Question

A small, rigid object carries positive and negative 4.00nC charges. It is oriented so that the positive charge has coordinates 1.20 mm,1.40 mm) and the negative charge is at the point (1.70 mm,−1.30 mm). (a) Find the electric dipole moment of the object.

Answer 1

The dipole moment of the object can be calculated by multiplying the charge of one of the charges by the distance between the charges. In this case, the positive charge is +4.00nC and the negative charge is -4.00nC. The dipole moment (μ) is then calculated as the product of the charge (+4.00nC) and the distance (2.75 mm).

Step-by-step explanation:

The dipole moment of the object can be calculated by multiplying the charge of one of the charges by the distance between the charges.

In this case, the positive charge is +4.00nC and the negative charge is -4.00nC. The distance between them can be found using the distance formula, which is the square root of the sum of the squares of the differences in the x and y coordinates:

d = √((1.70 mm - 1.20 mm)^2 + (-1.30 mm - 1.40 mm)^2)

d = √((0.50 mm)^2 + (-2.70 mm)^2)

= √((0.250 mm^2 + 7.290 mm^2))= √(7.540 mm^2)= 2.75 mm

The dipole moment (μ) is then calculated as the product of the charge (+4.00nC) and the distance (2.75 mm):

μ = Q * d= (4.00 nC) * (2.75 mm) = 11.0 nC·mm