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NEET UG Electric Charges and Fields — practice questions — page 7

Explore 189 free Electric Charges and Fields questions in NEET UG 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 181–189 of 189.

Four equal point charges q sit at the corners of a square of side d. The work needed to assemble this configuration from infinity isRead the answer and worked solution →An electric dipole of moment p is placed in a uniform field E. The work done to rotate it from the orientation θ = 0 (aligned with the field) to θ = 180° isRead the answer and worked solution →Electric field at any point inside a hollow charged conductor isRead the answer and worked solution →On an equipotential surface, the work done in moving a test charge between any two points isRead the answer and worked solution →A parallel-plate capacitor in vacuum has plate area A and separation d. Its capacitance isRead the answer and worked solution →The work done in bringing a charge q from infinity to a point where the potential is V isRead the answer and worked solution →True or false: electrostaticsRead the answer and worked solution →Electric potential is a:Read the answer and worked solution →NEET · Physics · Electric Charges & Fields · Force on Dipole in FieldRead the answer and worked solution →

Worked practice questions

Four equal point charges q sit at the corners of a square of side d. The work needed to assemble this configuration from infinity is

  1. (kq²/d)(2 + 1/√2)
  2. (kq²/d)(4 + √2)
  3. (kq²/d)(4 + 2/√2)
  4. (kq²/d)(1 + 1/√2)
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Correct answer: C. (kq²/d)(4 + 2/√2)

1. Pairs: 4 sides at distance d (4 pairs) and 2 diagonals at distance d√2 (2 pairs). 2. Total PE = k q²[(4/d) + (2/(d√2))] = (k q²/d)(4 + 2/√2). 3. The work done from infinity equals this PE since the field at infinity does no work. _Source: syllabus Physics Part I, Ch 2 Example 2.4_

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An electric dipole of moment p is placed in a uniform field E. The work done to rotate it from the orientation θ = 0 (aligned with the field) to θ = 180° is

  1. 2pE
  2. 0
  3. −2pE
  4. pE
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Correct answer: A. 2pE

1. PE in a uniform field: U(θ) = −p·E = −pE cos θ. 2. W = ΔU = U(180°) − U(0°) = (+pE) − (−pE) = 2pE. 3. The dipole goes from the most stable to the least stable orientation; work is positive on the system. _Source: syllabus Physics Part I, Ch 2 §2.8.3_

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Electric field at any point inside a hollow charged conductor is

  1. kq/r²
  2. zero
  3. the same as at the surface
  4. directed radially inward
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Correct answer: B. zero

1. syllabus result 1 of §2.9: inside a conductor in electrostatics, E = 0. 2. Any free charges have already redistributed so the interior field is null. 3. (A) would be true for a point charge in vacuum, not for the interior of a conductor. _Source: syllabus Physics Part I, Ch 2 §2.9 result 1_

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