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प्रश्न
A dipole with its charges, −q and + q located at the points (0, −b, 0) and (0, +b, 0) is present in a uniform electric field E. The equipotential surfaces of this field are planes parallel to the yz plane.
- What is the direction of the electric field E?
- How much torque would the dipole experience in this field?
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उत्तर
a. Given, the equipotentials of the external uniform electric field are planes parallel to the yz plane, the electric field `vec E = +- E hat i` that is, `vec E` is parallel to the x-axis.
A dipole in an external electric field along x-axis
b. The above diagram, the dipole moment, `vec p = q (2 b) hat j`
The torque on this dipole,
`vec tau = vec p xx vec E = (2 qb hat j) xx (+- E hat i) = (2 qbE)(hat j xx hat i)`
Since `hat j xx hat i = - bar k`
`vec tau = (+- 2 qbE)(-hat k) = (2 qbE)(+- hat k)`
So that the magnitude of the torque is τ = 2qbE.
If `vec E` is in the direction of the + x-axis, the torque `vec tau` is in the direction of - z-axis, while if `vec E` s in the direction of the - x-axis, the torque `vec tau` is in the direction of + z-axis.
संबंधित प्रश्न
Answer in brief:
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