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Current Question (ID: 16445)

Question:

$\text{The electric field at a point on the equatorial plane at a distance } r \text{ from the}$ $\text{centre of a dipole having dipole moment } \vec{P} \text{ is given by:}$ $(r \gg \text{ separation of two charges forming the dipole,}$ $\varepsilon_0 \text{ = permittivity of free space})$

Options:

1. $\vec{E} = \frac{\vec{P}}{4\pi\varepsilon_0 r^3}$
2. $\vec{E} = \frac{2\vec{P}}{\pi\varepsilon_0 r^3}$
3. $\vec{E} = \frac{\vec{P}}{4\pi\varepsilon_0 r^2}$
4. $\vec{E} = -\frac{\vec{P}}{4\pi\varepsilon_0 r^3}$

Solution:

$\text{Hint: At the equatorial plane, the direction of the electric field is opposite to}$ $\text{the direction of the dipole moment.}$ $\text{Step: Write the formula of the electric field at the equatorial plane.}$ $\text{The magnitude of the electric field at the equatorial plane is given by:}$ $E = \frac{K\vec{P}}{r^3} = \frac{\vec{P}}{4\pi\varepsilon_0 r^3}$ $\text{The direction of the electric field is opposite to the direction of the dipole}$ $\text{moment as shown in the figure.}$ $\text{Therefore electric field at the equatorial plane is given as:}$ $E = -\frac{K\vec{P}}{r^3} = -\frac{\vec{P}}{4\pi\varepsilon_0 r^3}$ $\text{Hence, option (4) as the correct answer.}$

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
}