Question

q1= - 4.30 nC and q2=+ 4.30 nC are separated by a distance of 3.00 mm, forming an electric dipole Find the magnitude of the electric dipole moment. (C/m)

Answer 1

The magnitude of the electric dipole moment for two charges of -4.30 nC and +4.30 nC separated by 3.00 mm is 12.9 × 10^-12 C·m.

Step-by-step explanation:

The electric dipole moment is a vector quantity that represents the strength and direction of an electric dipole. The dipole moment (μ) is given by the product of the charge magnitude (q) and the separation distance (d), and is defined as μ = qd. For charges q1 = -4.30 nC and q2 = +4.30 nC separated by a distance of 3.00 mm, the magnitude of the electric dipole moment can be calculated as: Convert nC to C: -4.30 nC = -4.30 × 10-9 C

Calculate the dipole moment: μ = q × d = (-4.30 × 10-9 C) × (3.00 × 10-3 m)

Magnitude of dipole moment: |μ| = 4.30 × 10-9 C × 3.00 × 10-3 m = 12.9 × 10-12 C·m

Therefore, the magnitude of the electric dipole moment is 12.9 × 10-12 C·m.