JEE Advanced Electromagnetism — practice questions — page 2

Explore 32 free Electromagnetism questions in JEE Advanced Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Electromagnetism questions and exercises with answers

Showing questions 31–32 of 32.

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

The electric flux through a closed surface enclosing a charge of +5 μC is (ε₀ = 8.85 × 10⁻¹² C²/N m²)

5 × 10⁵ N m²/C
1.77 × 10⁴ N m²/C
4.42 × 10⁻⁶ N m²/C
5.65 × 10⁵ N m²/C

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Correct answer: D. 5.65 × 10⁵ N m²/C

1. By Gauss's law, Φ = Q_enc / ε₀. 2. Substitute: Φ = 5 × 10⁻⁶ / 8.85 × 10⁻¹². 3. = 5.65 × 10⁵ N·m²/C. 4. Result is independent of the shape of the closed surface. _Source: syllabus Physics, Ch 1 "Electric Charges and Fields", §1.9_

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For a charge distribution with volume density ρ, the total charge Q equals

∫∫∫ ρ dV
∫ ρ dr
∫∫ ρ dA
ρ × surface area

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Correct answer: A. ∫∫∫ ρ dV

1. Volume charge density ρ has units C/m³. 2. Total charge is obtained by integrating ρ over the entire volume. 3. Q = ∫∫∫ ρ dV. 4. For a uniform distribution ρ × V; for varying ρ the full integral is needed. _Source: syllabus Physics, Ch 1 "Electric Charges and Fields", §1.14_

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