PRINT ER VERSION BACK NEXT ASSIGNMENT RESOURCES HW4 Ch24 Phys2 F17Chapter 24, Pr

Question

PRINT ER VERSION BACK NEXT ASSIGNMENT RESOURCES HW4 Ch24 Phys2 F17Chapter 24, Problem 022 In Figure (a), a particle of charge te is initially at coordinate z - 15 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 particle is then moved along a circular path around the dipole center until it is at coordinate z15nm. Figure (b) gives the work Wa done by the force moving the particle versus the angle that locates the particle. The scale of the vertical axis is set by Was-2.0 × 10-30 J. What is the magnitude of the dipole moment? 004 006 Chapter 24.Problem 014 016 Chapter 24, Problen Chapter 24 Problem 024 Chapter 24 Problem 026 036 044 Number Units 046 the tolerance is +/-2%

Explanation / Answer

We can calculate the electric field due to the electron at the dipole location.

E = kq/r^2 = 9E9*1.6022E-19/(15E-9)^2 = 6,408,800 V/m

Then the energy change deltaU = 2E-30J = 2Ep

==> dipole moment p = 2E-30/(2E) = 1.56E-37 C-m

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