Electric Charges and Fields | ISC Class 12 Physics Notes

This note covers electric charge, conductors and insulators, Coulomb’s law, electric fields and field lines, continuous charge distributions, electric dipoles, torque, electric flux, Gauss’s theorem, and the fields of a straight wire, plane sheet and thin spherical shell.

What is electric charge, and which properties does it obey?

Electric charge is the property responsible for electrical attraction and repulsion. Electrostatics studies forces and fields associated with static charges. Static means not moving or changing with time. Positive and negative are the two kinds of charge: like charges repel and unlike charges attract.

An electron is a negatively charged atomic particle; a proton is a positively charged particle in the nucleus, the central part of an atom. An atom has electrons surrounding its nucleus. An ion is an atom or group of atoms carrying a net charge.

Normally, materials are electrically neutral because their charges balance. Losing electrons makes a body positive; gaining them makes it negative. Rubbing glass with silk transfers electrons to the silk without creating charge.

How do conductors differ from insulators?

Most non-metals offer high resistance to the passage of electricity. A metal object held directly can lose charge through the body to earth; a wooden or plastic handle can prevent this connection.

How are charge, conservation and quantisation expressed?

In the International System of Units (SI), the SI unit of charge is the coulomb, symbol C. One coulomb passes through a wire in one second at a current of one ampere: 1 C = 1 A s. Current is charge transferred per unit time; A and s denote ampere and second.

Dimensions describe a quantity in terms of base quantities. Here M, L, T and A denote mass, length, time and electric current respectively. Charge has dimensions [TA]. A scalar has magnitude without direction; charge is a signed scalar.

• Additivity: total charge Q is the algebraic sum of individual charges q₁, q₂ and so on: Q = q₁ + q₂ + …. Their signs must be retained.
• Conservation: the total charge of an isolated system, one exchanging no charge with its surroundings, remains unchanged. Internal transfers redistribute charge.
• Quantisation: q = ne, where q is a body’s charge, n is an integer and e is the positive magnitude of the electron’s charge, approximately 1.6 × 10⁻¹⁹ C. Electron and proton charges are −e and +e.

For macroscopic charges enormous compared with e, quantisation has no practical consequence and can be ignored. For microscopic charges of a few tens or hundreds of e, it cannot be ignored.

How does Coulomb’s law describe the force between charges?

A point charge models a charged body whose size is very small compared with its separation from other charged bodies. Coulomb’s law gives the electrostatic force between two stationary point charges. The force acts along their joining line and follows an inverse-square dependence on separation.

Let q₁ and q₂ be the charges, r their separation and F the force magnitude. Separation has SI unit metre, m, and dimensions [L]. Force, a mechanical interaction, has SI unit newton, N, and dimensions [MLT⁻²]. Vertical bars around a quantity indicate its absolute magnitude.

F = k|q₁q₂|/r², where k = 1/(4πε₀). Here π is the ratio of a circle’s circumference to its diameter. The vacuum constant k is approximately 9 × 10⁹ N m² C⁻², with dimensions [ML³T⁻⁴A⁻²].

Permittivity is the medium-dependent quantity appearing in the electrostatic force law. The permittivity of free space, ε₀, is 8.854 × 10⁻¹² C² N⁻¹ m⁻². Its dimensions are [M⁻¹L⁻³T⁴A²]. Vacuum is the reference medium in this formula.

What do the vector form and medium dependence mean?

A vector has magnitude and direction; an arrow above a symbol denotes it. A hatted symbol denotes a dimensionless unit vector of magnitude one. Let r̂ point from q₁ towards q₂ and F⃗₂₁ be the force on q₂ due to q₁.

Then F⃗₂₁ = kq₁q₂r̂/r². A positive charge product gives repulsion; a negative product gives attraction. The force on q₁ due to q₂ is equal and opposite, agreeing with Newton’s third law.

A dielectric is an insulating material. Its permittivity ε and dimensionless relative permittivity, or dielectric constant K, satisfy ε = Kε₀. For charges in a uniform, isotropic linear dielectric, Coulomb’s formula uses ε instead of ε₀, provided the medium fills the intervening space.

