NCERT Solutions›Class 12 Physics›Chapter 1

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Chapter 1 · Class 12 Physics

Electric Charges and Fields

2 exercises23 questions solved

Exercise 1.1Coulomb's Law and Electric Field

What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?

Solution

Given: q₁ = 2 × 10⁻⁷ C, q₂ = 3 × 10⁻⁷ C, r = 30 cm = 0.3 m k = 9 × 10⁹ N m² C⁻² Using Coulomb's law: F = kq₁q₂/r² = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.3)² = (9 × 10⁹ × 6 × 10⁻¹⁴) / 0.09 = 54 × 10⁻⁵ / 0.09 = 6 × 10⁻³ N The force is repulsive (both charges are positive) and equals 6 × 10⁻³ N.

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Solution

Given: q₁ = 0.4 μC = 0.4 × 10⁻⁶ C, q₂ = –0.8 μC = –0.8 × 10⁻⁶ C, F = 0.2 N (a) Using F = k|q₁||q₂|/r²: 0.2 = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶) / r² 0.2 = (9 × 10⁹ × 0.32 × 10⁻¹²) / r² 0.2 = 2.88 × 10⁻³ / r² r² = 2.88 × 10⁻³ / 0.2 = 0.0144 r = 0.12 m = 12 cm (b) By Newton's third law, the force on the second sphere due to the first is equal in magnitude and opposite in direction. Force = 0.2 N (attractive, since charges are opposite).

Check that the ratio ke²/Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution

e = 1.6 × 10⁻¹⁹ C, me = 9.11 × 10⁻³¹ kg, mp = 1.67 × 10⁻²⁷ kg k = 9 × 10⁹ N m² C⁻², G = 6.67 × 10⁻¹¹ N m² kg⁻² Dimensional check: [ke²] = [N m² C⁻²][C²] = [N m²] = [kg m³ s⁻²] [Gmemp] = [N m² kg⁻²][kg][kg] = [N m²] = [kg m³ s⁻²] Ratio is dimensionless ✓ Value: ke² = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² = 9 × 10⁹ × 2.56 × 10⁻³⁸ = 23.04 × 10⁻²⁹ Gmemp = 6.67 × 10⁻¹¹ × 9.11 × 10⁻³¹ × 1.67 × 10⁻²⁷ = 6.67 × 9.11 × 1.67 × 10⁻⁶⁹ ≈ 101.5 × 10⁻⁶⁹ = 1.015 × 10⁻⁶⁷ Ratio = 23.04 × 10⁻²⁹ / 1.015 × 10⁻⁶⁷ ≈ 2.27 × 10³⁹ Significance: The electrostatic force between an electron and a proton is about 2.27 × 10³⁹ times stronger than the gravitational force between them.

Explain the meaning of the statement 'electric charge of a body is quantised'.

Solution

Electric charge of a body is always an integral multiple of a basic unit of charge e (= 1.6 × 10⁻¹⁹ C): q = ne, where n is an integer (positive or negative) and e is the charge of an electron/proton. This means: • Charge cannot take fractional values like 0.5e or 1.3e • The smallest possible charge is e = 1.6 × 10⁻¹⁹ C • Charge can only be gained or lost in discrete multiples of e Note: At the macroscopic level, since e is very small, charge appears to be continuous. But at the microscopic (atomic) level, quantisation is significant.

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Solution

When a glass rod is rubbed with silk: • The glass rod acquires positive charge • The silk cloth acquires equal negative charge This is consistent with conservation of charge because: • Before rubbing: both are electrically neutral (net charge = 0) • After rubbing: glass gets +q and silk gets –q • Total charge = +q + (–q) = 0 (unchanged) No new charge is created. Electrons are merely transferred from the glass rod to the silk cloth. The total charge of the system (glass + silk) remains zero throughout. This is the law of conservation of charge: the total charge of an isolated system remains constant.

