2}$ (often rounded to $9\times10^9$ for quick estimates), and $\varepsilon_0 \approx 8.85\times10^{-12}$ C²N⁻¹m⁻² is the permittivity of free space. Written as a vector, the force on charge 1 due to charge 2 is

\mathbf{F}_{12} = k\dfrac{q_1q_2}{r^2}\,\hat{r}_{21}

with $\hat{r}_{21}$ the unit vector pointing from charge 2 to charge 1. The sign of $q_1q_2$ does all the work: if it is positive (like charges) $\mathbf{F}_{12}$ points away from charge 2, a push; if negative (unlike charges) it points toward charge 2, a pull. By Newton’s third law, $\mathbf{F}_{21} = -\mathbf{F}_{12}$ always, exactly as with any other pair of forces.

Worked example

Force between two point charges

Two point charges, $q_1 = +3\,\mu\text{C}$ and $q_2 = -5\,\mu\text{C}$ , sit 20 cm apart in vacuum. Find the force between them.

Converting to SI ( $1\,\mu\text{C} = 10^{-6}\text{ C}$ ) and substituting directly:

F = \dfrac{(8.99\times10^9)(3\times10^{-6})(5\times10^{-6})}{(0.20)^2} \approx \textcolor{#e08a1e}{3.37\text{ N, attractive}}

Attractive, because the charges carry opposite signs — each pulls the other inward along the line joining them.

1.3 The Principle of Superposition and the Electric Field

Rather than recompute Coulomb’s law for every pair of charges whenever a new one is introduced, it is far more useful to assign a property to space itself: the electric field. Imagine placing a small, positive test charge $q_0$ at a point, small enough that it does not disturb the source charges producing the field. The field at that point is the force per unit test charge:

Electric field

\mathbf{E} = \dfrac{\mathbf{F}}{q_0}

measured in N/C (equivalently V/m). For a single source charge $Q$ , this is immediate from Coulomb’s law: $\mathbf{E} = k\dfrac{Q}{r^2}\hat{r}$ , pointing radially outward from $Q$ if it is positive, and radially inward if negative.

The genuinely powerful idea is the principle of superposition: because Coulomb’s law is linear in each source charge, the presence of other charges never alters any individual charge’s own contribution. So for several source charges $q_1, q_2, \ldots, q_n$ , the net field at any point is just the vector sum of the fields each one would produce on its own, as if the others were not there:

\mathbf{E} = \mathbf{E}_1 + \mathbf{E}_2 + \cdots + \mathbf{E}_n

This is exactly what the chapter’s interactive simulation puts numbers to. It fixes two source charges $q_1$ and $q_2$ 10 cm apart on a line, with a labelled point P sitting a further 15 cm beyond $q_2$ (25 cm from $q_1$ , all on the same axis), and uses $k=8.99\times10^9$ N·m²C⁻² throughout, so every value it displays is hand-checkable. Dragging either charge redraws a whole grid of field-direction arrows, the vector sum of both charges’ individual fields at every point in the plane, and recomputes the exact signed field at P.

Worked example

Superposition at the simulation's test point

With the simulation’s default charges, $q_1=+2\,\mu\text{C}$ at the origin and $q_2=-2\,\mu\text{C}$ 10 cm to its right, find the net field at P, 25 cm from $q_1$ along the same axis (taking rightward, away from $q_1$ , as positive).

$q_1$ is 25 cm from P and positive, so its field at P points further rightward (away from $q_1$ ):

E_1 = \dfrac{(8.99\times10^9)(2\times10^{-6})}{(0.25)^2} \approx +2.88\times10^5\text{ N/C}

$q_2$ is only 15 cm from P (P is beyond it) and negative, so its field at P points back toward $q_2$ , i.e. leftward, which is the negative direction here:

E_2 = \dfrac{(8.99\times10^9)(2\times10^{-6})}{(0.15)^2} \approx -7.99\times10^5\text{ N/C}

Superposition just adds the two signed numbers:

E_{\text{net}} = E_1+E_2 \approx \textcolor{#e08a1e}{-5.11\times10^5\text{ N/C}}

Negative, so the net field at P actually points back toward the charges (leftward), not away from them — the closer $q_2$ wins out over the more distant $q_1$ despite $q_1$ being the one P is further from “in front of”. Dragging either slider in the simulation reproduces this same number exactly, since it is computed by precisely this sum.

1.4 Electric Field Lines and Electric Flux

A field line is a curve drawn so that the tangent to it, at every point, gives the direction of $\mathbf{E}$ there. A few properties follow directly from this definition and from the field being single-valued at every point:

Field lines start on positive charge (or at infinity) and end on negative charge (or at infinity); they are never closed loops in electrostatics.
Two field lines can never cross. If they did, the field at the crossing point would have two different directions at once, which is meaningless.
Where lines are drawn closer together, the field is stronger; where they spread apart, it is weaker. Crowding is a visual stand-in for magnitude.

