Electric Charges and Fields

Electric charge is the property of matter because of which a body exerts and experiences electric forces. The Greek word elektron means amber; amber rubbed with wool attracts light objects, and that is the oldest electrical observation on record. Rubbing transfers electrons from one body to the other. The body that loses electrons is left with a deficit and becomes positively charged; the body that gains them becomes negatively charged. The signs are a convention that goes back to Benjamin Franklin: a glass rod rubbed with silk is positive and the silk negative, while a plastic or ebonite rod rubbed with fur or wool is negative and the fur positive. A polythene piece rubbed with wool is negative, which is the setting of NCERT exercise 1.11.

Three experimental facts carry the whole chapter. Like charges repel and unlike charges attract. The two rubbed bodies acquire equal and opposite charges, so the pair as a whole stays neutral. And a charged body attracts light uncharged objects, because it polarises them. A gold-leaf electroscope shows the presence of charge: when the knob is charged, the two leaves receive the same kind of charge and diverge, and the larger the charge, the larger the divergence.

Write these two lines whenever a question asks what charging means: a body is charged when it has an excess or a deficiency of electrons, and charging by friction is a transfer of electrons, not a creation of charge. A negatively charged body is very slightly heavier than the same body when neutral, because it has gained electrons; a positively charged body is very slightly lighter. Charge has no direction; it is a scalar, and its SI unit is the coulomb (C).

Conductors are materials that let electric charge pass through them easily; metals, the human body, the earth and salt solutions are conductors. Insulators such as glass, porcelain, plastic, nylon and dry wood offer high resistance to the passage of charge. The difference is in the electrons. In a metal the outer electrons of the atoms are free to move through the whole piece; in an insulator every electron stays bound to its own atom or molecule.

This changes what happens when you charge them. Charge given to a conductor spreads at once over its entire outer surface; charge given to an insulator stays exactly where it was put. It also explains a common demonstration: a metal rod held in the hand and rubbed with wool shows no charge, because whatever charge is produced leaks through the hand and the body to the earth. Hold the rod by a wooden or plastic handle and it stays charged. Connecting a body to the earth through a conductor is called earthing or grounding. The earth is so large that it can absorb or supply any amount of charge without its own potential changing, which is why the third pin of a household plug is connected to a metal plate buried in the ground.

Two consequences appear again and again in board questions. When two identical conducting spheres touch, the total charge divides equally between them, so balls carrying −Q and +3Q end up with +Q each and then repel (CBSE 2019). And the charge a metal sphere holds does not depend on whether it is hollow or solid (CBSE 2017), because in either case the charge sits only on the outer surface. Semiconductors such as silicon and germanium lie between the two classes: they conduct better than insulators but far worse than metals.

Three properties are examined, usually as a 1- or 2-mark question, and each is worth stating in one exact sentence.

Additivity. Charge is a scalar. The total charge of a system is the algebraic sum of the charges on its parts, signs included: a system with +5 μC, −3 μC and +1 μC has a total charge of +3 μC. Unlike mass, charge can be negative, so the total can be zero or smaller than any individual charge.

Conservation. The total charge of an isolated system is constant. Charge can be transferred from one body to another, but it is never created or destroyed. When a glass rod is rubbed with silk the rod becomes positive and the silk negative by exactly the same amount, so the total stays zero (NCERT exercise 1.5). In a nuclear process a neutron decays into a proton and an electron (with an antineutrino); the total charge is zero before and after.

Quantisation. All free charges are integral multiples of a basic unit e, the magnitude of the charge on an electron: q = ne with n = 0, ±1, ±2, ... and e = 1.6 × 10⁻¹⁹ C. So one coulomb contains 1/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons, and a charge of 2.4 × 10⁻¹⁹ C (1.5 e) is impossible. Faraday's laws of electrolysis first suggested quantisation; Millikan's oil-drop experiment (1912) established it. On the macroscopic scale e is so small compared with the charges we handle (a microcoulomb means about 6 × 10¹² electrons) that charge appears continuous, just as a heap of sand looks continuous though it is made of grains. That is why quantisation is ignored in macroscopic problems and charge is treated as a continuous variable (NCERT exercise 1.4).

