Question

The exact potential at a general point of a dipole with half-length l, before neglecting the \[l^2\cos^2\theta\] term, is:

विकल्प

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

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

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

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

Answer 1

Combining the two potentials: V = (q/4πε₀)[1/(r − l cos θ) − 1/(r + l cos θ)] = (q/4πε₀)(2l cos θ)/(r² − l²cos²θ). Using p = 2ql gives V = p cos θ/(4πε₀(r² − l²cos²θ)). For r >> l, the denominator reduces to r².