Question

The dipole moment of a dipole in a 300 NC electric field is initially perpendicular to the field, but it rotates to be in the same direction as the field. If the moment has a magnitude of 2x10⁻⁹ Cm, the work done by the field is __________.

Answer 1

The work done by the electric field on the dipole as it aligns with the field is -6 x 10^-7 J.

Step-by-step explanation:

The dipole moment of a dipole is a measure of its electric charge distribution and is represented by the vector quantity p. In this case, the dipole moment has a magnitude of 2x10^-9 Cm and is initially perpendicular to the applied electric field of 300 NC.

When an electric field is applied to a dipole, it exerts a torque on the dipole moment, causing it to rotate until it aligns with the field. This rotation results in work being done by the electric field on the dipole.

The work done by the electric field on the dipole is given by the formula:

W = -p · Ecosθ + p · Esinθ

In this equation, p is the dipole moment, E is the electric field, and θ is the angle between the dipole moment and the electric field. The negative sign in front of the first term indicates that work is done against the torque caused by the field's rotation.

Initially, when the dipole moment is perpendicular to the electric field (θ = 90°), we can simplify this equation to:

W = p · Esinθ |θ=0°| = p · E = 2x10^-9 Cm × 300 NC = 6x10^-6 Joules (J)

Therefore, when the dipole moment rotates to be in the same direction as the electric field (θ = 0°), it requires an initial work of 6x10^-6 Joules to align it with the field.