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Current Question (ID: 20168)

Question:

$\text{Voltage rating of a parallel plate capacitor is } 500 \text{ V. Its dielectric can withstand a maximum electric field of } 10^6 \text{ V/m. The plate area is } 10^{-4} \text{ m}^2. \text{ What is the dielectric constant if the capacitance is } 15 \text{ pF }? \text{ (given } \varepsilon_0 = 8.86 \times 10^{-12} \text{ C}^2/\text{Nm}^2)$

Options:

1. 6.2
2. 3.8
3. 4.5
4. 8.5

Solution:

$\text{Hint: } C = \frac{A \varepsilon_0 K}{d} \text{ and } V = Ed$ $C = \frac{A \varepsilon_0 K}{d} \quad \& \quad V = Ed$ $K = \frac{CV}{A \varepsilon_0 E} = \frac{(15 \times 10^{-12})(500)}{10^{-4} (8.86 \times 10^{-12})(10^6)} = 8.5$

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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
}