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Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]
  • a)
    √2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
  • b)
    qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
  • c)
    √2qa along +ve x direction
  • d)
    √2qa along +ve y direction
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Three point charges +q, –q and +q are placed at points (x = 0, y...
Three point charges +q, –2q and +q are placed
at points B (x = 0, y = a, z = 0),
O (x = 0, y = 0, z = 0) and A(x = a, y = 0, z = 0)
The system consists of two dipole moment
vectors due to (+q and –q) and again due to
(+q and –q) charges having equal
magnitudes qa units – one along 
 and other along   Hence, net dipole moment   along  at an angle 45° with positive X-axis. 
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Three point charges +q, –q and +q are placed at points (x = 0, y...
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Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer?
Question Description
Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer?.
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Here you can find the meaning of Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Three point charges +q, –q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are [2007]a)√2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)b)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)c)√2qa along +ve x directiond)√2qa along +ve y directionCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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