NCERT Solutions for Class 12 Physics Chapter 1: Electric Charges and Fields

Concept and context. The mutual electrostatic force between two stationary point charges is given by Coulomb’s law, discovered by the French physicist Charles Augustin de Coulomb in 1785 with his torsion balance. The law says that the force is directly proportional to the product of the magnitudes of the two charges, inversely proportional to the square of the distance between them, and acts along the line joining the two charges. Two small charged spheres whose radii are very small compared with their separation (here 30 cm) may be treated as point charges, so the law can be applied directly. Air is practically the same as vacuum for electrostatic purposes (its relative permittivity is about 1.0006), which is why NCERT lets us use the vacuum value of the constant k.

F = (1/4πε₀) · q₁q₂ / r²    with   1/4πε₀ = k = 9 × 10⁹ N m² C⁻²

Given: q₁ = 2 × 10⁻⁷ C, q₂ = 3 × 10⁻⁷ C, r = 30 cm = 0.30 m (always convert centimetres to metres before substituting, because k carries SI units).

Step-by-step solution

1. Write the law: F = k q₁q₂ / r². Each of the quantities is in SI units, so the answer will come out in newton.
2. Substitute the values: F = (9 × 10⁹) × (2 × 10⁻⁷) × (3 × 10⁻⁷) / (0.30)² = (9 × 10⁹ × 6 × 10⁻¹⁴) / (9 × 10⁻²).
3. Simplify the numerator and denominator: numerator = 54 × 10⁻⁵ = 5.4 × 10⁻⁴ ; denominator = 0.09 = 9 × 10⁻². Dividing gives F = 6 × 10⁻³ N.
4. Direction and nature: both charges are positive (like charges), so the force is repulsive: each sphere is pushed away from the other along the line joining their centres, and by Newton’s third law the two forces are equal in magnitude and opposite in direction.

F = 6 × 10⁻³ N (repulsive)

Physical significance. Although each charge (0.2 μC and 0.3 μC) is quite normal for laboratory electrostatics, the force of 6 mN is clearly measurable — about the weight of a 0.6 g object. This shows how strong the electric force is even for very small charges; a full 1 C placed 1 m from another 1 C would give 9 × 10⁹ N, which is why the coulomb is regarded as an enormous unit in electrostatics and μC and nC are used in practice.

Conclusion: the force between the two spheres is 6 × 10⁻³ N and it is repulsive, because both charges are positive; each sphere experiences this same magnitude of force along the line joining their centres, in accordance with Coulomb’s law and Newton’s third law.

Concept and context. This question uses Coulomb’s law “in reverse” — instead of finding the force from the distance, we find the distance from the measured force — and then tests understanding of Newton’s third law as applied to electrostatic forces. Coulomb’s force F = (1/4πε₀) |q₁q₂|/r² depends only on the magnitudes of the charges and the separation, so for two point charges the pair of forces is always equal in magnitude and opposite in direction (“action and reaction”).

Given: q₁ = 0.4 μC = 4 × 10⁻⁷ C; q₂ = −0.8 μC = −8 × 10⁻⁷ C; F = 0.2 N; k = 9 × 10⁹ N m² C⁻².

(a) Distance between the spheres

1. Use the magnitude form of Coulomb’s law (signs only decide attraction/repulsion): F = k|q₁||q₂| / r².
2. Rearrange for r: r² = k|q₁||q₂| / F.
3. Substitute: r² = (9 × 10⁹ × 4 × 10⁻⁷ × 8 × 10⁻⁷) / 0.2 = (9 × 10⁹ × 32 × 10⁻¹⁴) / 0.2 = (2.88 × 10⁻³) / 0.2 = 1.44 × 10⁻² m².
4. Take the square root: r = √(1.44 × 10⁻²) = 0.12 m = 12 cm.

r = 0.12 m (12 cm)

(b) Force on the second sphere due to the first

Because Coulomb’s law agrees with Newton’s third law, F₁₂ = −F₂₁. The force on sphere 2 due to sphere 1 therefore has the same magnitude, 0.2 N, but the opposite direction to the force on sphere 1 due to sphere 2. In vector form F₂₁ = (1/4πε₀)(q₁q₂/r²) r̂₂₁ and F₁₂ is obtained by simply interchanging the labels 1 and 2, which reverses r̂.

Since q₁ is positive and q₂ is negative the charges are unlike, so the force is attractive: the force on the second sphere is 0.2 N directed towards the first sphere along the line joining them, and the force on the first sphere is 0.2 N directed towards the second.

Conclusion: the two spheres are 12 cm apart, and the second sphere experiences a force of 0.2 N — equal in magnitude and opposite in direction to the force on the first sphere (attractive, directed towards the first sphere), as required by Newton’s third law.

Concept and context. The two most familiar inverse-square forces in physics are Coulomb’s electrostatic force F = k q₁q₂/r² and Newton’s gravitational force F = G m₁m₂/r². For an electron (charge −e, mass mₑ) and a proton (charge +e, mass mₚ) separated by a distance r, the electric force has magnitude ke²/r² and the gravitational force has magnitude G mₑmₚ/r². The ratio of the two, ke²/(G mₑmₚ), is therefore independent of r and measures the relative strength of the two forces. A ratio of two forces — both measured in newton — must be a pure number, i.e. dimensionless; we verify this explicitly.

