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Current Question (ID: 20038)

Question:

$\text{Find the electric field at point } P \text{ (as shown in the figure) on the perpendicular bisector of a uniformly charged thin wire of length } L \text{ carrying a charge } Q. \text{ The distance of point } P \text{ from the centre of the rod is } a = \frac{\sqrt{3}}{2}L.$

Options:

1. $\frac{\sqrt{3}Q}{4\pi \varepsilon_0 L^2}$
2. $\frac{Q}{3\pi \varepsilon_0 L^2}$
3. $\frac{2\sqrt{3}\pi \varepsilon_0 Q}{L^2}$
4. $\frac{Q}{4\pi \varepsilon_0 L^2}$

Solution:

$\textbf{Hint: } E = \frac{k\lambda}{a}(\sin \theta_1 + \sin \theta_2)$ $E = \frac{k\lambda}{a}(\sin \theta_1 + \sin \theta_2)$ $E = \frac{1}{4\pi \varepsilon_0} \times \frac{Q}{L} \times \frac{1}{\frac{\sqrt{3}L}{2}} \times \left(2 \sin \theta \right)$ $\tan \theta = \frac{\frac{L}{2}}{\frac{\sqrt{3}L}{2}} = \frac{1}{\sqrt{3}}$ $\sin \theta = \frac{1}{2}$ $E = \frac{1}{4\pi \varepsilon_0} \times \frac{2Q}{\sqrt{3}L^2} \times \left(2 \times \frac{1}{2} \right)$ $E = \frac{Q}{2\sqrt{3}\pi \varepsilon_0 L^2}$

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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
}