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JEE Main Equipotential Surfaces: related Physics practice

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Physics questions and exercises with answers

Showing questions 1–10 of 10.

Coulomb's law gives the force between two point charges q₁ and q₂ separated by distance r in vacuum asRead the answer and worked solution →Two point charges of +2 μC each are placed 3 m apart in vacuum. The force between them is (k = 9 × 10⁹ N m²/C²)Read the answer and worked solution →The electric field at distance r from an isolated point charge q isRead the answer and worked solution →The electric field of an infinite line of charge with linear charge density λ at perpendicular distance r isRead the answer and worked solution →The electric field just outside a uniformly charged infinite plane with surface charge density σ isRead the answer and worked solution →Gauss's law states that the total electric flux through a closed surface equalsRead the answer and worked solution →An electric dipole consists of two equal and opposite point charges ±q separated by distance 2a. Its dipole moment magnitude isRead the answer and worked solution →The electric field on the axial line of a short electric dipole at distance r isRead the answer and worked solution →The electric flux through a closed surface enclosing a charge of +5 μC is (ε₀ = 8.85 × 10⁻¹² C²/N m²)Read the answer and worked solution →For a charge distribution with volume density ρ, the total charge Q equalsRead the answer and worked solution →

Worked practice questions

Coulomb's law gives the force between two point charges q₁ and q₂ separated by distance r in vacuum as

  1. F = k q₁ q₂ / r²
  2. F = k q₁ q₂ / r
  3. F = k q₁ q₂ × r
  4. F = k q₁ q₂ × r²
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Correct answer: A. F = k q₁ q₂ / r²

1. Coulomb's law is the fundamental inverse-square law for electric force. 2. F = k q₁ q₂ / r² where k = 1/(4πε₀) ≈ 9 × 10⁹ N m² C⁻². 3. The force is attractive for opposite charges, repulsive for like. 4. It acts along the line joining the two charges. _Source: syllabus Physics, Ch 1 "Electric Charges and Fields", §1.5_

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Two point charges of +2 μC each are placed 3 m apart in vacuum. The force between them is (k = 9 × 10⁹ N m²/C²)

  1. 0.4 mN
  2. 4 mN
  3. 4 N
  4. 40 N
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Correct answer: B. 4 mN

1. F = k q₁ q₂ / r². 2. Substitute: F = 9 × 10⁹ × (2 × 10⁻⁶)² / 3². 3. = 9 × 10⁹ × 4 × 10⁻¹² / 9 = 4 × 10⁻³ N = 4 mN. 4. So the mutual force is 4 milli-newtons. _Source: syllabus Physics, Ch 1 "Electric Charges and Fields", §1.5_

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The electric field at distance r from an isolated point charge q is

  1. E = kq / r
  2. E = kq × r
  3. E = kq / r²
  4. E = kq × r²
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Correct answer: C. E = kq / r²

1. Electric field is force per unit test charge. 2. From Coulomb's law: E = F/q_test = k q / r². 3. Direction is radially outward for positive q, inward for negative. 4. SI unit is N/C or equivalently V/m. _Source: syllabus Physics, Ch 1 "Electric Charges and Fields", §1.6_

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More Physics topics

Electric Charges and FieldsMechanical Properties of Solids and FluidsSemiconductor ElectronicsWave OpticsCurrent ElectricityUnit 2A: Newton's Laws and Translational EquilibriumMoving Charges and MagnetismAtoms and Nuclei