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Current Question (ID: 16601)

Question:

$\text{Four point charges } -Q, -q, 2q \text{ and } 2Q \text{ are placed, one at each corner of the square.}$ $\text{The relation between } Q \text{ and } q \text{ for which the potential at the center of the square is zero, is:}$

Options:

1. $Q = -q$
2. $Q = -\frac{1}{q}$
3. $Q = q$
4. $Q = \frac{1}{q}$

Solution:

$\text{Hint: Use } V = \frac{kQ}{r}.$ $\text{Step: Find the potential at centre } \mathcal{C}. \Rightarrow V_c = V_1 + V_2 + V_3 + V_4$ $\Rightarrow V_c = -\frac{kQ}{r} + \frac{kq}{r} + \frac{2kQ}{r} + \frac{2kq}{r}$ $\text{If the potential at centre is zero, then}$ $\Rightarrow -Q - q + 2Q + 2q = 0$ $\Rightarrow Q = -q$ $\text{Hence, option (1) is the correct answer.}$

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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
}