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, 1.20 mm) and the negative charge is at the point (1.70 mm, -1.30 mm). Find the electric dipole moment of the object. Your response is off by a multiple of ten. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. The object is placed in an electric field E = (7.80 times 10 3 i - 4.90 imes 10 3 j) N/C. Find the torque acting on the object. Find the potential energy of the object-field system when the object is in this orientation. Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system.

Explanation / Answer

The electric dipole moment is a vector

p = ? q_i r_i

where q_i is the charge and r_i is the position vector of the i-th charge. Here r_1 = (-1.2, 1.20, 0) mm and r_2 = (1.7, -1.3, 0) mm and q_1 = + 3.0nC and q_2 = - 3.0nC

So the dipole moment vector is the sum:

p = 3.0(-1.2, 1.20, 0) - 3.0(1.7, -1.3, 0) in units of mm nC

p = (-8.7, 7.5, 0) mm nC

px = -8.7 x10^-12
py = 7.5 x10^-12
pz = 0.0

The cartesian components of the cross product giving the torque, (Tx, Ty, Tz), can be represented by the determinant of a matrix:

E = 7.8x10^3 i + -4.9 x 10^3 j

|.i....j....k....|
|.px.py.pz..|
|.Ex.Ey.Ez.|

Tx = pyEz - pzEy = 0
Ty = -(pxEz - pzEx) = 0
Tz = pxEy - pyEx = (42.63 - 58.5) = -16.2 x 10^-12x10^3 =-16.2x10^-9 Nm k

The potential energy can be found using the dot product

U = - p

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