Part 1 — Dimensional check

1. From Coulomb’s law, k = F r² / q², so the units of ke² are those of (force × length²): [ke²] = N m² = [M L T⁻²][L²] = [M L³ T⁻²].
2. From Newton’s law of gravitation, G = F r² / (m₁m₂), so the units of G mₑmₚ are also those of (force × length²): [G mₑmₚ] = N m² = [M L³ T⁻²].
3. Therefore [ke²/G mₑmₚ] = [M L³ T⁻²] / [M L³ T⁻²] = [M⁰ L⁰ T⁰] — dimensionless. (In SI units: N m² C⁻² × C² = N m², and N m² kg⁻² × kg² = N m², so the units cancel completely.)

Part 2 — Numerical value (constants from a Table of Physical Constants)

• Numerator: ke² = (9 × 10⁹)(1.6 × 10⁻¹⁹)² = 9 × 10⁹ × 2.56 × 10⁻³⁸ ≈ 2.3 × 10⁻²⁸ N m².
• Denominator: G mₑmₚ = (6.67 × 10⁻¹¹)(9.11 × 10⁻³¹)(1.67 × 10⁻²⁷) ≈ 1.0 × 10⁻⁶⁷ N m².
• Ratio: ke²/G mₑmₚ ≈ 2.3 × 10⁻²⁸ / 1.0 × 10⁻⁶⁷ ≈ 2.3 × 10³⁹ (NCERT Example 1.3 quotes ≈ 2.4 × 10³⁹ using slightly different rounding of the constants — i.e., about 10³⁹).

ke² / (G mₑmₚ) ≈ 2 × 10³⁹

Part 3 — What does the ratio signify?

• It is the ratio of the electrostatic force to the gravitational force between an electron and a proton at any distance r (r cancels because both forces vary as 1/r²).
• Its enormous value (~10³⁹) shows that electric forces are enormously stronger than gravitational forces at the atomic scale. For two protons the corresponding ratio is e²/(4πε₀Gmₚ²) ≈ 1.3 × 10³⁶. Inside a nucleus (10⁻¹⁵ m apart) two protons repel with ≈ 230 N but attract gravitationally with only ≈ 1.9 × 10⁻³⁴ N.
• Hence gravity is completely negligible in the structure of atoms and molecules, which is governed by electromagnetic forces. Yet gravity dominates in the cosmos because it is always attractive and cumulative, whereas electric charges come in both signs and large bodies are effectively neutral (electric forces cancel out).

Conclusion: the ratio ke²/G mₑmₚ is dimensionless and has the value of about 2 × 10³⁹; it signifies that the electrostatic attraction between an electron and a proton is about 10³⁹ times stronger than their gravitational attraction, so gravitational forces are utterly negligible on the atomic scale.

Concept and context. One of the three basic properties of electric charge — along with additivity and conservation — is quantisation. The word “quantised” means that a quantity can take only discrete values, in whole-number multiples of a smallest basic unit, instead of any value on a continuous scale. Quantisation of charge was first suggested by Faraday’s experimental laws of electrolysis and was demonstrated directly by Millikan’s oil-drop experiment in 1912.

(a) Meaning of quantisation of charge

• All free charges are integral multiples of a basic unit of charge e: q = ne, where n is any integer — positive, negative or zero (n = 0, ±1, ±2, ±3, …).
• The basic unit e is the magnitude of the charge on an electron or a proton: e = 1.602 × 10⁻¹⁹ C. By convention an electron has charge −e and a proton +e. If a body contains n₁ electrons and n₂ protons its total charge is n₂e + n₁(−e) = (n₂ − n₁)e; since n₂ − n₁ is an integer, the charge of any body is always an integral multiple of e.
• Consequently a body can gain or lose charge only in discrete steps of e — for example, it can have charge +2e or −5e but never +1.5e or −0.3e. Charge is “grainy” or “lumpy”, not smooth.

(b) Why quantisation can be ignored at the macroscopic level

• The step size e = 1.6 × 10⁻¹⁹ C is extremely small. In ordinary electrostatics we deal with charges of the order of a microcoulomb (10⁻⁶ C) or a millicoulomb (10⁻³ C). A charge of just 1 μC already contains 10⁻⁶ / 1.6 × 10⁻¹⁹ ≈ 6 × 10¹² (about 10¹³) times e.
• When n is such an enormous number, adding or removing one electron changes the charge by only about one part in 10¹³ — far below anything measurable — so the charge appears to vary continuously and may take any value.
• Analogy from NCERT: a dotted line viewed from a distance looks like a continuous line because the dots are so close together; likewise, so many tiny charges taken together look like a smooth, continuous charge distribution. A fluid looks continuous even though it consists of discrete molecules.
• Hence for macroscopic calculations we use continuous charge densities (σ, λ, ρ) and calculus, and quantisation has no practical consequence. At the microscopic level, where charges are only a few tens or hundreds of e and can be counted (e.g. in Millikan’s experiment or in atoms), the discrete nature must be taken into account.

Conclusion: charge is quantised because every free charge equals an integral multiple of the elementary charge e (q = ne); and because e is so tiny compared with the charges met in everyday (macroscopic) situations — typically ~10¹³ e or more — the charge then appears continuous, so quantisation is ignored at the large scale but not at the microscopic scale.

