Question

Using the exact general-point formula \[V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2 - l^2\cos^2\theta}\], the potential on the equatorial line (θ = 90°) is zero:

पर्याय

• Exactly, for any value of l, because \[\cos 90^\circ = 0\] makes the numerator zero

• Never exactly zero, due to the \[l^2\cos^2\theta\] term in the denominator

• Only when the half-length \[l\] equals zero

• Only approximately, and only when \[r \gg l\]

Answer 1

At θ = 90°, cos θ = 0, so the entire numerator vanishes identically regardless of the value of l or the denominator. Therefore V = 0 holds exactly at every point of the equatorial line, not merely as an approximation for r >> l.