Uniform means the material properties do not vary with position; isotropic means they do not depend on direction; linear means the induced electrical response is proportional to the applied field. Keeping charges and separation fixed, the force becomes its vacuum value divided by K.

How does electric force compare with gravitation?

Coulomb’s and Newton’s gravitational laws both give inverse-square forces along the joining line. Gravity attracts; electric force can attract or repel. Electrical attraction between an electron and proton is enormously stronger than gravitational attraction.

Worked example 1. Two small spheres carry +2 × 10⁻⁷ C and +3 × 10⁻⁷ C, separated by 30 cm in air. Find their mutual force, using k = 9 × 10⁹ N m² C⁻² and treating air as vacuum.

Formula: F = k|q₁q₂|/r². Substitute: r = 0.30 m; F = (9 × 10⁹)(2 × 10⁻⁷)(3 × 10⁻⁷)/(0.30)². Answer: 0.006 N, repulsive because both charges are positive.

How are electric field and superposition used?

A field assigns a physical quantity to each point in space. Gravitational, electric and magnetic fields are examples. The electric field intensity at a point is the force per unit positive test charge placed there without disturbing the source charges.

The source charge produces the field; a small test charge probes it without disturbing the source configuration. For test charge q₀ and force F⃗, E⃗ = F⃗/q₀ in the limit of vanishingly small q₀.

The SI unit of electric field is N C⁻¹; its dimensions are [MLT⁻³A⁻¹]. For a point source Q in vacuum, E = k|Q|/r², where E is field magnitude and r is distance from Q. The direction is radially outward for positive Q and inward for negative Q.

Equal field magnitudes at equal distances from a point charge express spherical symmetry. The field exists independently of the test charge. A charge q in an external field experiences F⃗ = qE⃗.

What does the superposition principle state?

Each pairwise electrostatic force is unaffected by the presence of other charges. The resultant force is their vector sum. Similarly, E⃗ = E⃗₁ + E⃗₂ + …, where E⃗₁ and E⃗₂ are the separate field contributions at the same observation point.

Determine directions first. Collinear fields add or subtract according to direction; otherwise add their components. A component is a vector’s projection along a chosen direction.

Worked example 2. Charges +10⁻⁸ C and −10⁻⁸ C are 0.10 m apart in vacuum. Find the field at their midpoint using k = 9 × 10⁹ N m² C⁻².

Formula: E₁ = k|q₁|/r²; E₂ = k|q₂|/r²; E = E₁ + E₂. Substitute: r = 0.05 m, so each field is (9 × 10⁹)(10⁻⁸)/(0.05)² = 3.6 × 10⁴ N C⁻¹. Answer: 72000 N C⁻¹, from the positive charge towards the negative charge.

How does a uniform field accelerate an electron?

A uniform field has the same magnitude and direction throughout the region. For particle mass m, Newton’s second law gives acceleration magnitude a = |q|E/m. Mass has unit kilogram, kg, and dimensions [M]; acceleration, the rate of change of velocity, has unit m s⁻² and dimensions [LT⁻²].

Worked example 3. An electron starts from rest and falls 1.5 cm in an upward uniform field of 2.0 × 10⁴ N C⁻¹. Find its fall time, neglecting gravity. Use e = 1.6 × 10⁻¹⁹ C and electron mass m = 9.11 × 10⁻³¹ kg.

Formula: a = eE/m; t = √(2h/a), where t is time in seconds and h is the fall distance. Substitute: h = 1.5 × 10⁻² m; t = √[2(1.5 × 10⁻²)(9.11 × 10⁻³¹)/((1.6 × 10⁻¹⁹)(2.0 × 10⁴))]. Answer: 0.0000000029 s (2.9 × 10⁻⁹ s). The electron’s force is downward, opposite to the field.

How is a continuous charge distribution described?

A continuous charge distribution smooths many microscopic charges into a macroscopic distribution. Each element contains many charges but is small enough to represent local conditions. Quantisation is ignored at this scale.