Concept and context. The law of conservation of charge states that the total (net) charge of an isolated system always remains constant: electric charge can neither be created nor destroyed, it can only be transferred from one body to another or redistributed within the system. (Charge-carrying particles may be created or destroyed, but the net charge does not change.) Everything seen in frictional electrification obeys this law.

Step-wise explanation

1. Before rubbing: both the glass rod and the silk cloth are electrically neutral — each atom has equal numbers of protons and electrons, so the positive and negative charges are exactly balanced. The total charge of the system (rod + silk) is zero.
2. During rubbing: friction brings the surfaces into intimate contact. In solids the electrons are the charges that can be transferred, because the outer electrons are less tightly bound than the protons locked in the nucleus. Some electrons from the glass rod are transferred to the silk cloth.
3. After rubbing: the glass rod, having lost electrons, has a deficit of negative charge, so it is positively charged. The silk, having gained the same number of electrons, has an excess of negative charge, so it is negatively charged. By Franklin’s convention the glass rod is positive and the silk is negative.
4. Accounting: the number of electrons gained by the silk equals the number lost by the glass, so charge on glass = +Q and charge on silk = −Q. The algebraic sum is (+Q) + (−Q) = 0, exactly the same as before rubbing. Hence “what one body gains in charge, the other loses” — no new charge is created and none is destroyed.
5. Supporting observation: if the electrified rod is touched to the silk with which it was rubbed, they cease to attract or repel light objects — the equal and opposite charges neutralise each other, confirming that the total charge was zero throughout.

Further points. Only a very small fraction of the total electrons of the bodies is transferred during rubbing. The same logic applies to any pair of bodies, e.g. plastic rod and cat’s fur or polythene and wool: one body becomes positive and the other negative by an equal amount. Another example of conservation is the decay of a neutron into a proton and an electron — the proton (+e) and electron (−e) have equal and opposite charges, so the total charge is zero both before and after.

Conclusion: rubbing merely transfers electrons from the glass rod to the silk, so the two bodies acquire equal and opposite charges; the net charge of the system remains zero before and after, which is exactly what the law of conservation of charge requires.

Concept and context. The force on a charge due to several other charges is found with the principle of superposition: the net force is the vector sum of the Coulomb forces due to each charge taken one at a time, each force being unaffected by the presence of the others. Here the geometry is highly symmetric — the test charge q₀ = 1 μC sits at the centre O of the square, which is equidistant from all four corners — so we can solve the problem either by direct calculation or by a pure symmetry argument.

Given: side of square = 10 cm = 0.10 m; diagonal AC = BD = 0.10 × √2 = 0.141 m, so OA = OB = OC = OD = diagonal/2 = 0.0707 m (r² = 5 × 10⁻³ m²). Charges: A = +2 μC, B = −5 μC, C = +2 μC, D = −5 μC. Test charge q₀ = +1 μC at O.

Step 1 — Forces due to the pair A and C (diagonal AC)

• Both A and C carry +2 μC, equal in magnitude, and both are at the same distance r from O. The test charge is positive, so each exerts a repulsive force: F_A = k(2 × 10⁻⁶)(1 × 10⁻⁶)/r² pushes q₀ along AO (away from A), and F_C of exactly the same magnitude pushes q₀ along CO (away from C).
• Since O lies midway between A and C, these two forces are equal in magnitude and exactly opposite in direction along the diagonal AC. Their resultant is therefore zero.

Step 2 — Forces due to the pair B and D (diagonal BD)

• B and D both carry −5 μC, so each attracts the positive test charge: one force is directed from O towards B and the other from O towards D, with equal magnitude k(5 × 10⁻⁶)(1 × 10⁻⁶)/r².
• Again O is the midpoint of BD, so these two forces are equal and opposite along BD and their resultant is zero.

Step 3 — Total force

F_net = (F_A + F_C) + (F_B + F_D) = 0 + 0 = 0

Numerical illustration (optional check). Magnitude of each of F_A, F_C = 9 × 10⁹ × 2 × 10⁻⁶ × 10⁻⁶ / (5 × 10⁻³) = 3.6 N; of each F_B, F_D = 9 × 10⁹ × 5 × 10⁻⁶ × 10⁻⁶ / (5 × 10⁻³) = 9.0 N. Although these individual forces are large, they cancel pairwise to give zero. Note: the answer is zero for any charge placed at the centre, because the net force depends only on the symmetry (opposite corners carry identical charges), not on the value of q₀.

Conclusion: the net force on the 1 μC charge at the centre of the square is zero, because the forces due to the identical charges at opposite corners (A, C and B, D) are equal and opposite and cancel in pairs by the principle of superposition.

Concept and context. An electric field line is a curve drawn so that the tangent to it at every point gives the direction of the net electric field at that point; an arrow on the line shows the sense of the field. Introduced by Michael Faraday (who called them “lines of force”), field lines give a pictorial map of the field: their relative density (closeness) shows its strength, and they are really curves in three-dimensional space. The properties of field lines come directly from the properties of the electric field itself.

