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#Coulomb's Law

46 public questions tagged with this topic.

Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the fo

Given: Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the force between them? These values define the system as per NCERT data. Formula: Using Coulomb’s law: F = k |q_1 q_2|/r². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 9 × 10⁹ Nm²/C², q_1 = 5 × 10⁻⁷ C, q_2 = 7 × 10⁻⁷ C, r = 0.5 m . |q_1 q_2| = 5 × 7 × 10⁻¹⁴= 35 × 10⁻¹⁴ C² . r² = (0.5)² = 0.25 m² . F = 9 × 10⁹ × frac35 × 10⁻¹⁴⁰.25 = 9 × 10⁹ × 1.4 × 10⁻¹²= 0.0126 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

Two charges q_1 = 2 μC and q_2 = -2 μC are 10 cm apart. What is the electric field at the midpoint between them?

Given: Two charges q_1 = 2 μC and q_2 = -2 μC are 10 cm apart. What is the electric field at the midpoint between them? These values define the system as per NCERT data. Formula: E_1 = k q_1/r² (towards q_2 ), E_2 = k |q_2|/r² (towards q_2 ). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Midpoint is 5 cm (0.05 m) from each charge. . E_1 = 9 × 10⁹ × frac2 × 10⁻⁶(0.05)² = 7.2 × 10⁶ N/C, E_2 = 7.2 × 10⁶ N/C . Net E = E_1 + E_2 = 7.2 × 10⁶+ 7.2 × 10⁶= 1.44 × 10⁷ N/C (towards q_2 ). Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.