In Figure (a), a particle of charge +e is initially at coordinate z = 25 nm on t

Question

In Figure (a), a particle of charge +e is initially at coordinate z = 25 nm on the dipole axis through an electric dipole, on the positive side of the dipole. (The origin of z is at the dipole center.) The prticle is then moved along a circular path around the dipole center until it is at coordinate z = -25 nm. Figure (b) gives the work W_a done by the force moving the particle versus the angle theta that locates the particle. The scale of the vertical axis is set by W_as = 2.0 x I0&-30 J. What is the magnitude of the dipole moment? Number Units

Explanation / Answer

We can calculate the electric field due to the electron at the dipole location.
E = kq/r^2 = 9E9*1.6022E-19/25E-9^2 = 2307168 V/m
the energy change deltaU = 2E-30J = 2Ep
==> dipole moment p = 2E-30/(2E) = 8.66E-37 C-m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.