(a) Why a field line cannot have sudden breaks

1. A field line is drawn so that its tangent at every point gives the direction of the field there. In any charge-free region the electric field is defined at every point and varies continuously from point to point (Coulomb’s law and superposition give a smooth field away from the charges). A line that stops abruptly would mean that the field suddenly vanishes (or loses its direction) at the break point, with a non-zero field just before it — which is physically impossible for a smoothly varying field.
2. Field lines can begin only on positive charges and end only on negative charges (or go to infinity for an isolated charge). A break in empty space would therefore imply the presence of a hidden source or sink of field lines — i.e., a charge — at that point, which contradicts the assumption that the region is charge-free.
3. Hence, in a charge-free region, electric field lines are continuous curves without any breaks, exactly as stated in property (ii) of field lines in NCERT.

(b) Why two field lines never cross

1. The tangent drawn to a field line at a point gives the direction of the net electric field at that point. At the point where two lines intersect we could draw two different tangents, i.e. the field would have two different directions at the same point.
2. But the electric field at a given point due to a given charge configuration has a unique magnitude and a unique direction (a test charge placed there can accelerate in only one direction). Having two directions at one point is absurd, so the lines cannot cross.
3. (Even at a point where the field is zero, such as a neutral point between two equal like charges, the lines do not cross: they just do not pass through that point.)

Conclusion: (a) a field line is continuous because the field in a charge-free region is defined and smoothly varying at every point and lines can start/end only on charges; (b) two field lines can never intersect because the field at the point of intersection would then have two different directions, which is impossible.

Concept and context. The electric field E at a point is the force per unit positive test charge placed there: E = F/q. For a point charge Q at distance r, E = (1/4πε₀)(Q/r²), directed away from a positive charge and towards a negative charge. For two charges the net field is the vector sum of the individual fields (superposition principle). Two equal and opposite charges form an electric dipole; the field between them points from the positive towards the negative charge.

Given: q_A = +3 μC = 3 × 10⁻⁶ C; q_B = −3 μC; AB = 20 cm = 0.20 m, so AO = OB = 0.10 m; k = 9 × 10⁹ N m² C⁻².

(a) Electric field at the midpoint O

1. Field due to +3 μC at A: magnitude E_A = k q/r² = 9 × 10⁹ × 3 × 10⁻⁶ / (0.10)² = 2.7 × 10⁴ / 10⁻² = 2.7 × 10⁶ N C⁻¹. A is positive, so the field at O points away from A, i.e. along AO → towards B.
2. Field due to −3 μC at B: magnitude E_B = 9 × 10⁹ × 3 × 10⁻⁶ / (0.10)² = 2.7 × 10⁶ N C⁻¹. B is negative, so the field at O points towards B, again along the direction A → B.
3. Resultant: both fields point in the same direction (from A to B), so they add: E = E_A + E_B = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N C⁻¹.

E at O = 5.4 × 10⁶ N C⁻¹, directed along AB (from the +3 μC charge towards the −3 μC charge)

(b) Force on a negative test charge

• q₀ = −1.5 × 10⁻⁹ C. Using F = q₀E: magnitude F = 1.5 × 10⁻⁹ × 5.4 × 10⁶ = 8.1 × 10⁻³ N.
• Direction: for a negative charge the force is opposite to the field. The field is along A → B, so the force on the negative test charge is directed towards A (i.e., along BA — towards the positive charge, as one expects, because the negative test charge is attracted by +3 μC and repelled by −3 μC, both effects pulling it towards A).

Conclusion: the electric field at the midpoint is 5.4 × 10⁶ N C⁻¹ directed from the positive charge (A) towards the negative charge (B), and the negative test charge of 1.5 × 10⁻⁹ C experiences a force of 8.1 × 10⁻³ N directed towards A.

Concept and context. Two equal and opposite point charges +q and −q separated by a small distance 2a form an electric dipole. The total charge of a system is the algebraic sum of its charges (charge is an additive scalar). The electric dipole moment is a vector, p = q × 2a, whose magnitude is the charge times the separation and whose direction is from the negative charge to the positive charge; its SI unit is coulomb-metre (C m).

Given: q_A = +2.5 × 10⁻⁷ C at A (0, 0, −15 cm); q_B = −2.5 × 10⁻⁷ C at B (0, 0, +15 cm). Both lie on the z-axis; separation AB = 15 + 15 = 30 cm = 0.30 m, so 2a = 0.30 m (a = 0.15 m), and the centre of the dipole is at the origin.

(i) Total charge

q_total = q_A + q_B = (+2.5 × 10⁻⁷) + (−2.5 × 10⁻⁷) = 0

Because charges add algebraically with proper signs, the equal positive and negative charges cancel: the system as a whole is electrically neutral. This does not mean its electric field is zero — the two charges are at different places, so their fields do not cancel exactly (they cancel only approximately at distances very large compared with 2a, where the field of the dipole falls off as 1/r³).

(ii) Dipole moment

1. Magnitude: p = q × 2a = (2.5 × 10⁻⁷ C) × (0.30 m) = 7.5 × 10⁻⁸ C m.
2. Direction: by definition the dipole moment points from the negative charge to the positive charge. The negative charge is at B (z = +15 cm) and the positive charge is at A (z = −15 cm), so the vector points from B to A, i.e. in the negative z-direction.
3. In vector form: p = 7.5 × 10⁻⁸ C m (−k̂) — i.e., 7.5 × 10⁻⁸ C m along the negative z-axis.

Conclusion: the total charge of the system is zero and its electric dipole moment has magnitude 7.5 × 10⁻⁸ C m, directed along the negative z-axis (from the −q at B to the +q at A).

