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) √(2)qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0) (b) q a along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0) (c) √(2)qa along +x direction (d) √(2)qa along +y direction.

Answer: (a). Split -2q at the origin into two -q charges, forming two dipoles. +q at (0, a, 0) with -q at origin: p₁ = qa j +q at (a, 0, 0) with -q at origin: p₂ = qa i p = qa(i + j), |p| = √(2)qa, along the line from (0,0,0) to (a, a, 0). So the answer is (a).

Three point charges +q, - 2q and + q are placed at points (x = 0, y =… | NEET 2007 (AIPMT)

Q49NEET 2007 · AIPMTMedium

Electric Field & Field Lines

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

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Answer: (a)

Split $-2q$ at the origin into two $-q$ charges, forming two dipoles.
$+q$ at $(0, a, 0)$ with $-q$ at origin: $\vec{p}_1 = qa\,\hat{j}$
$+q$ at $(a, 0, 0)$ with $-q$ at origin: $\vec{p}_2 = qa\,\hat{i}$
$\vec{p} = qa(\hat{i} + \hat{j})$ , $|\vec{p}| = \sqrt{2}qa$ , along the line from $(0,0,0)$ to $(a, a, 0)$ .
So the answer is (a).

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