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Current Question (ID: 16587)

Question:

$\text{The energy and capacity of a charged parallel plate capacitor are } E \text{ and } C \text{ respectively. If a dielectric slab of } E_r = 6 \text{ is inserted in it, then the energy and capacity become:}$ $\text{(Assuming the charge on plates remains constant)}$

Options:

1. $6E, 6C$
2. $E, C$
3. $\frac{E}{6}, 6C$
4. $E, 6C$

Solution:

$C_{\text{PPC}} = \frac{\varepsilon_0 \varepsilon_r A}{d} = C' = 6C$ $E_{\text{PPC}} = \frac{q^2}{2 \varepsilon_0 \varepsilon_r A} = E' = \frac{E}{6}$

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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
}