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Current Question (ID: 20101)

Question:

$\text{A parallel plate capacitor is made of two circular plates separated by a distance of } 5 \text{ mm and with a dielectric of dielectric constant } 2.2 \text{ between them.}$ $\text{When the electric field in the dielectric is } 3 \times 10^4 \text{ V/m, the charge density of the positive plate will be close to:}$

Options:

1. $3 \times 10^{-7} \text{ C/m}^2$
2. $3 \times 10^4 \text{ C/m}^2$
3. $6 \times 10^4 \text{ C/m}^2$
4. $6 \times 10^{-7} \text{ C/m}^2$

Solution:

$\text{Hint: } q = C \Delta V$ $E = \frac{\Delta V}{d}$ $\Delta V = Ed$ $q = C \Delta V$ $= \frac{\varepsilon_0 KA}{d} Ed$ $\sigma = \frac{q}{A} = \varepsilon_0 KE$ $= 8.85 \times 10^{-12} \times 2.2 \times 3 \times 10^4$ $= 6 \times 10^{-7} \text{ C/m}^2$

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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
}