The dipole moment of a dipole in a 300-N/C electric field is initially perpendicular to the field, but it rotates so it is in the same direction as the field.If the moment has a magnitude of 2 * 10^-9 CF.m the work done by the field is:

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Quizplus · Accepted Answer

The work done by the electric field on the dipole is given by the change in potential energy, which is $W = -\Delta U = -[U_f - U_i]$. The potential energy of a dipole in an electric field is $U = -pE\cos\theta$, where $p$ is the dipole moment, $E$ is the electric field strength, and $\theta$ is the angle between the dipole moment and the electric field direction. Initially, the dipole is perpendicular to the field ($\theta = 90^\circ$, $\cos90^\circ = 0$), so $U_i = 0$. Finally, the dipole is aligned with the field ($\theta = 0^\circ$, $\cos0^\circ = 1$), so $U_f = -pE$. Thus, $W = -(-pE - 0) = pE$. Given $p = 2 \times 10^{-9} C\cdot m$ and $E = 300 N/C$, $W = (2 \times 10^{-9} C\cdot m)(300 N/C) = 6 \times 10^{-7} J$, or $6 \times 10^{-7} J = 6 \times 10^{-7} J$.

The Dipole Moment of a Dipole in a 300-N/C Electric