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Current Question (ID: 16424)

Question:

$\text{Two-point charges } +8q \text{ and } -2q \text{ are located at } x = 0 \text{ and } x = L \text{ respectively. The location of a point on the } x\text{-axis at which the net electric field due to these two point charges is zero is:}$

Options:

1. $8L$
2. $4L$
3. $2L$
4. $\frac{L}{4}$

Solution:

$\text{Let } P \text{ be the observation point at a distance } r \text{ from } -2q \text{ and at } (L + r) \text{ from } +8q.$ $\text{Given Now, the net electric field intensity at } P = 0$ $\therefore \vec{E}_1 = \text{(Electric Field Intensity) at } P \text{ due to } +8q$ $\vec{E}_2 = \text{(Electric Field Intensity) at } P \text{ due to } -2q$ $|\vec{E}_1| = |\vec{E}_2|$ $\frac{k(8q)}{(L+r)^2} = \frac{k(2q)}{r^2}$ $\Rightarrow \frac{4}{(L+r)^2} = \frac{1}{(r)^2}$ $4r^2 = (L + r)^2$ $\Rightarrow 2r = L + r$ $\Rightarrow r = L$ $\text{So the distance of } P \text{ is at } \Rightarrow x = L + L = 2L \text{ from origin.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}