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Current Question (ID: 16472)

Question:

$\text{Consider a region inside where there are various types of charges but the total charge is zero. At points outside the region:}$ $\text{(a) the electric field is necessarily zero.}$ $\text{(b) the electric field is due to the dipole moment of the charge distribution only.}$ $\text{(c) the dominant electric field is } \propto \frac{1}{r^3}, \text{ for large } r, \text{ where } r \text{ is the distance from the origin in this region.}$ $\text{(d) the work done to move a charged particle along a closed path, away from the region, will be zero.}$ $\text{Which of the above statements are true?}$

Options:

1. $\text{(b) and (d)}$
2. $\text{(a) and (c)}$
3. $\text{(b) and (c)}$
4. $\text{(c) and (d)}$

Solution:

$\text{(4) Hint: Use Gauss law.}$ $\text{Step 1: Interpret the electric field at a large distance.}$ $\text{When there are various types of charges in a region, but the total charge is zero, the region can be supposed to contain a number of electric dipoles.}$ $\text{Therefore, at points outside the region (may be anywhere w.r.t. electric dipoles), the dominant electric field } \propto \frac{1}{r^3}, \text{ for large } r.$ $\text{Step 2: Find the work done in a closed path.}$ $\text{Further, as electric field is conservative, work done to move a charged particle along a closed path away from the region will be zero.}$

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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
}