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Current Question (ID: 20030)

Question:

$\text{Consider the force } F \text{ on a charge } q \text{ due to a uniformly charged}$ $\text{spherical shell of radius } R \text{ carrying charge } Q \text{ distributed uniformly}$ $\text{over it. Which one of the following statements is true for } F, \text{ if } q \text{ is}$ $\text{placed at a distance } r \text{ from the centre of the shell?}$

Options:

1. $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \text{ for } r > R$
2. $\frac{1}{4\pi\varepsilon_0} \frac{Qq}{R^2} > F > 0 \text{ for } r < R$
3. $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \text{ for all } r$
4. $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \text{ for } r < R$

Solution:

$\text{Hint: The electric field inside the spherical shell is zero.}$ $\text{Step: Find the value of the force inside and outside the spherical shell.}$ $\text{Inside the spherical shell, the electric field is zero and electrostatic}$ $\text{force is given by; } F = qE$ $\text{Therefore, for } r < R, F = 0$ $\text{Outside the spherical shell, the electric field is given by;}$ $E = \frac{kQ}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$ $\text{Therefore, the electrostatic force is given by; } F = qE$ $\Rightarrow F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \text{ for } r > R$ $\text{Hence, option (1) is the correct answer.}$

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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
}