Coulomb measured the force between charged spheres with a torsion balance (1785). The law: the force between two point charges at rest is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, and it acts along the line joining them. In magnitude, F = (1/4πε₀) · |q₁||q₂|/r². The constant 1/4πε₀ = 9 × 10⁹ N m² C⁻², and ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space. The law also defines the coulomb: 1 C is the charge which, placed 1 m from an equal charge in vacuum, repels it with a force of 9 × 10⁹ N.

Vector form (a regular 2-mark question). Let r₁ and r₂ be the position vectors of q₁ and q₂. Write r₂₁ = r₂ − r₁ for the vector from charge 1 to charge 2, and r̂₂₁ for the unit vector along it. The force on q₂ due to q₁ is F₂₁ = (1/4πε₀) · (q₁q₂/r₂₁²) r̂₂₁. If q₁q₂ > 0 the force is along r̂₂₁, that is, repulsive; if q₁q₂ < 0 it is along −r̂₂₁, attractive. Since r̂₁₂ = −r̂₂₁, F₁₂ = −F₂₁: the two forces are equal and opposite, in agreement with Newton's third law. Say in your answer that the vector form takes care of the direction and of both signs of charge automatically.

In a medium of dielectric constant (relative permittivity) K the force falls to F/K, because ε = Kε₀. Two charges that repel with 8 N in air repel with 8/80 = 0.1 N in water (K = 80). Coulomb's law holds for point charges at rest; for several charges the forces must be added vectorially, which is the next section. The law has the same 1/r² form as gravitation but is enormously stronger: for an electron and a proton the ratio of the electric to the gravitational force is about 2.3 × 10³⁹ (NCERT exercise 1.3), which is why gravity is ignored in atomic physics.

Coulomb's law gives the force between one pair. For three or more charges we need one more experimental fact, the principle of superposition: the force on any charge due to a number of other charges is the vector sum of the forces on it due to each of the others taken one at a time, and each pair force is unaffected by the presence of the remaining charges. For charges q₁, q₂, ..., q_n at r₁, ..., r_n, the force on q₁ is F₁ = F₁₂ + F₁₃ + ... + F_1n = (q₁/4πε₀) ∑ (q_i/r_1i²) r̂_1i, the sum running over i = 2 to n.

The exam versions are geometric. For charges at the corners of a square or an equilateral triangle, draw every force on the chosen charge as an arrow along the line joining the pair, resolve along two perpendicular directions, add the components, then find the magnitude and direction. In NCERT exercise 1.6, four charges at the corners of a square with equal charges on opposite corners give zero net force on a charge at the centre, because each diagonal pair produces equal and opposite forces. A charge at one corner of an equilateral triangle with charges at the other two feels two forces at 60° to each other, and the magnitude comes from F = √(F₁² + F₂² + 2F₁F₂ cos 60°).

The other standard question is the position where a third charge feels no net force. For like charges +q and +4q a distance d apart, the null point lies between them, nearer the smaller charge: kqQ/x² = k(4q)Q/(d − x)² gives d − x = 2x, so x = d/3 from q. For unlike charges the null point lies outside the pair, on the side of the smaller charge. The sign and size of the third charge never matter, because they cancel from the equation.

A charge does not need to touch another charge to push it; the field is how we describe this action at a distance. The electric field at a point is the force experienced by a unit positive test charge placed there: E(r) = F/q₀, taken in the limit q₀ → 0 so that the test charge does not disturb the charges producing the field. Its SI unit is N C⁻¹ (equivalently V m⁻¹) and it is a vector. Once E is known, the force on any charge q at that point is F = qE: along E for a positive charge and opposite to E for a negative one. A proton placed in a field along +x moves along +x; an electron moves along −x.

Field of a point charge. Put Q at the origin and the test charge q₀ at r. Coulomb's law gives F = (1/4πε₀) Qq₀/r² r̂, so E = (1/4πε₀) Q/r² r̂. It is radially outward for Q > 0 and inward for Q < 0, has the same magnitude at all points of a sphere centred on Q (spherical symmetry) and falls off as 1/r². Field of a system of charges. By superposition, E(r) = (1/4πε₀) ∑ (q_i/r_iP²) r̂_iP, the vector sum of the individual fields. In NCERT exercise 1.8, +3 μC and −3 μC placed 20 cm apart give a field at the midpoint made of two parallel contributions, 5.4 × 10⁶ N C⁻¹ in total, directed from the positive charge towards the negative one.

