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Current Question (ID: 16540)

Question:

$\text{A parallel plate air capacitor has a capacity of } C, \text{ the distance of separation between plates is } d \text{ and potential difference } V \text{ is applied between the plates.}$ $\text{The force of attraction between the plates of the parallel plate air capacitor is:}$

Options:

1. $\frac{C^2V^2}{2d}$
2. $\frac{CV^2}{2d}$
3. $\frac{CV^2}{d}$
4. $\frac{C^2V^2}{2d^2}$

Solution:

$\text{Force between plates of parallel capacitor,}$ $F = qE = q \left[ \frac{\sigma}{2\varepsilon_0} \right]$ $\therefore \text{Surface charge density } \sigma = \frac{q}{A}$ $\therefore F = q \left[ \frac{q}{2A\varepsilon_0} \right] \Rightarrow F = \frac{q^2}{2A\varepsilon_0}$ $\text{So, the net charge across a capacitor, } q = CV$ $F = \frac{C^2V^2}{2A\varepsilon_0} \times \left[ \frac{A\varepsilon_0}{d} \right]$ $\Rightarrow F = \frac{C^2V^2}{2d}$

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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
}