Four point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Solution

Side of square = 10 cm, so diagonal = 10√2 cm Distance from centre to each corner = (10√2)/2 = 5√2 cm = 5√2 × 10⁻² m The charge at the centre is q = 1 μC. Observation: The square has charges at corners: A(+2μC), B(–5μC), C(+2μC), D(–5μC) A and C are diagonally opposite (both +2μC) B and D are diagonally opposite (both –5μC) Force due to A and C: • FA on q: directed from A toward centre (repulsion) • FC on q: directed from C toward centre (repulsion) • A and C are diagonally opposite, so FA and FC are equal in magnitude but opposite in direction • Net force due to A and C = 0 Force due to B and D: • FB on q: directed from q toward B (attraction) • FD on q: directed from q toward D (attraction) • B and D are diagonally opposite, so FB and FD are equal in magnitude but opposite in direction • Net force due to B and D = 0 Total force on the charge at the centre = 0.

(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point?

Solution

(a) A field line is the path along which a positive test charge would move if free to do so. At every point, the electric field has a definite direction and magnitude. A sudden break would mean the field is zero or undefined at that point, which is physically impossible in a region with charges (unless we're at infinity or a point of zero field). Hence, field lines must be continuous curves. (b) If two field lines were to cross at a point, it would mean there are two directions of electric field at that single point. But the electric field at any point has a unique direction (as it is a vector). Having two different field directions at one point is a contradiction. Therefore, two field lines can never cross each other.

Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?

Solution

q_A = 3 × 10⁻⁶ C, q_B = –3 × 10⁻⁶ C, distance AB = 20 cm = 0.2 m O is the midpoint, so r = 0.1 m from each charge. (a) Electric field at O: Field due to q_A (positive) at O: E_A = kq_A/r² = (9 × 10⁹ × 3 × 10⁻⁶) / (0.1)² = 27000 / 0.01 = 2.7 × 10⁶ N/C Direction: from A toward B (away from +ve charge) Field due to q_B (negative) at O: E_B = k|q_B|/r² = (9 × 10⁹ × 3 × 10⁻⁶) / (0.1)² = 2.7 × 10⁶ N/C Direction: from O toward B (toward –ve charge) = same direction as E_A Total field E = E_A + E_B = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C Direction: from A to B (b) Test charge q = –1.5 × 10⁻⁹ C Force = qE = 1.5 × 10⁻⁹ × 5.4 × 10⁶ = 8.1 × 10⁻³ N Direction: opposite to E (from B to A), since charge is negative.

A system has two charges qA = 2.5 × 10⁻⁷ C and qB = –2.5 × 10⁻⁷ C located at points A: (0, 0, –15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

Solution

Total charge = q_A + q_B = 2.5 × 10⁻⁷ + (–2.5 × 10⁻⁷) = 0 Electric dipole moment: The dipole moment p = q × 2l where q = 2.5 × 10⁻⁷ C and 2l is the separation between the charges. 2l = distance from A(0,0,–15cm) to B(0,0,+15cm) = 30 cm = 0.30 m p = 2.5 × 10⁻⁷ × 0.30 = 7.5 × 10⁻⁸ C·m Direction: from negative charge (B is –ve? No — A is +ve, B is –ve) Wait: qA = +2.5 × 10⁻⁷ C at A(0,0,–15cm) and qB = –2.5 × 10⁻⁷ C at B(0,0,+15cm) Dipole moment direction: from –q to +q = from B to A = from (0,0,+15) to (0,0,–15) = in the –z direction Electric dipole moment p = 7.5 × 10⁻⁸ C·m, directed along –z axis (from B to A).

Q10

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

Solution

Given: p = 4 × 10⁻⁹ C·m, E = 5 × 10⁴ N/C, θ = 30° Torque on a dipole: τ = pE sin θ = 4 × 10⁻⁹ × 5 × 10⁴ × sin 30° = 4 × 10⁻⁹ × 5 × 10⁴ × 0.5 = 4 × 10⁻⁹ × 2.5 × 10⁴ = 10 × 10⁻⁵ = 1 × 10⁻⁴ N·m The torque acting on the dipole is 10⁻⁴ N·m.

