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www.GOALias.blogspot.comThis expression can alternately be understood also from Eq. (2.29).We apply Eq. (2.29) to the present system of two charges +q and –q. Thepotential energy expression then reads2r1 r24π ε 0× 2a(2.33)( θ) ( ) ( )U′= q[ V −V] −qHere, r 1and r 2denote the position vectors of +q and –q. Now, thepotential difference between positions r 1and r 2equals the work donein bringing a unit positive charge against field from r 2to r 1. Thedisplacement parallel to the force is 2a cosθ. Thus, [V(r 1)–V (r 2)] =–E × 2a cosθ . We thus obtain,U′( θ)2 2qq=−pEcosθ− =−pEg −(2.34)4π ε × 2a4π ε × 2a0 0We note that U′ (θ) differs from U(θ ) by a quantity which is just a constantfor a given dipole. Since a constant is insignificant for potential energy, wecan drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32).We can now understand why we took θ 0=π/2. In this case, the workdone against the external field E in bringing +q and – q are equal andopposite and cancel out, i.e., q [V (r 1) – V (r 2)]=0.Electrostatic Potentialand CapacitanceExample 2.6 A molecule of a substance has a permanent electricdipole moment of magnitude 10 –29 C m. A mole of this substance ispolarised (at low temperature) by applying a strong electrostatic fieldof magnitude 10 6 V m –1 . The direction of the field is suddenly changedby an angle of 60º. Estimate the heat released by the substance inaligning its dipoles along the new direction of the field. For simplicity,assume 100% polarisation of the sample.Solution Here, dipole moment of each molecules = 10 –29 C mAs 1 mole of the substance contains 6 × 10 23 molecules,total dipole moment of all the molecules, p = 6 × 10 23 × 10 –29 C m= 6 × 10 –6 C mInitial potential energy, U i= –pE cos θ = –6×10 –6 ×10 6 cos 0° = –6 JFinal potential energy (when θ = 60°), U f= –6 × 10 –6 × 10 6 cos 60° = –3 JChange in potential energy = –3 J – (–6J) = 3 JSo, there is loss in potential energy. This must be the energy releasedby the substance in the form of heat in aligning its dipoles.EXAMPLE 2.62.9 ELECTROSTATICS OF CONDUCTORSConductors and insulators were described briefly in Chapter 1.Conductors contain mobile charge carriers. In metallic conductors, thesecharge carriers are electrons. In a metal, the outer (valence) electronspart away from their atoms and are free to move. These electrons are freewithin the metal but not free to leave the metal. The free electrons form akind of ‘gas’; they collide with each other and with the ions, and moverandomly in different directions. In an external electric field, they driftagainst the direction of the field. The positive ions made up of the nucleiand the bound electrons remain held in their fixed positions. In electrolyticconductors, the charge carriers are both positive and negative ions; but67