Electric flux $\Phi$ through a surface measures how much of the field “passes through” it. For a flat area $\Delta A$ in a uniform field $\mathbf{E}$ , with $\theta$ the angle between $\mathbf{E}$ and the area’s outward normal:

\Delta\Phi = \mathbf{E}\cdot\Delta\mathbf{A} = E\,\Delta A\cos\theta

For an arbitrary (possibly curved, possibly closed) surface, this is summed, i.e. integrated, over every small patch: $\Phi = \oint \mathbf{E}\cdot d\mathbf{A}$ . Flux is a scalar, measured in N·m²/C, and it is exactly this quantity that Gauss’s law (§1.7) relates to enclosed charge.

1.5 The Electric Dipole

An electric dipole is a pair of equal and opposite charges, $+q$ and $-q$ , separated by a small distance $2a$ . Its strength and orientation are captured by a single vector, the dipole moment:

Dipole moment

\textcolor{#e08a1e}{\mathbf{p}} = q(2a)\,\hat{p}

directed from the negative to the positive charge, with magnitude $p=q\times2a$ . The water molecule and the ammonia molecule are everyday permanent dipoles for exactly this reason: their positive and negative charge centres do not coincide.

1.5.1 Field on the Axial Line

Take a point P at distance $r$ from the dipole’s centre, along the line through both charges, on the side nearer $+q$ . P is at distance $r-a$ from $+q$ (field pointing further outward, away from $+q$ ) and $r+a$ from $-q$ (field pointing back toward $-q$ , i.e. the same outward direction, since P lies beyond the midpoint). The two partially cancel, but the nearer $+q$ wins:

E_{\text{axial}} = kq\left[\dfrac{1}{(r-a)^2}-\dfrac{1}{(r+a)^2}\right] = kq\cdot\dfrac{4ar}{(r^2-a^2)^2}

Writing $4ar\cdot q = 2r(2aq) = 2rp$ turns this into a clean result in terms of $p$ alone:

E_{\text{axial}} = \dfrac{2kpr}{(r^2-a^2)^2}\;\xrightarrow{\,r\gg a\,}\;\dfrac{2kp}{r^3}

directed along $\mathbf{p}$ .

1.5.2 Field on the Equatorial Line

Now take P at perpendicular distance $r$ from the centre, on the line bisecting the dipole at right angles. Both charges are now equidistant from P, $\sqrt{r^2+a^2}$ , so the two fields have equal magnitude $kq/(r^2+a^2)$ . By symmetry, the components perpendicular to the dipole axis cancel exactly, while the components along the axis both point the same way opposite to $\mathbf{p}$ , and add:

E_{\text{equatorial}} = \dfrac{kp}{(r^2+a^2)^{3/2}}\;\xrightarrow{\,r\gg a\,}\;\dfrac{kp}{r^3}

directed opposite to $\mathbf{p}$ , and exactly half the axial magnitude at the same $r$ . Either way, notice the dipole field falls off as $1/r^3$ , a full power of $r$ faster than a single charge’s $1/r^2$ : at large distances the equal and opposite charges increasingly cancel each other out, leaving a much weaker residual field.

1.6 Torque on a Dipole in a Uniform External Field

Place a dipole in a uniform external field $\mathbf{E}$ , at angle $\theta$ between $\mathbf{p}$ and $\mathbf{E}$ . The force on $+q$ is $q\mathbf{E}$ and on $-q$ is $-q\mathbf{E}$ , equal in magnitude and opposite in direction since the field is the same at both locations (that sameness is what “uniform” buys you). The net force is therefore zero.

But the two equal, opposite forces act at two different points, separated by a perpendicular distance $2a\sin\theta$ , so they form a couple, a pure twisting effect with no net push. Its torque is force times perpendicular separation, $\tau = qE(2a\sin\theta) = pE\sin\theta$ , or in vector form:

Torque on a dipole

\boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}

The torque vanishes only when $\sin\theta=0$ , i.e. $\mathbf{p}$ aligned with $\mathbf{E}$ ( $\theta=0$ , a stable equilibrium) or directly against it ( $\theta=180^\circ$ , an unstable one). For every other orientation the dipole feels zero net force but a nonzero net torque that swings it toward alignment, exactly how a compass needle settles along a magnetic field, except here it is an electric dipole settling along $\mathbf{E}$ .

Worked example

Torque on a dipole at 30°

A dipole of moment $p=2\times10^{-8}\text{ C m}$ is held at $30^\circ$ to a uniform field $E=5\times10^4\text{ N/C}$ . Find the torque on it.

