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Current Question (ID: 16517)

Question:

$\text{The electric potential at a point in free space due to a charge } Q \text{ coulomb is } Q \times 10^{11} \text{ V. The electric field at that point is:}$

Options:

1. $4\pi\varepsilon_0 Q \times 10^{22} \text{ V/m}$
2. $12\pi\varepsilon_0 Q \times 10^{20} \text{ V/m}$
3. $4\pi\varepsilon_0 Q \times 10^{20} \text{ V/m}$
4. $12\pi\varepsilon_0 Q \times 10^{22} \text{ V/m}$

Solution:

$\text{Hint: Use } V = \frac{kQ}{R} = \frac{Q}{4\pi\varepsilon_0 R}$ $\text{Step: Find the electric field at the given point.}$ $\text{The relation between the electric potential and electric field at a point is:}$ $E = \frac{V}{R} \text{ where } E \text{ is the electric field, } V \text{ is the electric potential at a point}$ $\text{and } R \text{ is the distance of the point of observation from the charge.}$ $\text{At any point, the electric potential due to a charge } Q \text{ is:}$ $V = \frac{Q}{4\pi\varepsilon_0 R} \quad \cdots (1)$ $\text{At the same point, the Electric field } E \text{ is:}$ $E = \frac{Q}{4\pi\varepsilon_0 R^2} \quad \cdots (2)$ $\text{From equations (1) and (2), we get:}$ $\Rightarrow E = \frac{4\pi\varepsilon_0 Q}{V^2}$ $\Rightarrow E = \frac{4\pi\varepsilon_0 Q}{(Q \times 10^{11})^2} = 4\pi\varepsilon_0 Q \times 10^{22} \text{ V/m.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}