The symbol Δ means a small finite amount; d denotes an infinitesimal element used in integration. Let Δq be the charge in an element of length Δl, area ΔA or volume ΔV. Area and volume have units m² and m³, and dimensions [L²] and [L³].

Charge densities can vary with position; uniform density means constant density.

How is the field obtained from the density?

For an infinitesimal element, dq = λ dl, dq = σ dA or dq = ρ dV, according to the distribution. The symbol ∫ represents integration, the limiting sum over the chosen distribution.

Let s be the distance from a charge element to the observation point, and ŝ the unit vector pointing from that element towards the point. Its field contribution is dE⃗ = k dq ŝ/s². Add all contributions vectorially: E⃗ = k∫(dq ŝ/s²).

Distance and direction generally change between elements. The distribution’s geometry therefore matters as well as its density.

What do electric field lines show?

An electric field line is a curve whose tangent at each point gives the direction of the net electric field. A tangent is the local direction of the curve. Arrows distinguish the two possible directions along that tangent.

Lines represent a three-dimensional field. Crowded lines indicate stronger fields; widely spaced lines indicate weaker fields. Their absolute number depends on how many are chosen for the drawing.

• Lines start at positive charges and end at negative charges; for an isolated charge, they may start or end at infinity.
• In a charge-free region, lines can be taken as continuous curves without breaks.
• Two field lines cannot cross, because the field at a crossing would have two directions.
• Electrostatic field lines do not form closed loops.

The four drawings show outward radial arrows for a positive charge, inward radial arrows for a negative charge, curved outward lines around two positive charges, and lines directed from the positive to the negative member of an equal-and-opposite pair.

See Fig. 1.14 in your NCERT textbook

How is a uniform field represented?

Uniformly spaced parallel straight lines represent a uniform field. Between two oppositely charged parallel plates, draw the field arrows from the positive plate to the negative plate in the central region where edge effects are neglected.

Normal means perpendicular to a surface. Compare line densities across equal areas normal to the field, so orientation does not distort the comparison.

What is a dipole, and how is its axial field derived?

An electric dipole consists of equal and opposite point charges, +q and −q, separated by a distance 2l. Here q is the positive charge magnitude and l is half the separation. The joining line is the dipole axis; its midpoint is the dipole centre.

The dipole moment p⃗ is a vector directed from −q towards +q, with magnitude p = 2ql. The SI unit of dipole moment is C m, and its dimensions are [LTA]. Zero total charge does not imply a zero field because the two charges occupy different positions.

Derivation: Field at an external axial point

Take a point P on the axis, a distance r from the centre on the positive-charge side, with r greater than l. Let p̂ be the unit vector along the dipole moment.

1. The distance from +q to P is r − l. Its field has magnitude kq/(r − l)² and points along p̂.
2. The distance from −q to P is r + l. Its field has magnitude kq/(r + l)² and points opposite to p̂.
3. Subtract the opposing contributions: E⃗ = kq[1/(r − l)² − 1/(r + l)²]p̂.
4. Combining the fractions gives E⃗ = 4kqlr p̂/(r² − l²)². Substitute p = 2ql to express the result using dipole moment.

E⃗ = 2krp⃗/(r² − l²)² for the external axial point. The field is along the dipole moment on either external side of the axis.

When is the short-dipole approximation valid?

For r ≫ 2l, meaning the observation distance is much greater than the charge separation, l² can be neglected compared with r². The axial field becomes approximately E = 2kp/r³. This approximation concerns observation distance relative to dipole size.

The dipole’s inverse-cube decrease is faster than a point charge’s inverse-square decrease because the opposite charges’ fields nearly cancel at large distances.

How is the equatorial field of a dipole derived?

The equatorial plane is perpendicular to the dipole axis and passes through its centre. In a plane drawing, its intersection with the drawing is the perpendicular bisector of the dipole. Consider a point P in this plane at distance r from the centre.

Both charges are √(r² + l²) from P. Resolve their equal-magnitude fields into components parallel and perpendicular to the dipole axis.

