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Current Question (ID: 16489)

Question:

$\text{Four electric charges } +q, +q, -q \text{ and } -q \text{ are placed at the corners of a square of side } 2L \text{ (see figure). The electric potential at the point } A, \text{ mid-way between the two charges } +q \text{ and } +q \text{ is:}$

Options:

1. $\frac{1}{4\pi\varepsilon_0} \frac{2q}{L} \left( 1 + \frac{1}{\sqrt{5}} \right)$
2. $\frac{1}{4\pi\varepsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$
3. $\text{zero}$
4. $\frac{1}{4\pi\varepsilon_0} \frac{2q}{L} \left( 1 + \sqrt{5} \right)$

Solution:

$\text{Hint: The net potential will be the sum of potentials due to each charge.}$ $\text{Step: Find the net potential at point } A.$ $\text{Here, the distance between the point } A \text{ and } -q \text{ is given by:}$ $d = \sqrt{L^2 + 2L^2} = \sqrt{5L^2} = L\sqrt{5}$ $\text{The distance between the point } A \text{ and } -q \text{ is } L.$ $V_{\text{net}} = V_1 + V_2 + V_3 + V_4$ $V_{\text{net}} = 2V_{+q} + 2V_{-q}$ $V_{\text{net}} = \frac{1}{4\pi\varepsilon_0} \frac{2q}{L} - \frac{2q}{L\sqrt{5}}$ $\Rightarrow V_{\text{net}} = \frac{1}{4\pi\varepsilon_0} \times \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$ $\text{Hence, option (2) is the correct answer.}$

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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
}