\tau = pE\sin\theta = (2\times10^{-8})(5\times10^4)(\sin30^\circ) = \textcolor{#e08a1e}{5.0\times10^{-4}\text{ N m}}

1.7 Gauss's Law

Gauss’s law states that the total electric flux through any closed surface (a Gaussian surface) depends on nothing but the charge it encloses:

Gauss's law

\oint \mathbf{E}\cdot d\mathbf{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}

strikingly independent of the surface’s exact shape or size, of where inside it the charge sits, and of any charge outside the surface (external charge contributes flux entering one part of the surface and an equal amount leaving elsewhere, net zero). This is not an independent new law so much as a restatement of Coulomb’s law, made possible specifically because the field falls off as $1/r^2$ : surround a single point charge $q$ with an imaginary sphere of radius $r$ centred on it, and the flux through that sphere is

\Phi = E\cdot(4\pi r^2) = \dfrac{kq}{r^2}\cdot4\pi r^2 = 4\pi kq = \dfrac{q}{\varepsilon_0}

with the $r^2$ in the field and the $r^2$ in the sphere’s area cancelling exactly, leaving a result with no $r$ left in it at all. Superposing over many charges, and over surfaces of any shape, extends this to the general law above. The practical payoff, used throughout §1.8, is that for a sufficiently symmetric charge distribution, you can choose a Gaussian surface on which $\mathbf{E}$ is either constant in magnitude and parallel to the surface’s normal, or exactly perpendicular to it (zero flux), turning the surface integral into plain algebra.

1.8 Applications of Gauss's Law

1.8.1 Field of an Infinite Line Charge

An infinite straight line carries a uniform linear charge density $\lambda$ (C/m). By the symmetry of an infinite line, $\mathbf{E}$ can only point radially outward (or inward) from the line, with a magnitude depending only on the perpendicular distance $r$ , never on position along the line. This suggests a cylindrical Gaussian surface of radius $r$ and length $l$ , coaxial with the line.

The two flat end-caps contribute zero flux ( $\mathbf{E}$ runs parallel to them, perpendicular to their normal); only the curved side wall, where $\mathbf{E}$ is both constant in magnitude and everywhere parallel to the outward normal, contributes:

\Phi = E\cdot(2\pi r l) = \dfrac{q_{\text{enc}}}{\varepsilon_0} = \dfrac{\lambda l}{\varepsilon_0}

E = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{2k\lambda}{r}

falling off as $1/r$ , slower than a point charge’s $1/r^2$ , because charge keeps contributing from further and further along the line as $r$ grows.

1.8.2 Field of an Infinite Plane Sheet of Charge

A thin, infinite plane sheet carries a uniform surface charge density $\sigma$ (C/m²). By planar symmetry, $\mathbf{E}$ must point straight away from the sheet on both sides (toward it, if $\sigma$ is negative), with equal magnitude at equal distances. Choose a “pillbox”: a short cylinder of cross-sectional area $A$ straddling the sheet, its two flat faces parallel to it.

The curved side contributes nothing ( $\mathbf{E}$ runs parallel to it); each flat face contributes $EA$ , and both point outward through their respective face, so:

\Phi = 2EA = \dfrac{q_{\text{enc}}}{\varepsilon_0} = \dfrac{\sigma A}{\varepsilon_0}

Field of an infinite sheet

\textcolor{#e08a1e}{E} = \dfrac{\sigma}{2\varepsilon_0}

Notably, this does not depend on the distance from the sheet at all: close to an (idealised, infinite) charged sheet, the field is perfectly uniform.

Worked example

Field near a charged sheet

A large plane sheet carries a uniform surface charge density $\sigma = 4.0\times10^{-9}\text{ C/m}^2$ . Find the field near the sheet (away from its edges).

E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{4.0\times10^{-9}}{2(8.85\times10^{-12})} \approx \textcolor{#e08a1e}{226\text{ N/C}}

— From the NCERT exercises

Two of the chapter’s own classic exercise questions, with original worked solutions.

NCERT Exercise 1.11

An electric dipole with dipole moment 4×10⁻⁹ C m is aligned at 30° with the direction of a uniform electric field of magnitude 5×10⁴ N/C. Calculate the magnitude of the torque acting on the dipole.

Solution

Directly from §1.6’s result $\tau=pE\sin\theta$ :

\tau = (4\times10^{-9})(5\times10^4)(\sin30^\circ) = (2\times10^{-4})(0.5) = \textcolor{#e08a1e}{1\times10^{-4}\text{ N m}}

NCERT Exercise 1.15

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m². (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Solution

The radius is $R=1.2$ m, so the surface area is $4\pi R^2 = 4\pi(1.2)^2 \approx 18.10\text{ m}^2$ . The charge is just this area times the surface charge density:

Q = \sigma\cdot4\pi R^2 = (80.0\times10^{-6})(18.10) \approx \textcolor{#e08a1e}{1.45\times10^{-3}\text{ C}}

For part (b), Gauss’s law makes this almost no extra work at all: the total flux leaving any closed surface, including the sphere’s own surface, is just the enclosed charge divided by $\varepsilon_0$ , with no geometry left to compute:

\Phi = \dfrac{Q}{\varepsilon_0} = \dfrac{1.45\times10^{-3}}{8.85\times10^{-12}} \approx \textcolor{#e08a1e}{1.64\times10^{8}\text{ N m}^2\text{/C}}