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

Step-by-step 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=(1)/(4\pi \epsilon ) *(q)/(x-l)`

Potential at distance x along x axis due to - q

`v_2=(1)/(4\pi \epsilon ) *(-q)/(x+l)`

Total potential

v = v₁ + v₂

`v=(1)/(4\pi \epsilon ) *( (q)/(x-l)-(q)/(x+l))`

`v=(1)/(4\pi \epsilon ) *(2ql)/(x^2-l^2)`

`v=(1)/(4\pi \epsilon ) *(qs)/(x^2-((s)/(2)) ^2)`

Potential at distance y along y axis due to + q

`v_1=(1)/(4\pi \epsilon ) *(qs)/((y^2+(s^2)/(4))^(1)/(2) )`

Potential at distance y along y axis due to - q

`v_1=(1)/(4\pi \epsilon ) *(-qs)/((y^2+(s^2)/(4))^(1)/(2) )`

Total potential

v = v₁ + v₂

`v= 0`