Import Question JSON

Current Question (ID: 20031)

Question:

$\text{A cube of side } a \text{ has point charges } +Q \text{ located at each of its vertices}$ $\text{except at the origin where the charge is } -Q. \text{ The electric field at the}$ $\text{centre of cube is:}$

Options:

1. $\frac{-Q}{3\sqrt{3}\pi\varepsilon_0 a^2} \left( \hat{x} + \hat{y} + \hat{z} \right)$
2. $\frac{-2Q}{3\sqrt{3}\pi\varepsilon_0 a^2} \left( \hat{x} + \hat{y} + \hat{z} \right)$
3. $\frac{2Q}{3\sqrt{3}\pi\varepsilon_0 a^2} \left( \hat{x} + \hat{y} + \hat{z} \right)$
4. $\frac{Q}{3\sqrt{3}\pi\varepsilon_0 a^2} \left( \hat{x} + \hat{y} + \hat{z} \right)$

Solution:

$\text{Hint: We can replace } -Q \text{ charge at the origin by } +Q \text{ and } -2Q.$ $\text{Now due to } +Q \text{ charge at every corner of}$ $\text{cube. Electric field at center of cube is zero so now net electric field}$ $\text{at center is only due to } -2Q \text{ charge at origin.}$ $\vec{E} = \frac{kq \vec{r}}{r^3} = \frac{1}{4\pi\varepsilon_0} \frac{(-2Q) \frac{a}{2} (\hat{x}+\hat{y}+\hat{z})}{\left(\frac{a}{2}\sqrt{3}\right)^3}$ $\vec{E} = \frac{-2Q)(\hat{x}+\hat{y}+\hat{z})}{3\sqrt{3}\pi a^2 \varepsilon_0}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}