Q11

A polythene piece rubbed with wool is found to have a negative charge of 3 × 10⁻⁷ C. (a) Estimate the number of electrons transferred (from which to which?). (b) Is there a transfer of mass from wool to polythene?

Solution

(a) Charge on polythene = –3 × 10⁻⁷ C (negative) This means electrons have been transferred TO the polythene FROM the wool. Number of electrons: n = Q/e = 3 × 10⁻⁷ / 1.6 × 10⁻¹⁹ = (3/1.6) × 10¹² = 1.875 × 10¹² ≈ 1.88 × 10¹² electrons (b) Yes, there is a transfer of mass. Mass of each electron = 9.11 × 10⁻³¹ kg Mass transferred = n × me = 1.875 × 10¹² × 9.11 × 10⁻³¹ = 17.08 × 10⁻¹⁹ kg ≈ 1.7 × 10⁻¹⁸ kg This mass is extremely small and practically undetectable.

Q12

(a) Two insulating large non-conducting sheets of charge densities +σ and –σ are placed a certain distance apart. What is the electric field in each of the three regions: I (outside first sheet), II (between sheets), III (outside second sheet)? (b) What is the force between the sheets?

Solution

For an infinite sheet with surface charge density σ, the electric field on each side = σ/(2ε₀), directed away from the positive sheet and toward the negative sheet. (a) Let sheet 1 carry +σ and sheet 2 carry –σ, with region I to the left of sheet 1, region II between the sheets, and region III to the right of sheet 2. Sheet 1 (+σ): E₁ = σ/(2ε₀) pointing away from it (left in region I, right in region II) Sheet 2 (–σ): E₂ = σ/(2ε₀) pointing toward it (right in region I, left in region II, right in region III toward sheet 2) Region I (outside sheet 1): E₁ points left, E₂ points right → they cancel E = 0 Region II (between sheets): Both fields point in same direction (from +σ to –σ) E = σ/(2ε₀) + σ/(2ε₀) = σ/ε₀ Region III (outside sheet 2): E₁ points right, E₂ points left → they cancel E = 0 (b) Force per unit area between sheets: Sheet 2 is in the field of sheet 1: E₁ = σ/(2ε₀) Force per unit area = σ × E₁ = σ × σ/(2ε₀) = σ²/(2ε₀) The force is attractive (opposite charges).

Q13

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ N/C in Millikan's oil drop experiment. The density of the oil is 1.26 g cm⁻³. Find the radius of the drop (g = 9.81 m s⁻²; e = 1.6 × 10⁻¹⁹ C).

Solution

For the drop to be stationary: Electric force = Weight qE = mg Charge q = 12e = 12 × 1.6 × 10⁻¹⁹ = 1.92 × 10⁻¹⁸ C Weight mg = ρ × (4/3)πr³ × g Setting qE = ρ(4/3)πr³g: r³ = 3qE / (4πρg) = 3 × 1.92 × 10⁻¹⁸ × 2.55 × 10⁴ / (4π × 1.26 × 10³ × 9.81) = (3 × 1.92 × 2.55 × 10⁻¹⁴) / (4π × 1.26 × 10³ × 9.81) Numerator: 3 × 1.92 × 2.55 × 10⁻¹⁴ = 14.69 × 10⁻¹⁴ = 1.469 × 10⁻¹³ Denominator: 4 × 3.14159 × 1.26 × 10³ × 9.81 = 4 × 3.14159 × 12360.6 = 155,300 ≈ 1.553 × 10⁵ r³ = 1.469 × 10⁻¹³ / 1.553 × 10⁵ = 9.46 × 10⁻¹⁹ r = (9.46 × 10⁻¹⁹)^(1/3) ≈ 9.81 × 10⁻⁷ m ≈ 0.981 μm ≈ 1.0 μm

Q14

Which among the curves shown in a figure cannot possibly represent electrostatic field lines?