Derivation: Field on the perpendicular bisector

1. Each field magnitude is kq/(r² + l²), because the square of the charge-to-point distance is r² + l².
2. The components perpendicular to the dipole axis are equal and opposite, so they cancel.
3. The components parallel to the dipole axis point from +q towards −q. Each equals the individual field magnitude multiplied by l/√(r² + l²).
4. The two axial components add, giving magnitude 2kql/(r² + l²)³ᐟ². Their resultant direction is opposite to p⃗.

E⃗ = −kp⃗/(r² + l²)³ᐟ². For r ≫ 2l, the magnitude becomes approximately E = kp/r³, still opposite to the dipole moment.

The axial drawing labels the two opposite charges, separation 2a and observation point P. The equatorial drawing shows equal sloping field contributions at P and their resultant opposite to the dipole moment. The figure’s a is the half-separation called l here.

See Fig. 1.17 in your NCERT textbook

For the same dipole at equal large centre-to-point distances, the axial field magnitude is twice the equatorial magnitude.

Note: A dipole’s field depends on both distance and direction from its centre. Zero net charge must not be substituted for q in the single-point-charge field formula to conclude that the dipole field vanishes.

Why does a dipole experience torque in a uniform field?

A uniform external field exerts forces +qE⃗ and −qE⃗ on the dipole charges. Their resultant is zero, but their different application points allow a turning effect.

Torque, written τ⃗, measures this turning effect. The SI unit of torque is N m, and its dimensions are [ML²T⁻²]. A couple is a pair of equal, opposite, parallel forces acting along different lines.

Derivation: Torque on a dipole

Let θ be the angle between p⃗ and E⃗. Angles are dimensionless; degrees are indicated by °. The perpendicular separation of the force lines is the arm of the couple.

1. Each charge experiences force magnitude qE because the external field has equal magnitude at both charges.
2. The arm of the couple is 2l sin θ, where sin denotes the sine of the angle.
3. The torque magnitude is force multiplied by this perpendicular separation: τ = qE(2l sin θ).
4. Using p = 2ql gives τ = pE sin θ. Its direction is that of p⃗ × E⃗, perpendicular to their plane.

τ = pE sin θ; vectorially, τ⃗ = p⃗ × E⃗. Here × between vectors denotes the cross product, directed by the right-hand rule: curl the fingers from p⃗ towards E⃗; the thumb gives the torque direction.

Torque tends to align the dipole with the field. It vanishes for parallel or antiparallel alignment and is greatest for perpendicular alignment.

Worked example 4. A dipole of moment 4 × 10⁻⁹ C m makes 30° with a uniform field of 5 × 10⁴ N C⁻¹. Find its torque magnitude, using sin 30° = 1/2.

Formula: τ = pE sin θ. Substitute: τ = (4 × 10⁻⁹)(5 × 10⁴)(1/2). Answer: 0.0001 N m. The net force remains zero because the field is uniform.

Uniformity matters: in a non-uniform field the forces at the two ends need not cancel. A dipole parallel to a non-uniform field is drawn towards increasing field strength; one antiparallel to it is pushed towards decreasing field strength.

What is electric flux, and how does surface orientation affect it?

Electric flux, φ, describes the electric field through an oriented surface. Unlike liquid flux, it does not represent a physically observable substance flowing.

For liquid velocity v perpendicular to area A, the volume flow rate is vA. Velocity is displacement per unit time, measured in m s⁻¹, with dimensions [LT⁻¹]. Volume flow rate has unit m³ s⁻¹ and dimensions [L³T⁻¹].

An area vector A⃗ has magnitude A and points normal to the surface. Use the outward normal for a closed surface; specify the chosen normal for an open surface.

For uniform field E⃗ across a plane area, φ = EA cos θ, where θ is the angle between E⃗ and A⃗, and cos denotes cosine. Equivalently, φ = E⃗ · A⃗. The dot denotes the scalar product: multiply magnitudes and the cosine of their included angle.

