In the figure, a particle of elementary charge +e is initially at coordinate z=2

Question

In the figure, a particle of elementary charge +e is initially at coordinate z=20nm on the dipole axis (here a z axis) through an electric dipole, on the positive side of the dipole. (The origin of z is at the center of the dipole.) The particle is then moved along a circular path around the dipole center until it is at coordinate z=20nm, on the negative side of the dipole axis. The graph gives the work W(sub:a) done by the force moving the particle versus the angle that locates the particle relative to the positive direction of the z axis. The scale of the vertical axis is set by W(sub:as)=4.0E-30 J. What is the magnitude of the dipole moment?
0 (ros-01) "Al 2

Explanation / Answer

when particle is at z = 20 nm

suppose distance of dipole charge from origin is d.

PE of system = [k(q)(e) / (20-d)nm ] + [ k(-q)(e) / (20 + d)nm ]

= (9 x 10^9 x q x 1.6 x 10^-19) / (10^-9) [ 1/(20 -d) - 1/(20 + d) ]

= 1.44q [ 2d / (20^2 - d^2 ) ]

when charge e at z = 0

PE of system = [ k (q)(e) / sqrt(20^2 + d^2) ] + [ k (-q)(e) / sqrt(20^2 + d^2) ]

= [ k (q)(e) / sqrt(20^2 + d^2) ] - [ k (q)(e) / sqrt(20^2 + d^2) ]

= 0

initial PE = Wmax

Wmax = 2 vertical scale = 8 x 10^-30 ( i am liitlle confuse here whether it wil be 8 x 10^-30

or 2 x 10^-30 J)

1.44q [ 2d / (20^2 - d^2 ) ] = 8 x 10^-30

2.88qd = (20^2 - d^2) (8 x 10^-30 )

and d < < < 20

hence 20^2 - d^2 = 20^2

2.88qd = 400 x 8 x 10^-30

qd = 1.11 x 10^-27

(Or it will 0.278 x 10^-27 in other case when Wmax = 20 x 10^-30 J)

dipole moment = 2qd = 2.22 x 10^-27 Cm .........Ans

(or try 0.556 x 10^-27 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.