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Current Question (ID: 16468)

Question:

$\text{Two parallel infinite line charges with linear charge densities } +\lambda \text{ C/m and } +\lambda \text{ C/m are placed at a distance } R. \text{ The electric field mid-way between the two line charges is:}$

Options:

1. $\frac{\lambda}{2\pi\varepsilon_0 R} \text{ N/C}$
2. $\text{zero}$
3. $\frac{2\lambda}{\pi\varepsilon_0 R} \text{ N/C}$
4. $\frac{\lambda}{\varepsilon_0 R} \text{ N/C}$

Solution:

$\text{Hint: } E = \frac{\lambda}{2\pi\varepsilon_0 r}$ $\text{Step: Find the electric field mid-way between the two line charges.}$ $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ $\vec{E} = \vec{E}_1 - \vec{E}_2$ $\vec{E} = \frac{2K\lambda}{R/2} - \frac{2K\lambda}{R/2}$ $\vec{E} = \frac{\lambda}{\pi\varepsilon_0 R} - \frac{\lambda}{\pi\varepsilon_0 R}$ $\Rightarrow \vec{E} = 0$ $\text{Hence, option (2) is the correct answer.}$

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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
}