Physical significance. The field is a property of the charge configuration at every point of space, whether or not a test charge is there to feel it. Its real importance shows when charges move: if q₁ is displaced, q₂ does not feel the change instantly but only after a time r/c, because the change travels through the field at the speed of light. Time-dependent electric and magnetic fields carry energy and momentum and propagate as electromagnetic waves. So the field is not just a calculating device but a physical entity with its own dynamics.

Faraday introduced field lines as a picture of the field. A field line is a curve drawn so that its tangent at any point gives the direction of E there, and the lines are spaced so that the number crossing unit area held normal to them is proportional to the magnitude of E. Around a single point charge the lines are radial straight lines. The same number of lines cross every sphere centred on the charge, and the area of the sphere grows as r², so the density of lines, and with it the field, falls as 1/r². Crowded lines mean a strong field.

Properties to write, in this order. (i) Field lines start from positive charges and end at negative charges; if there is a single charge they start or end at infinity. (ii) In a charge-free region they are continuous curves without breaks. (iii) Two field lines never cross, because at the crossing point the field would have two directions, which is impossible. (iv) Electrostatic field lines do not form closed loops; this follows from the conservative nature of the electrostatic field. If a line closed on itself, a charge carried once round the loop would have net work done on it by the field, which cannot happen in an electrostatic field. Point (iv) was asked for 1 mark in CBSE 2014, and the continuity of lines is NCERT exercise 1.7.

Learn the four standard sketches: a single positive charge (lines radially outward), a single negative charge (inward), a dipole (lines leaving +q and curving round to −q, densest between the charges) and two equal positive charges (lines pushed apart, with a null point midway through which no line passes). Also learn the sketch of a point charge near an uncharged conducting plate (CBSE 2019): the lines leave the charge and meet the plate at right angles, because the field must be normal to the surface of a conductor. Field lines are a picture of a real vector field; they are not the paths of charges.

Think of the field lines as a flow. Electric flux measures how many of them pass through a surface. For a small plane area element define the area vector ΔS, of magnitude ΔS and direction along the normal to the surface. The flux through it is ΔΦ = E · ΔS = E ΔS cos θ, where θ is the angle between E and the normal. The flux is maximum (E ΔS) when the field is along the normal, that is, when the surface faces the field, and zero when the field lies in the plane of the surface (θ = 90°). The flux through a whole surface is the sum Φ = ∑ E · ΔS, which becomes a surface integral in the limit. Flux is a scalar. Its SI unit is N m² C⁻¹ (equivalently V m) and its dimensional formula is [M L³ T⁻³ A⁻¹].

For a closed surface the convention is that the outward normal is positive: flux leaving the surface is positive and flux entering is negative. A uniform field through a closed surface such as a cube gives zero net flux, because whatever enters one face leaves through the opposite one (NCERT exercise 1.15). A uniform field E = 5 × 10³ î N C⁻¹ through a square of side 10 cm parallel to the yz plane gives Φ = EA = 5 × 10³ × 0.01 = 50 N m² C⁻¹; if the plane is turned to make 30° with the x-axis, the normal makes 60° with the field and Φ = 50 cos 60° = 25 N m² C⁻¹ (CBSE 2014).

Two traps. The angle in the formula is between the field and the normal, not between the field and the plane. And for a non-uniform field, such as E = 50x î through a cylinder along the x-axis (CBSE 2013), evaluate the flux face by face: the two flat faces sit at different x and so have different E, the curved surface has zero flux because E is parallel to it, and the net flux is the difference between the two ends.

An electric dipole is a pair of equal and opposite point charges, +q and −q, separated by a distance 2a. Its dipole moment is p = q × 2a, a vector of magnitude 2qa directed along the axis from −q to +q. The SI unit is C m. The total charge of a dipole is zero, but its field is not, because the two charges are at slightly different distances from any field point. The field is proportional to p and falls off as 1/r³ at large distances, faster than the 1/r² field of a point charge.

