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Current Question (ID: 16442)

Question:

$\text{In a certain region of space, the electric field is along the } z\text{-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive } z\text{-direction, at the rate of } 10^5 \text{ NC}^{-1} \text{ per meter. What is the torque experienced by a system having a total dipole moment equal to } 10^{-7} \text{ C-m in the negative } z\text{-direction?}$

Options:

1. $10^{-2} \text{ N-m}$
2. $0$
3. $10^{-1} \text{ N-m}$
4. $0.01 \text{ N-m}$

Solution:

$\text{Hint: Since the electric field and dipole moment are along the same line, torque experienced by the dipole will be zero.}$ $\text{Step 1: Calculate force experienced by the dipole due to the non-uniform electric field.}$ $p = q \times dl = -10^{-7} \text{ C m}$ $\text{Rate of increase of electric field per unit length,}$ \frac{dE}{dl} = 10^{-5} \text{ N/Cm}$ $\text{Now, } \mathbf{F} = q \mathbf{E}$ $= q \times \frac{dE}{dl} \times dl$ $= p \times \frac{dE}{dl}$ $\Rightarrow \mathbf{F} = -10^{-7} \times 10^{-5} = -10^{-2} \text{ N}$ $\text{Step 2: Write torque exerted on the dipole by the electric field.}$ $\text{Here, force is in the negative } z\text{-direction, i.e., opposite to the direction of the electric field.}$ $\text{So, the angle between the electric field and the dipole moment is } 180^\circ$ $\text{Now, } \tau = p E \sin 180^\circ = 0$ $\text{Therefore, the torque experienced by the system is zero.}$

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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
}