Advertisements
Advertisements
प्रश्न
For a short dipole at a point P on the equatorial line, where r ≫ a, the electric field is given by:
पर्याय
\[E_{eq} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r}\]
\[E_{eq} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^3}\]
\[E_{eq} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^2}\]
\[E_{eq} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}\]
MCQ
Advertisements
उत्तर
Through the derivation involving the geometry of the equatorial line and the approximation r ≫ a, the electric field at the equatorial line simplifies to \[E_{eq} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^3}\]. This is exactly half the magnitude of the axial field at the same distance, with direction opposite to the dipole moment.
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