Concept and context. In a uniform electric field E a dipole of moment p = q·2a experiences forces +qE on the positive charge and −qE on the negative charge. These two forces are equal and opposite, so the net force is zero, but they act at different points and therefore form a couple that exerts a torque on the dipole. The torque tends to rotate the dipole until p is aligned with E, at which point the torque vanishes.

Derivation of the torque (for completeness). Let the dipole make angle θ with E. Magnitude of each force = qE. The perpendicular distance between the lines of action of the two forces (the arm of the couple) = 2a sin θ. Torque = force × arm = qE × 2a sin θ = (q·2a)E sin θ = pE sin θ. In vector form τ = p × E; its direction is perpendicular to the plane containing p and E (given by the right-hand screw rule) and it tends to rotate p towards E.

Given: p = 4 × 10⁻⁹ C m; E = 5 × 10⁴ N C⁻¹; θ = 30°, so sin 30° = 1/2.

Calculation

1. Use τ = pE sin θ.
2. Substitute: τ = (4 × 10⁻⁹) × (5 × 10⁴) × sin 30° = (4 × 10⁻⁹ × 5 × 10⁴) × 0.5.
3. Multiply: 4 × 5 = 20, 10⁻⁹ × 10⁴ = 10⁻⁵ → 20 × 10⁻⁵ = 2 × 10⁻⁴; × 0.5 = 1 × 10⁻⁴ N m.

τ = 1 × 10⁻⁴ N m

Conclusion: the torque on the dipole is 1 × 10⁻⁴ N m (half of the maximum value pE = 2 × 10⁻⁴ N m), acting so as to rotate the dipole towards alignment with the field.

Concept and context. Frictional electrification occurs by the transfer of electrons (the loosely bound charge carriers in solids) from one body to the other; protons stay fixed inside nuclei. A body that gains electrons becomes negatively charged; the body that loses them becomes positively charged. Because charge is quantised, the total charge on a body is q = ne, so the number of transferred electrons is n = q/e.

Given: charge on polythene = −3 × 10⁻⁷ C; e = 1.6 × 10⁻¹⁹ C; mass of an electron mₑ = 9.1 × 10⁻³¹ kg.

(a) Number of electrons transferred

1. Polythene has a negative charge, so it has gained electrons. Since wool and polythene were neutral and rubbed only against each other, the electrons must have come from the wool. Hence electrons are transferred from wool to polythene (wool becomes positively charged by an equal amount, +3 × 10⁻⁷ C).
2. Use q = ne ⇒ n = q/e = (3 × 10⁻⁷ C)/(1.6 × 10⁻¹⁹ C) = 1.875 × 10¹² ≈ 1.9 × 10¹² electrons.

(b) Is there a transfer of mass?

• Yes — electrons have mass, so when they move from the wool to the polythene a (very small) mass moves with them.
• Mass transferred m = n × mₑ = (1.875 × 10¹²) × (9.1 × 10⁻³¹ kg) ≈ 1.7 × 10⁻¹⁸ kg.
• This is utterly negligible — about 10⁻¹⁸ kg, which is billions of times smaller than anything a balance could detect (a typical piece of polythene weighs about 10⁻³ kg). So although mass is transferred in principle, there is no measurable transfer of mass in practice.

Conclusion: about 1.9 × 10¹² electrons are transferred from the wool to the polythene; this involves a transfer of mass of ≈ 1.7 × 10⁻¹⁸ kg, which is technically non-zero but entirely negligible.

Concept and context. Since the radii of the spheres are negligible compared with their separation, A and B behave as point charges and Coulomb’s law F = (1/4πε₀)q₁q₂/r² applies directly. Part (b) tests the dependence of F on q₁q₂ (directly proportional) and on r (inversely as the square) — it can be solved by simple scaling without recalculation.

Given: q₁ = q₂ = 6.5 × 10⁻⁷ C; r = 50 cm = 0.50 m; k = 9 × 10⁹ N m² C⁻².

(a) Force at the original separation

1. F = k q₁q₂ / r² = (9 × 10⁹) × (6.5 × 10⁻⁷)² / (0.50)².
2. (6.5 × 10⁻⁷)² = 42.25 × 10⁻¹⁴ = 4.225 × 10⁻¹³; numerator = 9 × 10⁹ × 4.225 × 10⁻¹³ = 3.8025 × 10⁻³; denominator = 0.25.
3. F = 3.8025 × 10⁻³ / 0.25 = 1.521 × 10⁻² N ≈ 1.5 × 10⁻² N (repulsive, since both charges are positive).

(b) Charges doubled, distance halved

• New charges: q₁′ = q₂′ = 2q; new distance r′ = r/2 = 25 cm.
• F′ = k (2q)(2q) / (r/2)² = k · 4q² / (r²/4) = 16 · k q²/r² = 16 F.
• F′ = 16 × 1.521 × 10⁻² N = 0.2434 N ≈ 0.24 N (repulsive).
• Check by direct substitution: F′ = 9 × 10⁹ × (1.3 × 10⁻⁶)² / (0.25)² = 9 × 10⁹ × 1.69 × 10⁻¹² / 0.0625 = 0.2434 N ✓.

Conclusion: the mutual repulsive force between the spheres is 1.5 × 10⁻² N in case (a); when each charge is doubled and the distance halved, the force becomes 16 times larger, i.e. ≈ 0.24 N.

