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Current Question (ID: 16470)

Question:

$X$ and $Y$ are large, parallel conducting plates close to each other. Each face has an area $A$. $X$ is given a charge $Q$. $Y$ is without any charge. Points $A$, $B$, and $C$ are as shown in the figure. The incorrect option is:

Options:

1. the field at $B$ is $\frac{Q}{2\varepsilon_0 A}$
2. the field at $B$ is $\frac{Q}{\varepsilon_0 A}$
3. the fields at $A$, $B$, and $C$ are of the same magnitude
4. the fields at $A$ and $C$ are of the same magnitude but in opposite directions

Solution:

$\text{Charge on each surface of } x = +\frac{Q}{2}$ $\text{Charge on the inner surface of } Y \text{ due to induction} = -\frac{Q}{2}$ $\text{Charge on the outer surface of } Y = +\frac{Q}{2}$ $\vec{E}_A = \text{Electric field due to } X \text{ and } Y = \frac{Q/2}{2\varepsilon_0 A} + \frac{Q/2}{2\varepsilon_0 A} + \frac{Q/2}{2\varepsilon_0 A} + \frac{Q/2}{2\varepsilon_0 A} = \frac{Q}{2\varepsilon_0 A} \text{ (Towards left)}$ $\vec{E}_B = \text{Electric field due to } X \text{ and } Y = \frac{Q/2}{2\varepsilon_0 A} + \frac{Q/2}{2\varepsilon_0 A} - \frac{Q/2}{2\varepsilon_0 A} - \frac{Q/2}{2\varepsilon_0 A} = \frac{Q}{2\varepsilon_0 A} \text{ (Towards right)}$ $\vec{E}_C = \text{Electric field due to } X \text{ and } Y = \frac{Q/2}{2\varepsilon_0 A} + \frac{Q/2}{2\varepsilon_0 A} + \frac{Q/2}{2\varepsilon_0 A} + \frac{Q/2}{2\varepsilon_0 A} = \frac{Q}{2\varepsilon_0 A} \text{ (Towards right)}$

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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
}