An electric dipole of dipole moment 4.5 x 10^-9 C·m is in a 250 N/C electric field.If the electric field points east, and the dipole moment vector points 37° west of south, what is the potential energy? A)-6.8 x 10^-8 J B)-9.0 x 10^-7 J C)0 J D)6.8 x 10^-8 J E)9.0 x 10^-7 J

Question

Quizplus · Accepted Answer

The potential energy of an electric dipole in an electric field is given by $U=-\vec{p}\cdot\vec{E}$, where $\vec{p}$ is the dipole moment and $\vec{E}$ is the electric field. Since the dipole moment vector points 37° west of south, we can break it into components parallel and perpendicular to the electric field: $p_{\parallel}=p\cos37°=-3.6	imes10^{-9}$ C·m and $p_{\perp}=p\sin37°=2.7	imes10^{-9}$ C·m. Then, the potential energy is $U=-(p_{\parallel}E+p_{\perp}E)=(3.6	imes10^{-9})(250)+(2.7	imes10^{-9})(250)=6.8	imes10^{-8}$ J, which is closest to choice D.

An Electric Dipole of Dipole Moment 4