The SI unit of electric flux is N m² C⁻¹, and its dimensions are [ML³T⁻³A⁻¹]. Flux is a signed scalar. Its sign records whether the field crosses in the chosen normal direction or against it.

How is flux found for a curved surface or varying field?

Divide the surface into small, approximately plane elements. For each element, dφ = E⃗ · dA⃗. Then add the contributions by integration: φ = ∫E⃗ · dA⃗. The field and the outward normal can both vary across the surface.

Worked example 5. A uniform field of 3 × 10³ N C⁻¹ points along the positive x-axis. Find the flux through a square of side 10 cm when its chosen normal is along that axis, then at 60° to it. Use cos 60° = 1/2.

Formula: A = b²; φ = EA cos θ, where b is the side length. Substitute: A = (0.10)² = 0.01 m². Answer: 30 N m² C⁻¹ for θ = 0°, and 15 N m² C⁻¹ for θ = 60°.

What does Gauss’s theorem state, and how is a Gaussian surface chosen?

Definition: Gauss’s theorem states that the total electric flux through a closed surface equals the net charge enclosed by that surface divided by the permittivity of free space.

Write φ = Qₑ/ε₀, where Qₑ is the algebraic sum of enclosed charges. In integral form, ∮E⃗ · dA⃗ = Qₑ/ε₀. The circle on ∮ indicates integration over a closed surface enclosing a volume.

The field includes contributions from inside and outside charges. Only enclosed charge enters the right side. External charges affect local fields but contribute zero net closed-surface flux.

What does zero net flux establish?

Zero flux establishes zero net enclosed charge, not the absence of charges or fields. Equal positive and negative charges may be enclosed; inward and outward flux may cancel.

A Gaussian surface is a closed mathematical surface, not necessarily a material one. The theorem holds for every closed shape and size; symmetry makes field calculations simpler.

• Choose a surface on which symmetry makes the field magnitude constant over the parts contributing to flux.
• Prefer parts where the field is normal to the surface, giving a simple product, or tangent to it, giving zero flux.
• Do not pass the surface through a discrete point charge, where that charge’s field is not defined.
• A Gaussian surface can pass through a continuous charge distribution. Count precisely the charge within its enclosed volume.

Worked example 6. A cylinder has radius 5 cm, length 20 cm, centre at the origin and axis along x. The field is 200 N C⁻¹ along +x for x > 0 and along −x for x < 0. Find outward flux and enclosed charge, using ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

Formula: A = πr²; φ = 2EA; Qₑ = ε₀φ. Substitute: each end contributes 200π(0.05)² = 1.57 N m² C⁻¹. The curved side contributes zero. Answer: total flux is 3.14 N m² C⁻¹ and enclosed charge is 2.78 × 10⁻¹¹ C.

Gauss’s law is often useful for easier calculations with symmetry. It follows from Coulomb’s inverse-square law.

How does Gauss’s law give the field of an infinite straight wire?

An infinitely long, straight, thin wire has uniform linear density λ in vacuum. Rotation around it or translation along it leaves its charge arrangement unchanged.

The field is radial and depends only on perpendicular distance r. Parallel components from corresponding charge elements on either side cancel.

Derivation: Cylindrical Gaussian surface

Choose a closed cylinder coaxial with the wire, meaning it has the same axis. Let its radius be r and its length be L. In this derivation L denotes a length in metres, rather than the dimensional symbol used inside square brackets.

1. The field is tangent to the two flat end faces, so each end has zero flux.
2. On the curved surface, the field is normal and has constant magnitude. The curved area is 2πrL.
3. For positive λ, the total outward flux is E(2πrL). The enclosed charge is λL.
4. Gauss’s law gives E(2πrL) = λL/ε₀. Cancelling L gives the radial field.

E = |λ|/(2πε₀r) for the magnitude. The field points away from a positive wire and towards a negative wire. Its magnitude decreases inversely with perpendicular distance from the wire.

A positively charged straight wire passes along the axis of a dashed cylinder. Its radius r and length l are labelled, with an outward field arrow at a point on the curved surface. The figure’s cylinder length l is L here.