Axial line (the line through both charges). Let P be at distance r from the centre O, on the side of +q. The field due to +q is q/[4πε₀(r − a)²] along p; the field due to −q is q/[4πε₀(r + a)²] opposite to p. P is nearer to +q, so the net field is along p: E = (q/4πε₀)[1/(r − a)² − 1/(r + a)²] = (q/4πε₀) · 4ar/(r² − a²)² = 2pr/[4πε₀(r² − a²)²]. For r ≫ a, E_axial = 2p/(4πε₀r³).

Equatorial line (the perpendicular bisector of the axis). P is now at distance √(r² + a²) from each charge, so the two fields have equal magnitudes q/[4πε₀(r² + a²)]. Their components perpendicular to the axis cancel; the components along the axis add, each being E_+q cos θ with cos θ = a/√(r² + a²). The net field is E = 2qa/[4πε₀(r² + a²)^3/2] = p/[4πε₀(r² + a²)^3/2], directed opposite to p. For r ≫ a, E_equatorial = −p/(4πε₀r³). At the same distance the axial field is twice the equatorial field, and the two point in opposite senses relative to p.

Physical significance. In molecules such as CO₂ and CH₄ the centres of positive and negative charge coincide, so the permanent dipole moment is zero; in polar molecules such as H₂O, NH₃ and HCl they do not coincide, and the molecule has a permanent dipole moment. A non-polar molecule develops an induced dipole moment when placed in an external field.

Place a dipole in a uniform field E with p at angle θ to E. The charge +q feels qE and −q feels −qE. The two forces are equal and opposite, so the net force is zero and the dipole does not translate. But they do not act along the same line, so they form a couple and produce a torque. Magnitude of torque = either force × perpendicular distance between their lines of action = qE × 2a sin θ = pE sin θ. In vector form τ = p × E, perpendicular to the plane of p and E; the torque tends to rotate the dipole so as to align p with E. This derivation was asked for 2 marks in CBSE 2020 and 2017.

The torque is zero at θ = 0 and θ = 180° and maximum (pE) at θ = 90°. At θ = 0, with p parallel to E, the dipole is in stable equilibrium: a small rotation produces a torque that brings it back. At θ = 180°, with p antiparallel to E, it is in unstable equilibrium: a small rotation produces a torque that turns it further away (CBSE 2008, 2010). NCERT exercise 1.10: a dipole of moment 4 × 10⁻⁹ C m at 30° to a field of 5 × 10⁴ N C⁻¹ feels τ = 4 × 10⁻⁹ × 5 × 10⁴ × sin 30° = 1 × 10⁻⁴ N m.

If the field is not uniform, the forces on the two charges differ in magnitude and the net force is no longer zero. When p is parallel to E the dipole is pulled in the direction of increasing field, and when antiparallel it is pushed towards decreasing field; in these two cases there is a net force but no torque. This is why a charged comb picks up bits of paper: the non-uniform field of the comb induces a dipole in the paper with its moment along the field, and the net force is towards the comb.

For a charged wire, sheet or solid we do not count electrons; we treat the charge as spread continuously and describe it by a density. Linear charge density λ = ΔQ/Δl (C m⁻¹) for a wire, surface charge density σ = ΔQ/ΔS (C m⁻²) for a sheet or the surface of a conductor, and volume charge density ρ = ΔQ/ΔV (C m⁻³) for charge filling a volume. Here ΔQ is the charge in a small length, area or volume element that is small on the laboratory scale but still contains a very large number of elementary charges, so the density is a smooth function of position. This is exactly the macroscopic limit in which quantisation can be ignored.

The field of a continuous distribution follows from Coulomb's law and superposition. Divide the body into small elements, treat the charge ρΔV of each as a point charge, and add the fields vectorially: E(r) = (1/4πε₀) ∑ (ρΔV/r'²) r̂', the sum running over all elements, with r' the distance from the element to the field point. In the limit the sum becomes an integral. For most shapes this integral is hard; for the symmetric ones we use Gauss's law instead, which is the point of the last two sections.

A uniformly charged conducting sphere of diameter 2.4 m with σ = 80 μC m⁻² carries Q = σ × 4πR² = 80 × 10⁻⁶ × 4π × (1.2)² = 1.45 × 10⁻³ C (NCERT exercise 1.21). A wire with λ = 0.1 μC m⁻¹ carries 0.1 μC on every metre of its length. Keep the three quantities apart: a question that gives σ for a sheet and λ for a wire is testing exactly whether you know which formula each one enters.