Concept and context. A charge q in a uniform electric field E experiences a constant force F = qE. The force is along E for a positive charge and opposite to E for a negative charge. The resulting acceleration is a = qE/m = (q/m)E. Therefore a particle moving with constant velocity along x while subject to a constant sideways force follows a parabolic track, bending towards the plate that attracts it; the larger the transverse acceleration (i.e. the larger the ratio q/m), the greater the bending for the same time/distance.

Reading the figure. The upper plate carries positive charge (+ + + +) and the lower plate negative charge (− − − −). The electric field between plates is therefore directed from the upper plate to the lower plate (downward, −y direction, since field lines run from + to −). All three particles enter from the left moving along +x.

Part 1 — Signs of the charges

1. Tracks 1 and 2 bend upwards, i.e. towards the positive plate, which means the force on them is opposite to the downward field. A force opposite to E acts only on a negative charge. Hence particles 1 and 2 are negatively charged.
2. Track 3 bends downwards, towards the negative plate, i.e. in the direction of the field. A force along E acts on a positive charge. Hence particle 3 is positively charged.

Part 2 — Highest charge-to-mass ratio

• For a particle that enters with speed v along x, the time to cross a plate length L is t = L/v, and the transverse deflection is y = ½ a t² = ½ (q/m) E (L/v)². For particles with similar speeds, y ∝ q/m.
• In the figure the track of particle 3 shows the greatest deflection (its curve departs the most from the straight line). Therefore particle 3 has the highest charge-to-mass ratio (q/m).

Conclusion: particles 1 and 2 are negative and particle 3 is positive; particle 3, which is deflected the most, has the highest charge-to-mass ratio.

Concept and context. Electric flux through a small area element ΔS is defined as Δφ = E · ΔS = E ΔS cos θ, where ΔS is the area vector (magnitude = area, direction = normal to the surface) and θ is the angle between E and the normal to the surface. Flux is proportional to the number of field lines cutting the surface; it is a scalar, with unit N C⁻¹ m² (= V m). For a plane surface in a uniform field, φ = E S cos θ.

Given: E = 3 × 10³ N/C along +x; side = 10 cm = 0.10 m ⇒ area S = (0.10)² = 1.0 × 10⁻² m².

(a) Plane of the square parallel to the yz plane

• A plane parallel to the yz-plane has its normal along the x-axis. The field E is also along x, so the normal is parallel to E: θ = 0° and cos θ = 1.
• φ = E S cos 0° = 3 × 10³ × 1.0 × 10⁻² × 1 = 30 N m² C⁻¹.
• This is the maximum possible flux through the square because all the field lines strike it head-on (perpendicular).

(b) Normal makes 60° with the x-axis

• Now θ = 60° (angle between the normal and E), cos 60° = ½.
• φ = E S cos 60° = 3 × 10³ × 1.0 × 10⁻² × ½ = 15 N m² C⁻¹.
• The tilted square presents a smaller projected area (S cos 60° = ½S) normal to the field, so only half as many field lines pass through it.

Conclusion: the flux is 30 N m² C⁻¹ when the square is parallel to the yz-plane (normal along E) and 15 N m² C⁻¹ when its normal makes 60° with the x-axis.

Concept and context. For a closed surface the area vector of every element is taken along the outward normal. Hence field lines that enter the surface give negative flux (θ > 90°) and lines that leave give positive flux (θ < 90°). Gauss’s law states φ = ∮E·dS = q_enclosed/ε₀, so for a closed surface the net flux depends only on the enclosed charge. A uniform field has no sources inside the cube: every line that enters also leaves.

Given: E = 3 × 10³ î N/C (uniform); cube of side 20 cm = 0.20 m, faces parallel to the coordinate planes, area of each face S = (0.20)² = 0.04 m².

Face-by-face calculation

φ_net = +120 − 120 + 0 + 0 + 0 + 0 = 0

• The four faces parallel to the field (±y, ±z) have the field running along them, so no lines cross them (cos 90° = 0).
• The field lines enter through the left face (−120) and leave through the right face (+120); the numbers are equal and opposite, so the net flux is zero.
• Gauss’s law check: φ = 0 ⇒ q_enclosed = ε₀φ = 0 — the cube encloses no net charge, which is consistent with a uniform field having no sources inside.

Conclusion: the net electric flux through the cube is zero: +120 N m² C⁻¹ leaves through one face and −120 N m² C⁻¹ enters through the opposite face, in agreement with Gauss’s law since no charge is enclosed.

Concept and context. Gauss’s law relates the net flux through any closed surface to the charge enclosed: φ = ∮E·dS = q/ε₀, where q is the algebraic sum of all charges inside the surface and ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻². Even though the black box’s contents are unknown, measuring the field on its surface (and thus the net flux) tells us the net charge inside.

(a) Net charge inside the box

1. From Gauss’s law: q = ε₀ φ.
2. Substitute: q = (8.854 × 10⁻¹² C² N⁻¹ m⁻²)(8.0 × 10³ N m² C⁻¹) = 7.08 × 10⁻⁸ C.
3. So the net charge inside is ≈ 7.1 × 10⁻⁸ C = 0.07 μC — positive, because the net flux is outward (positive).

(b) Zero net flux — can we conclude there is no charge?

