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PhysicsMediumClass 12

Coulomb's Law — Force between charges

Electrostatics

18

JEE Qs

8%

Hard

60

min

Master the vector form of Coulomb's Law and the Principle of Superposition for multiple charges, as most problems involve finding net force on a system of charges.

🧮 Key Formulas

F = k * |q1 * q2| / r^2
k = 1 / (4 * pi * epsilon_0)
k = 9 * 10^9 Nm^2/C^2 (in vacuum)
epsilon_0 = 8.854 * 10^-12 C^2/(Nm^2)
F_vector = k * q1 * q2 / |r_vector|^3 * r_vector
F_medium = F_vacuum / epsilon_r (or K), where epsilon_r is the dielectric constant of the medium

✅ Key Points for JEE

  • 1Coulomb's Law applies to point charges. For extended charge distributions, integration or symmetry arguments (often with Gauss's Law) are required.
  • 2The electrostatic force is a central force, always acting along the line joining the two charges. Like charges repel, unlike charges attract.
  • 3The Principle of Superposition is crucial: The net force on a charge due to multiple other charges is the vector sum of individual forces due to each charge.
  • 4The presence of a dielectric medium reduces the electrostatic force between charges by a factor equal to its dielectric constant (epsilon_r or K).

⚠️ Common Mistakes

  • Forgetting that electrostatic force is a vector quantity and performing scalar addition instead of vector addition when multiple forces are present.
  • Incorrectly using the sign of charges directly in the magnitude formula, leading to confusion about force direction. Determine magnitude using |q1q2| and then assign direction based on charge types.
  • Mistaking the 'r' in the vector form F_vector = k * q1 * q2 / |r_vector|^3 * r_vector, often leading to incorrect powers of distance in the denominator.

📝 Practice Questions

See all

Q38.A capacitor, C1 = 6μ F is charged to a potential difference of V0 = 5 V using a 5 V battery. The battery is removed and another capacitor, C2 = 12μ F is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the 2025 (29 Jan Shift 2) JEE Main Previous Year Paper charges (q1 and q2) on the capacitors C1 and C2 when equilibrium condition is reached. (1) q1 = 10μC, q2 = 20μC (2) q1 = 30μC, q2 = 15μC (3) q1 = 20μC, q2 = 10μC (4) q1 = 15μC, q2 = 30μC

2025·NumericalMedium

Q44.A point charge causes an electric flux of −2 × 104Nm2C−1 to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is : (Given ϵ0 = 8.85 × 10−12C2 N−1 m−2 ) (1) 15.7 × 10−8C (2) 17.7 × 10−8C (3) −15.7 × 10−8C (4) −17.7 × 10−8C

2025·MCQEasy

Q45.An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σo . Choose the correct option from the following. (1) Potential energy and torque both are maximum. (2) Torque on dipole is zero and net force is directed away from the sheet. (3) Torque on dipole is zero and net force acts (4) Potential energy of dipole is minimum and torque towards the sheet. is zero.

2025·MCQMedium

Q26.An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 106 m/s. If the magnitude of the electric field between the plates is 9.1 V/cm , then the vertical component of velocity of electron is (mass of electron = 9.1 × 10−31 kg and charge of electron = 1.6 × 10−19C ) (1) 0 (2) 1 × 106 m/s (3) 16 × 106 m/s (4) 16 × 104 m/s

2025·MCQMedium

Q38.A line charge of length ' a ' is kept at the center of an edge BC of a cube ABCDEFGH having edge length ' 2 a ' as shown in the figure. If the density of line charge is λ C per unit length, then the total electric flux through all the faces of the cube will be . (Take, ϵ0 as the free space permittivity) (1) λa (2) λa 2ϵ0 4ϵ0 (3) λa (4) λa 16ϵ0 8ϵ0

2025·MCQHard

Q44.A parallel-plate capacitor of capacitance 40μ F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are (1) 4 mC and 0.2 J (2) 8 mC and 2.0 J (3) 2 mC and 0.4 J (4) 2 mC and 0.2 J 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

2025·MCQMedium

NCERT Chapters

  • Class 12 Physics Ch 1: Electric Charges and Fields