Solution

The following cannot represent valid electrostatic field lines: (a) Field lines that form closed loops — Electrostatic field lines never form closed loops (unlike magnetic field lines). This is because electrostatic fields are conservative (∮E·dl = 0). (b) Field lines that show sudden breaks — Field lines are continuous; they start from positive charges and end at negative charges (or go to infinity). Sudden breaks imply regions with no field, which is unphysical. (c) Field lines that intersect each other — Two field lines crossing means the electric field has two directions at that point, which is impossible. (d) Field lines that don't emerge/enter perpendicularly from/into conductors — Static field lines must be perpendicular to the surface of a conductor.

Q15

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ N/C per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10⁻⁷ C·m in the negative z-direction?

Solution

Given: dE/dz = 10⁵ N/C per m (field increases along +z) Dipole moment p = 10⁻⁷ C·m in the –z direction Force on a dipole in a non-uniform field: F = p(dE/dz) along direction of p But here p is along –z and dE/dz is along +z: F = p × dE/dz = 10⁻⁷ × 10⁵ = 10⁻² N Since p is along –z and E is along +z, the force on p is: F = –p(dE/dz) ẑ × ... The net force on the dipole: F = (p·∇)E For p = –p₀ẑ and E = Eẑ: F = –p₀ (dE/dz) ẑ = –10⁻⁷ × 10⁵ ẑ = –10⁻² ẑ N Magnitude of force = 10⁻² N, directed in –z direction. Torque: τ = p × E Since p is along –z and E is along +z, they are antiparallel: sin θ = sin 180° = 0 Torque = pE sin 180° = 0 The torque is zero (p and E are antiparallel).

Q16

(a) A conductor A with a cavity as shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Solution

(a) In electrostatic equilibrium, the electric field inside a conductor is zero. Using Gauss's law, consider a Gaussian surface just inside the conductor (within the material): ∮E·dA = Q_enclosed/ε₀ Since E = 0 everywhere inside the conductor: Q_enclosed = 0 Therefore, no charge resides inside the cavity walls, and all charge Q appears on the outer surface. (b) When conductor B (charge q) is placed inside the cavity: • The inner surface of A develops a charge –q (by induction, to make E = 0 inside the conductor material) • Since A has total charge Q, and inner surface has –q, the outer surface must have Q + q • Total charge on outer surface of A = Q + q (c) To shield a sensitive instrument from external electrostatic fields: • Enclose the instrument in a metallic (conducting) enclosure — this is called a Faraday cage. • The conductor redistributes charges on its surface, maintaining zero electric field inside. • Electric field inside a closed metallic shell is always zero regardless of external fields.

Q17

Answer the following: (a) A comb run through one's dry hair attracts small bits of paper. Why? What happens if the hair is wet or if it is a rainy day? (b) A charged rod attracts an uncharged conductor. However, a charged rod attracts even a conducting uncharged sphere. Explain.

Solution

(a) When a comb is run through dry hair, it gets charged by friction (usually negatively). When brought near small bits of paper: • The paper is a dielectric (insulator) • The nearby face of each paper piece gets induced charges of opposite sign • This induced charge is closer to the comb, so the attractive force > repulsive force • Net result: the paper pieces are attracted If hair is wet or it is a rainy day: • Moisture makes the surface conducting • Charges leak away quickly • The comb cannot retain its charge • So the comb will not attract paper bits (or attract very feebly) (b) When a charged rod is brought near an uncharged conductor (sphere): • Free electrons in the conductor redistribute (electrostatic induction) • The near side gets charges opposite to the rod (attraction) • The far side gets same sign charges as rod (repulsion) • Since the near side is closer, attractive force > repulsive force • Net force: the conductor is attracted toward the charged rod

Q18

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Solution

In a uniform electric field, the trajectory of a charged particle is parabolic (similar to projectile motion in a gravitational field). • A positively charged particle accelerates in the direction of the electric field (E field direction) • A negatively charged particle accelerates opposite to E From the figure (typical arrangement): • Particle 1: curves in direction of E → positive charge (+) • Particle 2: travels straight (uncharged or path along E) → based on figure, could be positive • Particle 3: curves opposite to E → negative charge (–) For charge-to-mass ratio (q/m): The curvature of the path ∝ acceleration ∝ qE/m = (q/m)E A more curved path means higher q/m ratio. The particle with the sharpest curve (highest deflection) has the highest charge-to-mass ratio. From the figure: Particle 3 (most curved) has the highest q/m.

