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Question
At an axial point located on the side of the negative charge (θ = 180°), the potential due to the dipole is:
Options
\[V = 0\]
\[V = -\frac{1}{4\pi\varepsilon_0}\frac{p}{r}\]
\[V = +\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}\]
\[V = -\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}\]
MCQ
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Solution
Since cos 180° = −1, the general formula V = p cos θ/(4πε₀r²) gives V = −p/(4πε₀r²). The potential is negative on the −q side of the axial line and positive on the +q side.
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