See Fig. 1.26 in your NCERT textbook

The entire wire produces the field. Infinite length is crucial; the result is approximately true around a long finite wire’s central portions, where end effects may be ignored.

How is the field of a uniformly charged infinite plane sheet found?

For a thin infinite plane sheet of uniform density σ in vacuum, symmetry makes the field perpendicular to the sheet, with equal magnitudes on both sides.

Derivation: A Gaussian box through the sheet

Choose a closed box crossing the sheet, with two equal faces of area A parallel to it. The remaining faces are perpendicular to the sheet. A cylinder with its flat faces parallel to the sheet also works.

1. The field is tangent to the side faces, giving no flux through those faces.
2. For positive σ, the field points outward through both parallel end faces. Each contributes flux EA.
3. The total flux is 2EA and the enclosed charge is σA.
4. Gauss’s law gives 2EA = σA/ε₀. Cancelling A gives the field on either side.

E = |σ|/(2ε₀) for the magnitude. The field points away from a positively charged sheet and towards a negatively charged sheet. It is independent of distance from an ideal infinite sheet.

A rectangular box passes through a plane labelled with surface charge density σ. Opposite end faces are numbered 1 and 2. Field arrows point away from the sheet through these two faces.

See Fig. 1.27 in your NCERT textbook

What changes for finite sheets and two plates?

For a finite large planar sheet, this expression is approximately true in the middle regions, away from the ends.

For two ideal infinite parallel sheets with equal and opposite surface densities, superposition makes the fields add between them and cancel outside. Between the sheets, the magnitude is |σ|/ε₀ and the direction runs from positive to negative. Large parallel plates approximate this away from their edges.

How does the field vary inside and outside a thin spherical shell?

A uniformly charged thin spherical shell has radius R, charge Q and density σ in vacuum: Q = 4πR²σ. Its radial field depends only on centre-to-point distance r.

Derivation: Concentric spherical Gaussian surfaces

1. Outside the shell, choose a sphere of radius r > R with the same centre. Its area is 4πr².
2. For positive Q, the field is outward normal and constant in magnitude over that sphere, giving flux E(4πr²).
3. The enclosed charge is Q. Therefore E(4πr²) = Q/ε₀ and the external field is the same as that of Q concentrated at the centre.
4. Inside, choose a sphere with r < R. It encloses no charge. Symmetry again gives flux E(4πr²), so the internal field is zero.

E = k|Q|/r² outside; E = 0 throughout the interior. Outside a negatively charged shell, the direction is radially inward. The zero interior result includes points away from the centre.

Both drawings show a shell centred at O with radius R. The first dashed Gaussian sphere passes through an external point P; the second dashed sphere passes through an internal point P. The observation radius is labelled r.

See Fig. 1.28 in your NCERT textbook

What does the field-versus-distance graph show?

Immediately outside the surface, the field magnitude is k|Q|/R² = |σ|/ε₀. The ideal surface distribution produces a discontinuity: the inner limiting field is zero, while the outer limiting field is non-zero for a charged shell.

Put radial distance r horizontally and field magnitude E vertically. Draw E = 0 for r < R. At R, mark the jump to k|Q|/R² on the outside, then draw an inverse-square decrease for r > R.

The field “on the surface” denotes the exterior limiting value here. The ideal infinitely thin surface has different inner and outer limits.

Both zero enclosed charge and spherical symmetry establish the zero interior field. Without symmetry, zero flux alone would be insufficient.

Glossary

• Electric charge — A property of matter responsible for electrical attraction and repulsion, occurring in positive and negative forms.
• Quantisation — The property that a body’s charge is an integral multiple of the elementary charge magnitude.
• Conservation of charge — The principle that the total electric charge of an isolated system remains unchanged.
• Point charge — An idealisation concentrating a body’s charge at one point when its size is negligible compared with separation.
• Superposition — The principle that the resultant electric force or field is the vector sum of independent individual contributions.
• Electric field intensity — Force per unit positive test charge at a point, measured without disturbing the source configuration.
• Permittivity — A property of the medium appearing in the electrostatic force law and determining its proportionality factor.
• Electric dipole — A pair of equal and opposite point charges separated by a finite distance.
• Dipole moment — A vector equal in magnitude to charge magnitude times separation, directed from negative to positive charge.
• Electric flux — The surface integral of the electric field’s normal component, with sign set by surface orientation.
• Gaussian surface — A closed mathematical surface selected for applying Gauss’s theorem to an enclosed charge distribution.

