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Current Question (ID: 16585)

Question:
$\text{A parallel plate condenser has a uniform electric field } E (\text{V/m}) \text{ in the space between the plates. If the distance between the plates is } d (\text{m}) \text{ and area of each plate is } A (\text{m}^2), \text{ the energy (joule) stored in the condenser is:}$
Options:
  • 1. $\frac{1}{2} \varepsilon_0 E^2$
  • 2. $\varepsilon_0 E A d$
  • 3. $\frac{1}{2} \varepsilon_0 E^2 A d$
  • 4. $\frac{E^2 A d}{\varepsilon_0}$
Solution:
$\text{The energy stored in the condenser}$ $U = \frac{1}{2} C V^2$ $U = \frac{1}{2} \left( \frac{A \varepsilon_0}{d} \right) (E d)^2 \text{ (so,)} C = \frac{A \varepsilon_0}{d} \text{ and } V = E d$ $U = \frac{1}{2} \varepsilon_0 E^2 A d$

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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
}