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Current Question (ID: 16508)

Question:

$\text{Consider a uniform electric field in the } z\text{-direction. The potential is constant:}$ $\text{a. in all space}$ $\text{b. for any } x \text{ for a given } z$ $\text{c. for any } y \text{ for a given } z$ $\text{d. on the } x-y \text{ plane for a given } z$

Options:

1. $\text{(a), (b), (c)}$
2. $\text{(a), (c), (d)}$
3. $\text{(b), (c), (d)}$
4. $\text{(c), (d)}$

Solution:

$\text{Hint: The electric field is always perpendicular to the equipotential surfaces.}$ $\text{Step 1: Find the direction of the electric field.}$ $\text{Here, the electric field always remains in the direction in which the potential decreases steepest. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.}$ $\text{Step 2: Find the equipotential surfaces.}$ $\text{The electric field in the } z\text{-direction suggests that equipotential surfaces are in the } x-y \text{ plane. Therefore, the potential is a constant for any } x \text{ for a given } z, \text{ for any } y \text{ for a given } z \text{ and, on the } x-y \text{ plane for a given } z.$ $\text{Note: The shape of equipotential surfaces depends on the nature and type of distribution of charge e.g., point charge leads to produce spherical surfaces whereas line charge distribution produces cylindrical equipotential surfaces.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}