Common errors and misconceptions

• Misconception: Electric forces and fields can be added as unsigned magnitudes. Correct: They are vectors. Determine directions first, then add collinear contributions with signs or resolve non-collinear contributions into components.
• Misconception: Zero total charge makes a dipole’s field zero. Correct: The charges occupy different positions. Their fields generally do not cancel; the distant field decreases as the inverse cube of distance.
• Misconception: A uniform field cannot turn a dipole because its net force is zero. Correct: The two forces can form a couple, giving torque pE sin θ.
• Misconception: The angle in the flux formula is measured from the surface plane. Correct: Measure it between the electric field and the surface normal. A field parallel to the surface gives zero flux.
• Misconception: Zero flux through a closed surface proves zero field everywhere. Correct: It proves zero net enclosed charge. Non-zero inward and outward contributions can cancel.
• Misconception: A charged sheet and the exterior surface of a charged shell have the same field formula. Correct: An isolated infinite sheet gives |σ|/(2ε₀) on either side; a uniform shell gives |σ|/ε₀ immediately outside and zero inside.

Exam-style questions with model answers

Q1. Explain quantisation of charge and why it can be ignored for macroscopic charges. [2 marks]

1. Quantisation means q = ne, where q is the charge, e is the elementary charge magnitude and n is an integer.
2. Macroscopic charges contain enormous numbers of elementary charges. The individual steps are then too small to matter in practical calculations, so charge appears continuous.

Q2. Point charges +10⁻⁸ C and −10⁻⁸ C are 0.10 m apart in vacuum. Calculate the field at their midpoint and explain its direction. Use k = 9 × 10⁹ N m² C⁻². [4 marks]

1. The midpoint is at distance r = 0.05 m from each charge. The field magnitude due to either charge is found using E = k|q|/r².
2. The positive charge contributes (9 × 10⁹)(10⁻⁸)/(0.05)² = 3.6 × 10⁴ N C⁻¹, directed away from itself towards the negative charge.
3. The negative charge contributes the same magnitude, 3.6 × 10⁴ N C⁻¹, directed towards itself. At the midpoint this is the same direction as the first contribution.
4. By superposition, the resultant magnitude is 7.2 × 10⁴ N C⁻¹. Its direction is along the joining line from the positive charge towards the negative charge.

Q3. A dipole in vacuum consists of −q and +q separated by 2l, with q positive. Derive its electric field at an axial point a distance r > l from its centre, on the positive-charge side. Give the limit for r ≫ 2l. Use k = 1/(4πε₀), where ε₀ is vacuum permittivity. [5 marks]

1. The point is r − l from +q and r + l from −q. Define dipole moment magnitude p = 2ql, directed from negative to positive charge.
2. The positive charge produces field magnitude kq/(r − l)², directed along the dipole moment because the observation point lies beyond the positive charge.
3. The negative charge produces field magnitude kq/(r + l)² in the opposite direction. Superposition therefore requires subtraction of these two contributions.
4. The resultant magnitude is kq[1/(r − l)² − 1/(r + l)²] = 4kqlr/(r² − l²)² = 2kpr/(r² − l²)², along the dipole moment.
5. For r ≫ 2l, neglect l² compared with r². The field magnitude becomes approximately 2kp/r³, with its direction still along the dipole moment.

