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Three point charges +q, –2q 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
  • 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 + x direction
  • d)
    √2qa along + y direction
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Three point charges +q, –2q and +q are placed at points (x = 0, ...
This consists of two dipoles, –q and +q with dipole moment along with the +y-direction and –q and +q along the x-direction.


Along the direction 45° that is along OP, where P is (+a, +a, 0).
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Most Upvoted Answer
Three point charges +q, –2q and +q are placed at points (x = 0, ...
R, and s are placed at the vertices of an equilateral triangle. The distance between each charge is equal to d. The total electric potential at the center of the triangle is given by the equation:

V = k * (q/r + q/s + r/q + r/s + s/q + s/r)

where k is the Coulomb's constant.

To simplify the equation, we can note that the charges q, r, and s are all the same distance away from the center of the triangle, so we can rewrite the equation as:

V = k * (2q/d + 2r/d + 2s/d)

Factoring out the common factor of 2/d, we get:

V = 2k/d * (q + r + s)

Since the charges q, r, and s are all placed at the vertices of an equilateral triangle, their magnitudes are equal, so q = r = s.

Substituting this into the equation, we get:

V = 2k/d * (3q)

Simplifying further, we get:

V = 6kq/d

Therefore, the total electric potential at the center of the equilateral triangle is equal to 6 times the product of the Coulomb's constant, the charge magnitude, and the distance between the charges.
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Three point charges +q, –2q 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 area)√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 + x directiond)√2qa along + y directionCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Three point charges +q, –2q 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 area)√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 + x directiond)√2qa along + y directionCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three point charges +q, –2q 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 area)√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 + x directiond)√2qa along + y directionCorrect answer is option 'A'. Can you explain this answer?.
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