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Current Question (ID: 20047)

Question:

$\text{Two charges, } q_1 \text{ and } q_2, \text{ separated by a distance } d, \text{ are placed in a medium with a dielectric constant } K. \text{ If these charges are placed in the air, the equivalent separation at which they experience the same force is:}$

Options:

1. $d\sqrt{K}$
2. $\frac{K}{\sqrt{d}}$
3. $\frac{2d}{\sqrt{K}}$
4. $\frac{1.5d}{\sqrt{K}}$

Solution:

$\text{Hint: } F = \frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}$ $\text{Step 1: Find the force when charges are placed in the air.}$ $\text{The force between two charges separated by a distance } r_{eq} \text{ placed in the air is given by:}$ $F = \frac{q_1 q_2}{4 \pi \varepsilon_0 r_{eq}^2}$ $\text{Step 2: Find the force when charges are placed in the dielectric medium.}$ $\text{The force between two charges separated by a distance } d \text{ placed in a dielectric medium of dielectric constant } K \text{ is given by:}$ $F' = \frac{q_1 q_2}{4 \pi K \varepsilon_0 d^2}$ $\text{Step 3: Find the equivalent distance at which charges experience the same force.}$ $\text{The equivalent distance } (r_{eq}) \text{ at which they experience the same force is:}$ $\frac{q_1 q_2}{4 \pi \varepsilon_0 K d^2} = \frac{q_1 q_2}{4 \pi \varepsilon_0 r_{eq}^2}$ $\Rightarrow r_{eq}^2 = K d^2$ $\Rightarrow r_{eq} = d\sqrt{K}$ $\text{Therefore, the equivalent distance } (r_{eq}) \text{ is } d\sqrt{K}. \text{ Hence, option (1) is the correct answer.}$

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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
}