Q.The dipole moment of a water molecule is 6.3×10^-30 Cm. A sample of water contains 10^21 molecules, whose dipole moments are all oriented in an electric field of strength 2.5×10^5 N/C. Calculate the work to be done to rotate the dipoles from their initial orientation θ1 = 0 to one in which all the dipoles are perpendicular to the field, θ2 = 90°.

[!TLDR] Work per molecule from θ=0 to 90 degrees is pE(cos0-cos90)=pE; multiplying by N=10^21 molecules …

For a single dipole rotated from $\theta_1=0^\circ$ to $\theta_2=90^\circ$ in field E, the work done (section 8.6.5) is $w=pE(\cos\theta_1-\cos\theta_2)=pE(\cos0^\circ-\cos90^\circ)=pE(1-0)=pE$. For $N=10^{21}$ identical, identically-oriented dipoles all rotated together, the total work is $W=NpE=10^{21}\times6.3\times10^{-30}\times2.5\times10^5$. Computing: $10^{ …

Apply w=pE(cosθ1-cosθ2) for one dipole, then multiply by the number of molecules N since …

A common slip is to forget the cos90°=0 term and simply use w=pE without checking it against the general formula, o …