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Current Question (ID: 16438)

Question:

$\text{Three-point charges } +q, -2q \text{ and } +q \text{ are placed at points}$ $\text{(} x = 0, y = a, z = 0 \text{)}, \text{(} x = 0, y = 0, z = 0 \text{)} \text{ and } \text{(} x = a, y = 0, z = 0 \text{)},$ $\text{respectively. The magnitude and direction of the electric dipole moment vector}$ $\text{of this charge assembly are:}$

Options:

1. $\sqrt{2}qa \text{ along } +y \text{ direction}$
2. $\sqrt{2}qa \text{ along the line joining points } (x = 0, y = 0, z = 0) \text{ and }$ $\text{(} x = a, y = a, z = 0 \text{)}$
3. $qa \text{ along the line joining points } (x = 0, y = 0, z = 0) \text{ and }$ $\text{(} x = a, y = a, z = 0 \text{)}$
4. $\sqrt{2}qa \text{ along } -y \text{ direction}$

Solution:

$\text{Given,}$ $\text{Three point charges } +q, -2q, +q \text{ are placed at } A(0,a,0), B(0,0,0) \text{ and } C(a,0,0)$ $\text{respectively.}$ $\text{We know,}$ $\text{Electric dipole moment is a vector quantity directed from the negative}$ $\text{charge to the positive charge.}$ $\text{Now, choose the three coordinate axes as } x, y \text{ and } z \text{ and plot the charges}$ $\text{with the given coordinates as shown.}$ $O \text{ is the origin at which } -2q \text{ charge is placed. The system is equivalent to two}$ $\text{dipoles along with } x \text{ and } y \text{ directions respectively. The dipole moments of}$ $\text{two dipoles are shown in the figure.}$ $\text{The resultant dipole moment will be directed along with OP where } P \equiv (a, a, 0).$ $\text{The magnitude of the resultant dipole moment is:}$ $p' = \sqrt{p^2 + p^2}$ $= \sqrt{(qa)^2 + (qa)^2}$ $= \sqrt{2} \, qa$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}