• No. Zero net flux only means that the algebraic sum of the charges inside is zero (q_net = φε₀ = 0). The box may contain equal amounts of positive and negative charge — for example an electric dipole (+q and −q) or a neutral atom/molecule — whose total is zero even though individual charges exist.
• Gauss’s law gives only the net charge; it cannot tell how many individual charges are present or where they are. Likewise, the field lines leaving through one part of the surface can be exactly balanced by lines entering through another part, giving zero net flux while non-zero fields exist on the surface.

Conclusion: (a) the net charge inside the box is ≈ 7.1 × 10⁻⁸ C (positive); (b) zero net flux does not prove the absence of charges — only that equal amounts of positive and negative charge (zero net charge) are enclosed.

Concept and context. The direct calculation of flux through a square when the charge is above it would need integration. The hint uses Gauss’s law with symmetry: put the charge at the centre of a cube of edge 10 cm — then the given square is one of the six identical faces. The charge sits at distance 5 cm (= half the edge) above the centre of the square, i.e. exactly at the centre of the cube.

Given: q = +10 μC = 1.0 × 10⁻⁵ C; ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

1. Total flux through the closed cube: by Gauss’s law φ_total = q/ε₀ = 1.0 × 10⁻⁵ / 8.854 × 10⁻¹² = 1.13 × 10⁶ N m² C⁻¹.
2. Symmetry: the charge is at the cube’s centre, so every face is equivalent and receives an equal share of the flux.
3. Flux through one face (the given square): φ = φ_total/6 = q/(6ε₀) = 1.13 × 10⁶/6 = 1.88 × 10⁵ N m² C⁻¹.

φ_square = q / 6ε₀ ≈ 1.9 × 10⁵ N m² C⁻¹

Conclusion: the electric flux through the square is q/6ε₀ ≈ 1.9 × 10⁵ N m² C⁻¹, being one-sixth of the total flux q/ε₀ through the imaginary cube.

Concept and context. A Gaussian surface is any imaginary closed surface chosen for applying Gauss’s law: the net electric flux through it equals q_enclosed/ε₀. The law is true for a closed surface of any shape and size — the flux depends only on the charge inside, not on the size or shape of the surface, nor on where inside the charge sits. This is a consequence of the inverse-square form of Coulomb’s law: the field falls as 1/r² while the area rises as r², so the number of field lines crossing is the same for every enclosing surface.

Given: q = 2.0 μC = 2.0 × 10⁻⁶ C (inside the cube, at its centre); edge of cube = 9.0 cm (irrelevant to the result!); ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

1. Apply Gauss’s law: φ = q / ε₀.
2. Substitute: φ = (2.0 × 10⁻⁶) / (8.854 × 10⁻¹²) = 0.2259 × 10⁶ = 2.26 × 10⁵ N m² C⁻¹.
3. The flux is outward (positive) because the charge is positive. By symmetry each of the six faces carries one-sixth of it: q/6ε₀ ≈ 3.8 × 10⁴ N m² C⁻¹.

φ = q/ε₀ = 2.26 × 10⁵ N m² C⁻¹

• Why the 9.0 cm edge is not needed: the answer would be the same for a cube of any edge length (or for a sphere), as long as the charge is enclosed — a deliberate “distractor” in the question.

Conclusion: the net electric flux through the cubic Gaussian surface is q/ε₀ = 2.26 × 10⁵ N m² C⁻¹ (outward), independent of the size of the cube.

Concept and context. Gauss’s law says φ = q/ε₀ for any closed surface enclosing the charge q. The flux through a sphere centred on a point charge is φ = E × 4πr² = (q/4πε₀r²)(4πr²) = q/ε₀ — the r² in the field cancels the r² in the area, so the flux is independent of the radius. A negative flux means field lines enter the surface, so the enclosed charge is negative.

Given: φ = −1.0 × 10³ N m² C⁻¹ at r = 10.0 cm; ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

(a) Flux when the radius is doubled (to 20 cm)

• The charge enclosed is unchanged (same point charge at the centre). Hence by Gauss’s law the flux remains the same: φ = −1.0 × 10³ N m² C⁻¹.
• Physically: although the field at the larger surface is four times weaker (E ∝ 1/r²), the area is four times larger (S ∝ r²), so the product ES — the number of field lines crossing — is constant.

(b) Value of the point charge

1. q = ε₀ φ = (8.854 × 10⁻¹²)(−1.0 × 10³) = −8.854 × 10⁻⁹ C.
2. q ≈ −8.9 × 10⁻⁹ C ≈ −8.9 nC (negative, since the flux is inward).

Conclusion: (a) the flux through the doubled sphere is still −1.0 × 10³ N m² C⁻¹ because the enclosed charge is unchanged; (b) the point charge is q = ε₀φ ≈ −8.9 × 10⁻⁹ C (negative).

Concept and context. The charge on a conductor resides on its surface, and the sphere has spherical symmetry. By Gauss’s law, at any point outside a uniformly charged sphere or shell the field is the same as if the entire charge were concentrated at the centre: E = q/(4πε₀r²) (radial). The field direction tells the sign of the charge — radially outward for positive and radially inward for negative charge.

Given: R = 10 cm = 0.10 m; r = 20 cm = 0.20 m (point is outside the sphere, since r > R); |E| = 1.5 × 10³ N C⁻¹, directed radially inward; k = 9 × 10⁹ N m² C⁻².

