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Current Question (ID: 16538)

Question:

$\text{The electric potential difference between two parallel plates is } 2000 \text{ V. If the plates are separated by } 2 \text{ mm, what is the magnitude of the electrostatic force on a charge of } 4 \times 10^{-6} \text{ C located midway between the plates?}$

Options:

1. $4 \text{ N}$
2. $6 \text{ N}$
3. $8 \text{ N}$
4. $1.5 \times 10^{-6} \text{ N}$

Solution:

$\text{Hint: } \left| \vec{E} \right| = \frac{V}{d}$ $\text{Step 1: Find the electric field between plates.}$ $E = \frac{\Delta v}{d} = \frac{2000}{2 \times 10^{-3}} = 10^6 \text{ N/C}$ $\text{Step 2: Find the electric force on the charge.}$ $F_e = qE = 4 \times 10^{-6} \times 10^6 = 4 \text{ N}$

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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
}