Question

Point charges q1=− 4.50 nC and q2=+ 4.50 nC are separated by distance 3.30 mm , forming an electric dipole. Find the magnitude of the electric dipole moment.

Answer 1

The magnitude of the electric dipole moment for two charges of − 4.50 nC and + 4.50 nC separated by 3.30 mm is 1.485 × 10⁻⁻ C·m.

Step-by-step explanation:

The magnitude of the electric dipole moment is given by the product of the charge magnitude and the separation distance between the charges. With charges q1 = − 4.50 nC and q2 = + 4.50 nC separated by 3.30 mm, the dipole moment can be calculated using the formula p = qd, where q represents the magnitude of one of the charges and d is the separation distance. Here, q = 4.50 nC or 4.50 × 10⁻⁹ C and d = 3.30 mm or 3.30 × 10⁻⁳ m.

The magnitude of the dipole moment (p) is therefore:
p = 4.50 × 10⁻⁹ C × 3.30 × 10⁻⁳ m
p = 14.85 × 10⁻⁼ C·m
p = 1.485 × 10⁻⁻ C·m.

The electric dipole moment is a measure of the strength and orientation of an electric dipole. It is calculated by multiplying the magnitude of one of the charges by the distance between the charges:

p = qd

Using the given information, the magnitude of the electric dipole moment can be calculated as:

p = (4.50 nC)(3.30 mm) = 14.85 nC·mm

Therefore, the magnitude of the electric dipole moment is 14.85 nC·mm.