Import Question JSON

Current Question (ID: 16422)

Question:
$\text{The electrostatic field due to a charged conductor just outside the conductor is:}$
Options:
  • 1. $\text{zero and parallel to the surface at every point inside the conductor.}$
  • 2. $\text{zero and is normal to the surface at every point inside the conductor.}$
  • 3. $\text{parallel to the surface at every point and zero inside the conductor.}$
  • 4. $\text{normal to the surface at every point and zero inside the conductor.}$
Solution:
$\text{Hint: The electric field is zero inside a conductor. Just outside a conductor, the electric field lines are perpendicular to its surface.}$ $\text{Explanation: The electric field inside a conductor is always zero in electrostatic equilibrium. This is because charges redistribute themselves on the surface of the conductor to cancel any internal field.}$ $\text{The electric field just outside the conductor is perpendicular (normal) to the surface at every point. This is due to the fact that surface charges on a conductor create an electric field that is directed outward (or inward for negative charges), normal to the surface.}$ $\text{At the surface of the conductor, the electric field has no parallel component because any parallel component would cause the charges to move along the surface. In electrostatic equilibrium, charges have stopped moving, so the field must be purely normal.}$ $\text{Therefore the field is normal to the surface at every point and zero inside the conductor.}$ $\text{Hence, option (4) is the correct answer.}$

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
}