Exercise 1.2Gauss's Law and Applications

What is the net flux of the uniform electric field E = 3 × 10³ î N/C through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Solution

For a uniform electric field (E = 3 × 10³ î N/C) passing through a closed surface: The field lines entering the cube on one side exit from the opposite side with the same magnitude. Using Gauss's law: Φ = Q_enclosed/ε₀ Since the field is uniform (no charge enclosed in the cube): Net flux = 0 Alternatively: Flux through right face (+x) = E × A = 3 × 10³ × (0.2)² = 3 × 10³ × 0.04 = 120 N·m²/C Flux through left face (–x) = –120 N·m²/C (field enters this face) Flux through all other faces = 0 (E is perpendicular to normal) Net flux = 120 – 120 + 0 = 0

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ N·m²/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box?

Solution

(a) Using Gauss's law: Φ = Q_enclosed/ε₀ Q_enclosed = ε₀Φ = 8.85 × 10⁻¹² × 8.0 × 10³ = 70.8 × 10⁻⁹ C ≈ 7.08 × 10⁻⁸ C ≈ 0.071 μC (b) No, we cannot conclude there are no charges inside. If the net outward flux is zero, it means the net charge inside is zero (Q_enclosed = ε₀ × 0 = 0). However, there could be equal and opposite charges inside the box — they would produce zero net flux while still existing. For example: +q and –q charges inside → net charge = 0 → net flux = 0, but charges do exist.

A point charge +10 μC is at a distance 5 cm directly above the centre of a square of side 10 cm. What is the magnitude of the electric flux through the square?

Solution

Strategy: Imagine the square is the bottom face of a cube of side 10 cm, with the charge at the center of the cube. If the charge were at the center of a complete cube of side 10 cm: • The charge is at the center only if it is 5 cm from each face • The charge is 5 cm above the square of side 10 cm — this is consistent with the charge being at the center of a cube of side 10 cm Total flux from point charge q = q/ε₀ = 10 × 10⁻⁶ / 8.85 × 10⁻¹² = 1.13 × 10⁶ N·m²/C By symmetry, the cube has 6 faces, and the flux is equally distributed. Flux through the square (one face) = Total flux / 6 = 1.13 × 10⁶ / 6 ≈ 1.88 × 10⁵ N·m²/C ≈ 1.88 × 10⁵ N m²/C

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

Solution

(a) Radius r = 1.2 m, surface charge density σ = 80.0 × 10⁻⁶ C/m² Surface area = 4πr² = 4π × (1.2)² = 4 × 3.14159 × 1.44 = 18.096 m² Charge Q = σ × A = 80 × 10⁻⁶ × 18.096 = 1447.6 × 10⁻⁶ C ≈ 1.45 × 10⁻³ C (b) Total electric flux: Φ = Q/ε₀ = 1.45 × 10⁻³ / 8.85 × 10⁻¹² = 1.638 × 10⁸ N·m²/C ≈ 1.64 × 10⁸ N m²/C

An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.

Solution

Electric field due to infinite line charge: E = λ/(2πε₀r) Given: E = 9 × 10⁴ N/C, r = 2 cm = 0.02 m Linear charge density: λ = E × 2πε₀r = 9 × 10⁴ × 2π × 8.85 × 10⁻¹² × 0.02 = 9 × 10⁴ × 2 × 3.14159 × 8.85 × 10⁻¹² × 0.02 = 9 × 10⁴ × 1.113 × 10⁻¹² × 2π × 0.02 ... Simplifying: λ = E × 2πε₀r = (E × r) / (1/2πε₀) = E × r × (1/k) × (1/2) Using k = 9 × 10⁹: E = 2kλ/r → λ = Er/(2k) = (9 × 10⁴ × 0.02)/(2 × 9 × 10⁹) = 1800 / (18 × 10⁹) = 100 × 10⁻⁹ C/m = 10⁻⁷ C/m = 0.1 μC/m

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