Q4. Derive the torque on a dipole with charges ±q separated by 2l in a uniform field of magnitude E. Let θ be the angle between its dipole moment and the field. State the resultant force. [5 marks]

1. The forces on +q and −q are +qE⃗ and −qE⃗. Since the field is uniform, these forces are equal and opposite and the resultant force is zero.
2. When their lines of action differ, the two forces form a couple. Its torque magnitude equals either force magnitude multiplied by the perpendicular distance between the force lines.
3. For charge separation 2l at angle θ to the field, this perpendicular distance is 2l sin θ. Thus the torque magnitude is τ = qE(2l sin θ).
4. The dipole moment has magnitude p = 2ql and points from negative to positive charge. Substitution gives τ = pE sin θ, with SI unit N m.
5. Vectorially, τ⃗ = p⃗ × E⃗. Its direction is perpendicular to the plane of the dipole and field, and the turning effect tends to align the dipole with the field.

Q5. State Gauss’s theorem and use it to obtain the field magnitude of a uniformly charged infinite plane sheet in vacuum with positive surface charge density σ. Let ε₀ be vacuum permittivity. [4 marks]

1. Gauss’s theorem states that the net electric flux through a closed surface is its net enclosed charge divided by ε₀.
2. Choose a Gaussian box crossing the sheet, with two faces of equal area A parallel to it. Symmetry makes the field perpendicular to the sheet and equal in magnitude on both sides.
3. The side faces contribute zero flux. Each parallel face contributes EA, so the total flux is 2EA. The enclosed charge is σA.
4. Therefore 2EA = σA/ε₀, giving E = σ/(2ε₀). It points away from the positive sheet on both sides and is independent of distance.

Q6. A thin spherical shell of radius R carries uniformly distributed positive charge Q in vacuum. Using Gauss’s theorem, derive its field for r < R and r > R, where r is distance from its centre. Describe the field-magnitude graph and exterior surface limit. Let ε₀ be vacuum permittivity. [5 marks]

1. Spherical symmetry makes the field radial and constant in magnitude over a concentric spherical Gaussian surface of radius r. Its total outward flux is E(4πr²).
2. For r > R, the Gaussian sphere encloses Q. Therefore E(4πr²) = Q/ε₀ and E = Q/(4πε₀r²), directed radially outward.
3. For r < R, the Gaussian sphere encloses no charge. Hence E(4πr²) = 0, giving zero field throughout the shell’s interior.
4. Immediately outside the shell, the limiting field magnitude is Q/(4πε₀R²). Immediately inside it is zero, so the ideal surface has a discontinuity in field.
5. The graph stays on the zero-field axis throughout the interior, jumps to the exterior limiting value at R, and then decreases as the inverse square of r outside.

Key takeaways

• Electric field is force per unit positive test charge at a specified point, and a negative charge experiences force opposite to its direction.
• Dipole moment points from negative to positive charge; distant axial and equatorial fields decrease as the inverse cube of distance.
• A dipole in a uniform field has zero resultant force but can experience torque of magnitude pE sin θ.
• Electric flux depends on the angle between field and surface normal, with the outward normal used for closed surfaces.
• Gauss’s theorem relates total closed-surface flux to net enclosed charge; symmetry is needed to extract a simple field expression.
• An ideal infinite wire gives inverse-distance field, an infinite sheet gives constant field, and a uniform thin shell has zero interior field.

Test yourself

Why can two electric field lines not cross?

Their crossing would assign two different field directions to the same point, whereas the net field has a unique direction.

Where does the equatorial dipole field point relative to its dipole moment?

It points opposite to the dipole moment because the axial components add towards the negative charge.

When is the approximation E = 2kp/r³ valid for an axial dipole field?

It applies when distance r from the centre is much greater than the dipole’s charge separation.

Does zero closed-surface flux mean there are no charges inside?

No. It means the algebraic total is zero; equal amounts of positive and negative charge can still be enclosed.

Why is flux through the end faces of a Gaussian cylinder around an infinite wire zero?

The radial field lies parallel to those end faces, so its component along each face’s normal vanishes.

Why can zero enclosed charge establish zero field inside a uniform thin spherical shell?

Spherical symmetry makes the field radial with equal magnitude across a concentric Gaussian sphere, so zero flux requires zero field.