Question

A dipole is oriented along the x axis. The dipole moment is p (= qs). (Assume the center of the dipole is located at the origin with positive charge to the right and negative charge to the left.)
Calculate exactly the potential V (relative to infinity) at a location x, 0, 0 on the x axis and at a location 0, y, 0 on the y axis, by superposition of the individual 1/r contributions to the potential. (Use the following as necessary: q, ε0, x, s and y.)

Answer 1

Answer:

Explanation:

dipole moment = qs = q x s

= charge x charge separation

charge = q

separation between charge = s

half separation l = s / 2

dipole has two charges + q and - q separated by distance s .

Potential at distance x along x axis due to + q

$v_1=\frac{1}{4\pi \epsilon } \times\frac{q}{x-l}$

Potential at distance x along x axis due to - q

$v_2=\frac{1}{4\pi \epsilon } \times\frac{-q}{x+l}$

Total potential

v = v₁ + v₂

$v=\frac{1}{4\pi \epsilon } \times( \frac{q}{x-l}-\frac{q}{x+l})$

$v=\frac{1}{4\pi \epsilon } \times\frac{2ql}{x^2-l^2}$

$v=\frac{1}{4\pi \epsilon } \times\frac{qs}{x^2-(\frac{s}{2}) ^2}$

Potential at distance y along y axis due to + q

$v_1=\frac{1}{4\pi \epsilon } \times\frac{qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }$

Potential at distance y along y axis due to - q

$v_1=\frac{1}{4\pi \epsilon } \times\frac{-qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }$

Total potential

v = v₁ + v₂

$v= 0$