Question

At an axial point located on the side of the negative charge (θ = 180°), the potential due to the dipole is:

पर्याय

• \[V = 0\]

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

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

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

Answer 1

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.