Chapter 1: Electric Charges and Fields Application Questions | physics Class 12 CBSE

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Chapter 1: Electric Charges and Fields Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A polythene piece rubbed with wool acquires a negative charge of 3 × 10⁻⁷ C. Estimate the number of electrons transferred. Does the wool gain or lose electrons?

Since the polythene gains a negative charge, it gains electrons (electrons are transferred from wool to polythene). The number of electrons transferred is n = Q/e = (3 × 10⁻⁷ C) / (1.6 × 10⁻¹⁹ C) ≈ 1.875 × 10¹² electrons.
The wool loses electrons (is the donor) and therefore acquires an equal and opposite positive charge of 3 × 10⁻⁷ C, consistent with the law of conservation of charge — the total charge of the wool-polythene system remains zero.
This enormous number (~1.875 × 10¹²) represents only a tiny fraction of the total electrons in the material, since a cubic centimetre of copper alone contains about 2.5 × 10²⁴ electrons, confirming that rubbing transfers only a negligible fraction of the available charges.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

Two small charged spheres A (charge q) and B (charge q') repel each other with force F at distance r. Sphere A then touches an identical uncharged sphere C, and B touches another identical uncharged sphere D. A and B are then brought to distance r/2. Find the new force between them.

When A (charge q) touches identical uncharged sphere C, charges redistribute equally. Each sphere gets q/2 by symmetry and additivity of charge. Similarly, when B (charge q') touches D, each of B and D gets q'/2.
The new charges are q_A = q/2 and q_B = q'/2, and the new separation is r/2. The new force F' = (1/4πε₀)(q/2)(q'/2)/(r/2)² = (1/4πε₀)(qq'/4)/(r²/4) = (1/4πε₀)(qq'/r²).
Since original F = (1/4πε₀)(qq'/r²), we get F' = F. The force remains unchanged! The reduction in charges by a factor of 4 is exactly compensated by the reduction in separation by a factor of 2 (which increases force by 4), so the net force is the same.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

With reference to Gauss's law, a cubic Gaussian surface of side 9 cm has a point charge of 2.0 μC at its centre. Calculate the net electric flux through the surface. What would the flux be if the charge were placed at a corner of the cube instead of the centre?

By Gauss's law, the total electric flux through any closed surface depends only on the total charge enclosed. Here, q = 2.0 × 10⁻⁶ C enclosed by the cube, so φ = q/ε₀ = (2.0 × 10⁻⁶)/(8.854 × 10⁻¹²) ≈ 2.26 × 10⁵ N m² C⁻¹.
If the charge were placed at a corner of the cube, it would be shared among 8 identical cubes meeting at that corner (by symmetry). The charge effectively enclosed by this single cube would be q/8.
The flux through one cube would then be φ = (q/8)/ε₀ = (2.0 × 10⁻⁶)/(8 × 8.854 × 10⁻¹²) ≈ 2.82 × 10⁴ N m² C⁻¹, which is one-eighth of the flux when the charge is at the centre, demonstrating that Gauss's law depends only on enclosed charge, not on the charge's position within the surface.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

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

The torque on a dipole in a uniform electric field is τ = pE sinθ, where θ is the angle between the dipole moment p and the field E. Here, τ = (4 × 10⁻⁹)(5 × 10⁴) sin 30° = (4 × 10⁻⁹)(5 × 10⁴)(0.5) = 10⁻⁴ N m.
The torque is maximum when sinθ = 1, i.e., when θ = 90°, so the dipole axis is perpendicular to the electric field. Maximum torque = pE = (4 × 10⁻⁹)(5 × 10⁴) = 2 × 10⁻⁴ N m.
This torque at 30° is exactly half the maximum possible torque, consistent with sin 30° = 0.5. The torque always acts to align the dipole parallel to the field (stable equilibrium at θ = 0°), and the net translating force on the dipole in this uniform field is zero.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

An electric field E = 3 × 10³ î N/C is uniform. Calculate (a) the flux through a 10 cm × 10 cm square whose plane is parallel to the yz-plane, and (b) the flux if the normal to the plane makes 60° with the x-axis.

When the plane is parallel to the yz-plane, its outward normal is along the x-axis (î direction). The area is ΔS = 0.1 × 0.1 = 0.01 m². Flux φ = E · ΔS = E ΔS cosθ = (3 × 10³)(0.01) cos 0° = 30 N m² C⁻¹.
When the normal makes 60° with the x-axis (and hence with E), flux φ = E ΔS cos 60° = (3 × 10³)(0.01)(0.5) = 15 N m² C⁻¹.
The net flux through a closed cube of side 20 cm with faces parallel to coordinate planes in this field is zero, because for every face where flux enters, an equal amount leaves from the opposite face — consistent with Gauss's law since no charge is enclosed.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

An infinite line charge produces an electric field of 9 × 10⁴ N/C at a perpendicular distance of 2 cm. Using this scenario, calculate the linear charge density of the wire.

For an infinitely long wire with linear charge density λ, the electric field at perpendicular distance r is E = λ/(2πε₀r). Rearranging, λ = E × 2πε₀r.
Substituting E = 9 × 10⁴ N/C, r = 0.02 m, and 2πε₀ = 2π × 8.854 × 10⁻¹² ≈ 5.56 × 10⁻¹¹: λ = (9 × 10⁴)(5.56 × 10⁻¹¹)(0.02)/(1) — more directly, λ = E × 2πε₀r = 9 × 10⁴ × (1/(2 × 9 × 10⁹)) × 0.02 = 9 × 10⁴/(9 × 10⁹) × 0.02 = 10⁻⁵ × 0.02 = 1 × 10⁻⁷ C/m.
Thus λ = 0.1 μC/m = 10⁻⁷ C/m. This illustrates that the field from a line charge decreases as 1/r (not 1/r²) — at double the distance (4 cm), the field would be halved to 4.5 × 10⁴ N/C, confirming the linear dependence on 1/r.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C directed radially inward, determine the net charge on the sphere and explain what the inward direction indicates.