Gauss's law: the electric flux through any closed surface S is 1/ε₀ times the total charge enclosed by S. Φ = ∮ E · dS = q_enclosed/ε₀. The circle on the integral sign means the surface is closed, and the closed surface chosen for applying the law is called a Gaussian surface.

Verification for a point charge. Take a sphere of radius r centred on a charge q. At every point of the sphere E is radial with magnitude q/(4πε₀r²), parallel to the outward normal, so the flux through an element ΔS is E ΔS. Adding over the whole sphere, Φ = E × 4πr² = [q/(4πε₀r²)] × 4πr² = q/ε₀. The r² cancels: the flux is the same through a sphere of any radius and, because field lines are continuous in empty space, the same through any closed surface enclosing q. This cancellation is a consequence of the inverse-square law; Gauss's law is Coulomb's law in another form. Doubling the radius of a Gaussian sphere leaves the flux unchanged (NCERT exercise 1.19; CBSE 2009, 2016).

Points the examiner looks for. (i) The law holds for any closed surface, whatever its shape or size. (ii) The charge on the right side is the total charge enclosed, with sign; charges outside contribute nothing to the net flux, because every line they send in also comes out. Charges −2Q at (a, 0) and +Q at (4a, 0) give a flux of −2Q/ε₀ through a sphere of radius 3a about the origin, since only −2Q is inside (CBSE 2013). (iii) The E on the left side is the total field due to all charges, inside and outside, even though only the enclosed charge fixes the flux. (iv) Zero flux means zero net enclosed charge; it does not mean zero field and it does not mean no charges. A dipole inside a cube gives zero flux (CBSE 2015). (v) The law is useful for calculating E only when symmetry lets you choose a surface on which E is constant and either normal or parallel to the surface. A charge at the centre of a cube sends q/(6ε₀) through each face by symmetry (CBSE 2012); a charge at a corner sends q/(8ε₀) through the cube, since the cube covers one-eighth of the space around it.

Three derivations, each a regular 5-mark question, follow one recipe: use symmetry to say what E looks like, choose a Gaussian surface that matches it, evaluate the flux, set it equal to the enclosed charge over ε₀, and solve for E.

Infinitely long straight wire, linear charge density λ. By symmetry the field is radial, perpendicular to the wire, with the same magnitude at all points at distance r. Take a coaxial cylinder of radius r and length l as the Gaussian surface. On the curved surface E is along the outward normal, so its flux is E × 2πrl; on the two flat ends E is perpendicular to the normal, so the flux is zero. The enclosed charge is λl. Gauss's law gives E × 2πrl = λl/ε₀, so E = λ/(2πε₀r), radially outward for positive λ and inward for negative λ. The graph of E against r is a rectangular hyperbola (CBSE 2018).

Infinite plane sheet, surface charge density σ. By symmetry the field is perpendicular to the sheet, pointing away from it on both sides for positive σ, with the same magnitude at equal distances. Take a cylinder (pillbox) of cross-section A whose axis is normal to the sheet and which the sheet cuts in half. The flux through the curved surface is zero; through each flat end it is EA, so the total flux is 2EA. The enclosed charge is σA. So 2EA = σA/ε₀ and E = σ/(2ε₀), independent of the distance from the sheet, directed away from the sheet for σ > 0 and towards it for σ < 0 (CBSE 2012). For two parallel sheets carrying +σ and −σ the fields add between the sheets to σ/ε₀ and cancel outside (NCERT exercise 1.23).

Uniformly charged thin spherical shell, radius R, charge q, so σ = q/(4πR²). Outside (r > R): take a concentric Gaussian sphere of radius r; E is radial and constant on it, the flux is E × 4πr² and the enclosed charge is q, so E = q/(4πε₀r²) = σR²/(ε₀r²). The shell behaves as if all its charge were at the centre. Inside (r < R): the concentric Gaussian sphere encloses no charge, so E × 4πr² = 0 and E = 0 everywhere inside. On the surface E = σ/ε₀. The E–r graph is zero up to R, jumps to its maximum q/(4πε₀R²) at R and then falls as 1/r² (CBSE 2008).