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Solutions

Overview

Step-by-step NCERT solutions for Electric Charges and Fields (Chapter 1, CBSE Class 12 Physics) — every question and answer worked out in full, not just the final result. You can also read the Electric Charges and Fields textbook chapter.

Solved

What these solutions cover

All 23 questions in Electric Charges and Fields are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

  1. 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?
  2. 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?
  3. Check that the ratio ke² / G m_e m_p is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
  4. Explain the meaning of the statement 'electric charge of a body is quantised'. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
  5. 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.
  6. Four point charges q_A = 2 μC, q_B = -5 μC, q_C = 2 μC, and q_D = -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?
  7. 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?
  8. Two point charges q_A = 3 μC and q_B = -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?
  9. A system has two charges q_A = 2.5 × 10⁻⁷ C and q_B = -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?
  10. 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⁻¹. Calculate the magnitude of the torque acting on the dipole.
  11. 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?
  12. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10⁻⁷ C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
  13. Figure 1.30 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?
  14. Consider a uniform electric field E = 3 × 10³ î N/C.
    • (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
    • (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
  15. What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
  16. 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? Why or Why not?
  17. A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
  18. A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
  19. A point charge causes an electric flux of -1.0 × 10³ N m²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
    • (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
    • (b) What is the value of the point charge?
  20. 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 and points radially inward, what is the net charge on the sphere?
  21. 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?
  22. An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.
  23. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10⁻²² C/m². What is E:
    • (a) in the outer region of the first plate,
    • (b) in the outer region of the second plate, and
    • (c) between the plates?
Sample

Worked examples

3 questions from Electric Charges and Fields, solved step by step.

Exercises

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?

Given: q₁ = 2 × 10⁻⁷ C, q₂ = 3 × 10⁻⁷ C, r = 30 cm = 0.30 m

k = 9 × 10⁹ N m² C⁻²

Using Coulomb's law:

F = k q₁ q₂ / r²

F = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.30)²

F = (9 × 10⁹ × 6 × 10⁻¹⁴) / 0.09

F = (5.4 × 10⁻⁴) / 0.09

F = 6.0 × 10⁻³ N

Since both charges are positive, the force is repulsive.

Answer: F = 6.0 × 10⁻³ N (repulsive)

Exercises

Explain the meaning of the statement 'electric charge of a body is quantised'. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

(a) Quantisation of electric charge means that the charge on any body is always an integral multiple of the elementary charge e = 1.6 × 10⁻¹⁹ C:

q = ne, where n = 0, ±1, ±2, ±3, ...

Charge cannot take arbitrary values; it can only change in discrete steps of e. A body cannot carry a charge of, say, 1.5e or 0.7e.

(b) At macroscopic scales, the charges involved are enormously large compared to e. For example, 1 μC contains about 6 × 10¹² elementary charges. At this scale, adding or removing one electron changes the total charge by only 1 part in 6 × 10¹² — an imperceptibly tiny step. The granularity of charge is completely invisible, and charge behaves as if it were a continuous variable. Quantisation can therefore be safely ignored in macroscopic problems.

Answer: (a) q = ne (n integer); charge changes only in discrete units of e = 1.6 × 10⁻¹⁹ C. (b) Macroscopic charges are so large (~10¹² e) that the discrete step size is negligible.

Exercises

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.

When a glass rod is rubbed with silk, electrons are transferred from the glass rod to the silk cloth. The glass rod loses electrons and becomes positively charged; the silk gains the same number of electrons and becomes equally negatively charged.

No new charge is created in this process. The positive charge gained by the glass equals in magnitude the negative charge gained by the silk. The algebraic sum of the total charge of the system (glass rod + silk) remains zero throughout — the same as before rubbing.

This is fully consistent with conservation of charge: the total charge of an isolated system remains constant. Rubbing merely redistributes existing charges between the two bodies; it neither creates nor destroys net charge.

Answer: Rubbing transfers electrons from glass to silk (or vice versa). Equal and opposite charges appear on each, keeping the total charge of the system unchanged (zero). This is consistent with conservation of charge.

All 23 questions in Electric Charges and Fields are solved step by step in the Solutions PDF above.

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