1. Since the point lies outside the sphere, use E = |q| / (4πε₀ r²) ⇒ |q| = 4πε₀ r² E = r² E / k.
2. Substitute: |q| = (0.20)² × (1.5 × 10³) / (9 × 10⁹) = (0.04 × 1.5 × 10³) / (9 × 10⁹) = 60 / (9 × 10⁹) = 6.67 × 10⁻⁹ C.
3. The field points radially inward, so the charge is negative: q = −6.7 × 10⁻⁹ C (≈ −6.7 nC).

q = −6.7 × 10⁻⁹ C

• The sphere’s radius (10 cm) is needed only to confirm that the point 20 cm from the centre lies outside the sphere; the result does not depend on R as long as r > R.

Conclusion: the net charge on the conducting sphere is ≈ −6.7 × 10⁻⁹ C — negative, because the field outside points radially inward.

Concept and context. Surface charge density σ is charge per unit area, σ = ΔQ/ΔS (unit C m⁻²). For a uniformly charged surface the total charge is q = σ × (surface area). For a sphere of radius R, area = 4πR². The total flux through any closed surface enclosing the sphere is, by Gauss’s law, q/ε₀ (equivalently, flux = E × 4πR² with E = σ/ε₀ just outside the surface).

Given: diameter = 2.4 m ⇒ R = 1.2 m; σ = 80.0 μC/m² = 8.0 × 10⁻⁵ C m⁻²; ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

(a) Charge on the sphere

1. Surface area A = 4πR² = 4 × 3.1416 × (1.2)² = 4 × 3.1416 × 1.44 = 18.10 m².
2. Charge q = σ A = (8.0 × 10⁻⁵ C m⁻²)(18.10 m²) = 1.448 × 10⁻³ C ≈ 1.45 × 10⁻³ C (1.45 mC).

(b) Total flux leaving the surface

1. By Gauss’s law φ = q/ε₀ = (1.448 × 10⁻³)/(8.854 × 10⁻¹²) = 1.635 × 10⁸.
2. Total flux ≈ 1.63 × 10⁸ N m² C⁻¹ (outward, since q is positive).

Conclusion: the sphere carries a charge of 1.45 × 10⁻³ C, and the total electric flux leaving its surface is 1.63 × 10⁸ N m² C⁻¹.

Concept and context. An infinitely long thin straight wire carrying uniform linear charge density λ (charge per unit length, C m⁻¹) has cylindrical symmetry, so its field is radial and depends only on the perpendicular distance r. Applying Gauss’s law to a coaxial cylindrical surface of radius r and length l: flux passes only through the curved part (the end caps have E parallel to their surfaces → zero flux), so E × 2πrl = λl/ε₀, giving E = λ/(2πε₀ r). (The infinite length is essential so that E can be taken normal to the curved surface.)

Given: E = 9 × 10⁴ N C⁻¹; r = 2 cm = 2 × 10⁻² m; k = 1/4πε₀ = 9 × 10⁹ N m² C⁻².

1. Start with E = λ/(2πε₀ r) = 2kλ/r (because 1/4πε₀ = k ⇒ 1/2πε₀ = 2k).
2. Rearrange: λ = E r / (2k) = (9 × 10⁴ × 2 × 10⁻²) / (2 × 9 × 10⁹) = (1.8 × 10³) / (1.8 × 10¹⁰).
3. Therefore λ = 1.0 × 10⁻⁷ C m⁻¹ = 0.1 μC m⁻¹ (positive if the field points away from the wire).

λ = 10⁻⁷ C m⁻¹

Conclusion: the linear charge density of the line charge is λ = 1.0 × 10⁻⁷ C m⁻¹ (0.1 μC/m).

Concept and context. The field of a single uniformly charged infinite plane sheet with surface charge density σ is, by Gauss’s law, E = σ/2ε₀, directed away from the sheet if σ > 0 and towards it if σ < 0, and independent of the distance from the sheet. Two large parallel plates close together can be treated as two such infinite sheets, and the net field is found by the principle of superposition: E = E₁ + E₂ (vector sum). The inner faces carry +σ and −σ.

Given: σ = 17.0 × 10⁻²² C m⁻² on each inner face (opposite signs); ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

Field of each plate: E₀ = σ/2ε₀ = 17.0 × 10⁻²² / (2 × 8.854 × 10⁻¹²) = 9.6 × 10⁻¹¹ N C⁻¹.

• (a) and (b): outside the plates the two sheet fields have equal magnitude σ/2ε₀ but opposite directions, so they cancel: E = 0 in both outer regions.
• (c) Between the plates: both fields point from the positive plate towards the negative plate, so they add: E = σ/2ε₀ + σ/2ε₀ = σ/ε₀ = 17.0 × 10⁻²² / 8.854 × 10⁻¹² = 1.92 × 10⁻¹⁰ N C⁻¹ ≈ 1.9 × 10⁻¹⁰ N C⁻¹, directed from the positive to the negative plate.
• This is the basis of the parallel-plate capacitor (Chapter 2): a uniform field confined between the plates.

Conclusion: the field is zero in the outer region of the first plate (a) and of the second plate (b), and σ/ε₀ ≈ 1.9 × 10⁻¹⁰ N C⁻¹ between the plates, directed from the positively charged to the negatively charged plate.