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Current Question (ID: 16405)

Question:

$\text{An infinite number of charges, each of charge } 1 \, \mu\text{C}, \text{ are placed on the } x\text{-axis}$ $\text{with co-ordinates } x = 1, 2, 4, 8, \ldots \infty. \text{ If a charge of } 1 \, \text{C is kept at the}$ $\text{origin, then what is the net force acting on } 1 \, \text{C charge?}$

Options:

1. $9000 \, \text{N}$
2. $12000 \, \text{N}$
3. $24000 \, \text{N}$
4. $36000 \, \text{N}$

Solution:

$\text{Recall the formula for the sum of infinite geometric progression.}$ $\text{Write the summation of forces on the } 1 \, \text{C charge due to all other charges.}$ $\text{The total force acting on } 1 \, \text{C charge is given by:}$ $F = \frac{1}{4\pi\varepsilon_0} \left( \frac{1 \times 1 \times 10^{-6}}{(1)^2} + \frac{1 \times 1 \times 10^{-6}}{(2)^2} + \frac{1 \times 1 \times 10^{-6}}{(4)^2} + \frac{1 \times 1 \times 10^{-6}}{(8)^2} + \ldots \right)$ $\text{Use the formula for the summation of infinite geometric progression to calculate the net force.}$ $F = \frac{10^{-6}}{4\pi\varepsilon_0} \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots \right) = 9 \times 10^9 \times 10^{-6} \left( \frac{1}{1 - \frac{1}{4}} \right)$ $F = 9 \times 10^9 \times 10^{-6} \times \frac{4}{3} = 9 \times 10^3 \times \frac{4}{3} = 12000 \, \text{N}$ $\text{Hence, option (2) is the correct answer.}$

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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
}