A small, rigid object carries positive and negative 3.00 nC charges. It is orien

Question

A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates (-1.20 mm, 1.60 mm) and the negative charge is at the point (1.50 mm, -1.30 mm). Find the electric dipole moment of the object. 10.4e-12 the response you submitted has the wrong sign. C middot m i + 10.2e-12 your response differs from the correct answer by more than 10%. Double check your calculations. C middot m j the object is placed in an electric field E = (7.80 times 10^3 I - 4.90 times 10^3 j) N/C. Find the torque acting on the object. 30.2e-8 the response you submitted has the wrong sign. N middot m k Find the potential energy of the object-field system when the object is in this orientation. 159e-9 your response differs from the correct answer by more than 10%. Double check your calculations. J Assuming the orientation of the object can change; find the difference between the maximum and the minimum potential energies of the system. 326e-9 your response differs from the correct answer by more than 10%. Double check your calculations. J

Explanation / Answer

a) dipole moment, P = q*( (x2 - x1)i + (y2-y1) j)

= 3*10^-9*((-1.2) - 1.5)i + (1.6 - (-1.3))j)*10^-3

= 3*10^-12*(-2.7 i + 2.9 j)

= (-8.1*10^-12 i + 8.7*10^-12 j) C.m

b) T = P cross E

= (-8.1*10^-12 i + 8.7*10^-12 j) cross (7.8*10^3 i - 4.9*10^3 j)

= -8.1*10^-12*(-4.9*10^3) k + 8.7*10^-12*7.8*10^3 (-k)

= -2.82*10^-8 N.m k

c) U = -P.E

= -(-8.1*10^-12 i + 8.7*10^-12 j).(7.8*10^3 i - 4.9*10^3 j)

= -((-8.1*10^-12*7.8*10^3) + 8.7*10^-12*(-4.9*10^3) )

= 1.06*10^-7 J

d) diffrnce between maximum and minimum PE = 2*|P|*|E|

= 2*sqrt(8.1^2 + 8.7^2)*10^-12*sqrt(7.8^2 + 4.9^2)*10^3

= 2.19*10^7 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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