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Current Question (ID: 16448)

Question:

$\text{Two point dipoles of dipole moment } \vec{p}_1 \text{ and } \vec{p}_2 \text{ are at a distance } x \text{ from each other and } \vec{p}_1 \parallel \vec{p}_2. \text{ The force between the dipole is:}$

Options:

1. $\frac{1}{4\pi\varepsilon_0} \frac{4p_1p_2}{x^4}$
2. $\frac{1}{4\pi\varepsilon_0} \frac{3p_1p_2}{x^3}$
3. $\frac{1}{4\pi\varepsilon_0} \frac{6p_1p_2}{x^4}$
4. $\frac{1}{4\pi\varepsilon_0} \frac{8p_1p_2}{x^4}$

Solution:

$\text{Force on } \vec{p}_2 \text{ due to } \vec{p}_1 \text{ is}$ $\vec{F} = \vec{F}_{-q} + \vec{F}_{+q} = -q_2 \vec{E}_A + q_2 \vec{E}_B = -q_2 \frac{2kp_1}{\left(x - \frac{d}{2}\right)^3} \hat{i} + q_2 \frac{2kp_1}{\left(x + \frac{d}{2}\right)^3} \hat{i}$ $= -2kp_1 q_2 \left[\frac{\left(x - \frac{d}{2}\right)^2 + \left(x + \frac{d}{2}\right)^2}{\left(x^2 - \frac{d^2}{4}\right)^3}\right] \hat{i}$ $\text{As } x \gg \frac{d}{2}$ $\text{so, } \left(x - \frac{d}{2}\right) \approx x$ $\therefore \vec{F} = -\frac{2kp_1 q_2 d \left[x^2 + x^2 + x^2\right]}{x^6} \hat{i} = -\frac{6kp_1 p_2}{x^4} \hat{i}$ $= -\frac{1}{4\pi\varepsilon_0} \frac{6p_1p_2}{x^4} \hat{i} \quad \text{(As) } p_2 = q_2 d$

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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
}