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प्रश्न
At an axial point of an electric dipole (θ = 0° or 180°), the potential is:
विकल्प
\[V = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}\] for both θ = 0° and θ = 180°
\[V = \pm\frac{1}{4\pi\varepsilon_0}\frac{p}{r}\]
\[V = \pm\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}\], positive for θ = 0° and negative for θ = 180°
\[V = 0\] for both θ = 0° and θ = 180°
MCQ
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उत्तर
Substituting into V = p cos θ/(4πε₀r²): cos 0° = +1 gives V = +p/(4πε₀r²), while cos 180° = −1 gives V = −p/(4πε₀r²). The potential is positive on the +q side of the axis and negative on the −q side.
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