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Current Question (ID: 16401)

Question:

$\text{The force between two charges } 0.06 \text{ m apart is } 5 \text{ N. If each charge is moved}$ $\text{towards the other by } 0.01 \text{ m, then the force between them will become:}$

Options:

1. $7.20 \text{ N}$
2. $11.25 \text{ N}$
3. $22.50 \text{ N}$
4. $45.00 \text{ N}$

Solution:

$\text{Hint: Electrostatic Force between two charges is calculated by Coulomb's}$ $\text{law which states that the magnitude of the electrostatic force between two}$ $\text{point charges is directly proportional to the product of the magnitude of}$ $\text{charges and inversely proportional to the distance between them.}$ $\text{Step 1: Write the required formula to solve this question.}$ $\text{i.e. } \mathbf{F} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 \times q_2}{r^2}$ $\text{Step 2: Use the relation to find the required force i.e. electrostatic force}$ $\text{between two point charges is inversely proportional to the distance between}$ $\text{them.}$ $\mathbf{F} \propto \frac{1}{r^2} \Rightarrow \frac{F_1}{F_2} = \left( \frac{r_2}{r_1} \right)^2$ $\Rightarrow \frac{5}{F_2} = \left( \frac{0.04}{0.06} \right)^2 \Rightarrow F_2 = 11.25 \text{ N}$ $\text{(Caution: If each charge is moved}$ $\text{towards the other by } 0.01 \text{ m, then the distance between two charges}$ $\text{becomes } (0.06 - 0.02) = 0.04 \text{ m.)}$

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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
}