Question

At an axial point of an electric dipole (θ = 0° or 180°), the potential is:

Options

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

Answer 1

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.