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Agniveer Air Force Electric Charges and Fields — practice questions

Explore 12 free Electric Charges and Fields questions in Agniveer Air Force Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Electric Charges and Fields questions and exercises with answers

Showing questions 1–12 of 12.

Coulomb's law gives the electrostatic force between two point charges as:Read the answer and worked solution →SI unit of electric charge is the coulomb (C). 1 C is approximately:Read the answer and worked solution →Electric field E at a point due to a point charge q at distance r is:Read the answer and worked solution →Electric flux Φ through a closed surface enclosing total charge Q_enc (Gauss's law) equals:Read the answer and worked solution →An infinite uniform line of linear charge density λ produces an electric field at perpendicular distance r equal to:Read the answer and worked solution →Just outside a large uniformly-charged conducting sheet with surface charge density σ, the magnitude of the electric field is:Read the answer and worked solution →Two equal point charges q each are placed at distance d. The electrostatic PE of the system (zero at infinity) is:Read the answer and worked solution →Electric potential V due to a point charge q at distance r is:Read the answer and worked solution →The electric field at the centre of a uniformly-charged thin spherical SHELL of radius R is:Read the answer and worked solution →Torque on a dipole of moment p placed in a uniform field E at angle θ is:Read the answer and worked solution →Two large parallel conducting plates carry equal and opposite surface charge density σ. The electric field BETWEEN the plates is:Read the answer and worked solution →Judge a dipole sitting in a fieldRead the answer and worked solution →

Worked practice questions

Coulomb's law gives the electrostatic force between two point charges as:

  1. F = k q₁ q₂ / r
  2. F = k q₁ q₂ / r²
  3. F = k (q₁ + q₂) / r²
  4. F = k q₁ q₂ r²
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Correct answer: B. F = k q₁ q₂ / r²

1. F = k q₁ q₂ / r², where k = 1 / (4πε₀) ≈ 9 × 10⁹ N m² C⁻². 2. Inverse square in r, same as Newtonian gravity. 3. Force is along the line joining the charges; repulsive for like signs. _Source: syllabus Physics Ch 1 §1.5 Coulomb's Law_

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SI unit of electric charge is the coulomb (C). 1 C is approximately:

  1. 6.25 × 10¹⁸ electron charges
  2. 6.25 × 10⁹ electron charges
  3. 1.6 × 10⁻¹⁹ electron charges
  4. 1 electron charge
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Correct answer: A. 6.25 × 10¹⁸ electron charges

1. Electron charge e ≈ 1.6 × 10⁻¹⁹ C. 2. So 1 C / e = 1 / 1.6 × 10⁻¹⁹ ≈ 6.25 × 10¹⁸ electron charges. 3. Charge is quantised: Q = n e for integer n. _Source: syllabus Physics Ch 1 §1.4 Quantisation of Charge_

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Electric field E at a point due to a point charge q at distance r is:

  1. E = k q / r
  2. E = k q / r²
  3. E = k q² / r²
  4. E = k q² / r
Show answer and solution

Correct answer: B. E = k q / r²

1. E = F / q_test = k q / r², independent of the test charge. 2. Direction: radially outward from positive q (inward for negative). 3. SI unit: N C⁻¹ = V m⁻¹. _Source: syllabus Physics Ch 1 §1.6 Electric Field_

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