For a conducting sphere, the electric field outside (r > R) is equivalent to that of a point charge at the centre: E = q/(4πε₀r²). Here, r = 0.20 m and E = 1.5 × 10³ N/C.
Solving for q: q = E × 4πε₀r² = (1.5 × 10³) × (0.20)²/(9 × 10⁹) = (1.5 × 10³ × 0.04)/(9 × 10⁹) = 60/(9 × 10⁹) ≈ −6.67 × 10⁻⁹ C ≈ −6.67 nC.
The field is directed radially inward, which indicates that the sphere carries a negative charge. For a positive charge, the field would point radially outward. The magnitude of the charge is approximately 6.67 nC, and the sphere is negatively charged.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A system of two charges qA = 2.5 × 10⁻⁷ C and qB = −2.5 × 10⁻⁷ C is located at A: (0, 0, −15 cm) and B: (0, 0, +15 cm). Find the total charge of the system and the electric dipole moment.

Total charge of the system = qA + qB = 2.5 × 10⁻⁷ + (−2.5 × 10⁻⁷) = 0. The system is electrically neutral overall, forming an electric dipole.
The dipole has charge magnitude q = 2.5 × 10⁻⁷ C and separation 2a = distance from A to B = 30 cm = 0.30 m. The dipole moment magnitude is p = q × 2a = (2.5 × 10⁻⁷)(0.30) = 7.5 × 10⁻⁸ C m.
By convention, the dipole moment vector points from the negative charge (qB = −2.5 × 10⁻⁷ C at B: +15 cm) toward the positive charge (qA = +2.5 × 10⁻⁷ C at A: −15 cm), i.e., in the −z direction. So p = 7.5 × 10⁻⁸ (−k̂) C m.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

With reference to Coulomb's law, four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, qD = −5 μC are located at the corners of a square ABCD of side 10 cm. Find the force on a 1 μC charge at the centre of the square.

The centre O of the square is equidistant from all four corners, with distance r = (10√2)/2 = 5√2 cm = 7.07 × 10⁻² m. The charges at opposite corners are equal: qA = qC = 2 μC and qB = qD = −5 μC.
The force on the central charge due to qA and qC are equal in magnitude but opposite in direction (they are at diagonally opposite corners), so they cancel. Similarly, the forces due to qB and qD are equal and opposite, so they also cancel.
Total force on the central charge = 0. This result follows from the symmetry of the configuration: for each charge at one corner, there is an equal charge at the diagonally opposite corner, and their forces on the central charge are equal and opposite, leading to complete cancellation by the superposition principle.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A point charge causes an electric flux of −1.0 × 10³ N m² C⁻¹ through a spherical Gaussian surface of radius 10 cm centred on the charge. Find the charge. If the radius is doubled, what is the new flux?

By Gauss's law, φ = q/ε₀. Therefore q = φε₀ = (−1.0 × 10³)(8.854 × 10⁻¹²) = −8.854 × 10⁻⁹ C ≈ −8.85 nC. The negative sign of flux indicates that the field lines are directed inward (toward the charge), confirming the charge is negative.
If the radius of the Gaussian surface is doubled (to 20 cm), by Gauss's law the flux depends only on the enclosed charge, not on the size or shape of the surface. The enclosed charge is still q = −8.85 nC (same point charge at centre).
Therefore, the flux through the doubled sphere is still φ = q/ε₀ = −1.0 × 10³ N m² C⁻¹, unchanged. This is a key implication of Gauss's law: flux through any closed surface enclosing the same charge is the same, regardless of the surface size.

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Chapter sections

Chapter 1: Electric Charges and Fields | Application Questions

A polythene piece rubbed with wool acquires a negative charge of 3 × 10⁻⁷ C. Estimate the number of electrons transferred. Does the wool gain or lose electrons?

Two small charged spheres A (charge q) and B (charge q') repel each other with force F at distance r. Sphere A then touches an identical uncharged sphere C, and B touches another identical uncharged sphere D. A and B are then brought to distance r/2. Find the new force between them.

With reference to Gauss's law, a cubic Gaussian surface of side 9 cm has a point charge of 2.0 μC at its centre. Calculate the net electric flux through the surface. What would the flux be if the charge were placed at a corner of the cube instead of the centre?

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

An electric field E = 3 × 10³ î N/C is uniform. Calculate (a) the flux through a 10 cm × 10 cm square whose plane is parallel to the yz-plane, and (b) the flux if the normal to the plane makes 60° with the x-axis.

An infinite line charge produces an electric field of 9 × 10⁴ N/C at a perpendicular distance of 2 cm. Using this scenario, calculate the linear charge density of the wire.

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C directed radially inward, determine the net charge on the sphere and explain what the inward direction indicates.

A system of two charges qA = 2.5 × 10⁻⁷ C and qB = −2.5 × 10⁻⁷ C is located at A: (0, 0, −15 cm) and B: (0, 0, +15 cm). Find the total charge of the system and the electric dipole moment.

With reference to Coulomb's law, four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, qD = −5 μC are located at the corners of a square ABCD of side 10 cm. Find the force on a 1 μC charge at the centre of the square.

A point charge causes an electric flux of −1.0 × 10³ N m² C⁻¹ through a spherical Gaussian surface of radius 10 cm centred on the charge. Find the charge. If the radius is doubled, what is the new flux?

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