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Current Question (ID: 16458)

Question:
$\text{A square surface of a side } L \text{ (m) is in the plane of the paper. A uniform electric field } \vec{E} \text{ (V/m), also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is:}$
Options:
  • 1. $\frac{EL^2}{2\varepsilon_0}$
  • 2. $\frac{EL^2}{2}$
  • 3. $\text{zero}$
  • 4. $EL^2$
Solution:
$\text{Hint: } \phi_E = \int \vec{E} \cdot d\vec{S}$ $\text{Step: Find the electric flux in SI units associated with the surface.}$ $\text{The electric flux is given by:}$ $\text{Electric flux, } \phi_E = \int \vec{E} \cdot d\vec{S}$ $\phi_E = \int E dS \cos \theta$ $\text{The electric flux (} \phi_E \text{) is a measure of the number of field lines crossing a surface. In the given case, the field lines are parallel to the surface, i.e., the angle between the area vector and the field is } 90^\circ. \text{ Hence, the electric flux crossing the surface will be zero.}$ $\phi_E = \int E dS \cos 90^\circ = 0.$ $\text{Hence, option (3) is the correct answer.}$

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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
}