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Current Question (ID: 16469)

Question:
$\text{A point charge } q \text{ is placed at a distance } \frac{a}{2} \text{ directly above the centre of a square of side } a. \text{ The electric flux through the square (i.e., one face) is:}$
Options:
  • 1. $\frac{q}{\varepsilon_0}$
  • 2. $\frac{q}{\pi \varepsilon_0}$
  • 3. $\frac{q}{4 \varepsilon_0}$
  • 4. $\frac{q}{6 \varepsilon_0}$
Solution:
$\text{Hint: Use Gauss's law. Imagine a cubic Gaussian surface.}$ $\text{Step 1: Imagine a cubic Gaussian surface enclosing the charge.}$ $\text{An imaginary cube can be made by considering charge } q \text{ at the centre and given square is one of its faces.}$ $\text{Step 2: Apply Gauss's law and hence by symmetry, find the flux through one face of the cube (given square).}$ $\text{So flux from the given square (i.e. one face) } \varphi = \frac{q}{6 \varepsilon_0}$

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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
}