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The electric potential at a point P due to a dipole (at distance r from the centre, at angle θ from the dipole axis) is given by:

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Question

The electric potential at a point P due to a dipole (at distance r from the centre, at angle θ from the dipole axis) is given by:

Options

  • \[V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p\cos\theta}{r^2}\]

  • \[V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^3}\]

  • \[V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r}\]

  • \[V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p\sin\theta}{r}\]

MCQ
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Solution

For a point at distance r from the centre of a dipole at angle θ from the dipole axis (where r ≫ a), the electric potential is \[V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p\cos\theta}{r^2}\]. This shows that the potential depends on both the dipole moment and the angular orientation, and falls off as 1/r² rather than 1/r for a point charge.

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