Import Question JSON

Current Question (ID: 16572)

Question:

$\text{Two thin dielectric slabs of dielectric constants } K_1 \text{ and } K_2 (K_1 < K_2) \text{ are inserted between plates of a parallel capacitor, as shown in the figure.}$ $\text{The variation of the electric field } E \text{ between the plates with distance } d \text{ as measured from the plate } P \text{ is correctly shown by:}$

Options:

1. $\text{Graph 1}$
2. $\text{Graph 2}$
3. $\text{Graph 3}$
4. $\text{Graph 4}$

Solution:

$\text{Hint: The net electric field inside a dielectric will be less than the external electric field.}$ $\text{Step: Find the electric field inside the two dielectrics.}$ $\text{The electric field } \vec{E} \text{ inside the dielectric is given by:}$ $\vec{E} = \frac{\vec{E}_0}{K}$ $\text{For the dielectric } K_1 \text{ the electric field } \vec{E}_1 \text{ is given by:}$ $\vec{E}_1 = \frac{\vec{E}_0}{K_1}$ $\text{For the dielectric } K_2 \text{ the electric field } \vec{E}_2 \text{ is given by:}$ $\vec{E}_2 = \frac{\vec{E}_0}{K_2}$ $\text{The correct representation of the electric field is given by:}$ $\text{Since, } K_2 > K_1; E_2 < E_1 < E_0$ $\text{the graph (3) will be the right graph, of the electric field inside the dielectrics will be less than the electric field outside the dielectrics.}$ $\text{The electric field inside the dielectric could not be zero.}$ $\text{As, } K_2 > K_1 \text{ the drop in electric field for } K_2 \text{ dielectric must be more than } K_1.$ $\text{Hence, option (3) is the correct answer.}$

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
}