Import Question JSON

Current Question (ID: 16400)

Question:

$\text{Two charged particles } P \text{ and } Q \text{ are } 0.10 \text{ m apart. The charge on } P \text{ is } 1.50 \times 10^{-7} \text{ C and the charge on } Q \text{ is } 1.50 \times 10^{-7} \text{ C.}$ $\text{Particle } P \text{ experiences an electrostatic force of magnitude } F \text{ because it is near to the charge on particle } Q.$ $\text{The distance between the two particles is increased to } 0.20 \text{ m.}$ $\text{The charge on } P \text{ increases to } 4.50 \times 10^{-7} \text{ C and the charge on } Q \text{ increases to } 6.00 \times 10^{-7} \text{ C.}$ $\text{What is the magnitude of the force that particle } P \text{ experiences now?}$

Options:

1. $\frac{F}{4}$
2. $12F$
3. $6F$
4. $3F$

Solution:

$\text{Hint: } F = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r^2}$ $\text{Step 1: Find the initial force between the charges.}$ $F = \frac{1}{4\pi\varepsilon_0} \frac{1.5 \times 10^{-7} \times 1.5 \times 10^{-7}}{(0.10)^2} = 2.25k \times 10^{-12} \text{ N} \quad \cdots (1)$ $\text{Step 2: Find the final force between the charges.}$ $F' = \frac{1}{4\pi\varepsilon_0} \frac{4.5 \times 10^{-7} \times 6.0 \times 10^{-7}}{(0.20)^2} = 6.75k \times 10^{-12} \text{ N} \quad \cdots (2)$ $\text{From the equation } (1) \text{ and } (2) \text{ we get;}$ $\Rightarrow F' = 3F$ $\text{Hence, option } (4) \text{ is the correct answer.}$

Import JSON File

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
}