Editing Content :: Admin Scraper

Chapter

Content type

Position

Value

Electrostatic Potential and Capacitance This 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. The potential energy expression then reads q2 U′ ( )= q[V ( 1V r ](2) (2.33) 4π ε0× 2a Here, r1and r d2note the position vectors of +q and –q. Now, the potential difference between positions r an1 r equ2ls the work done in bringing a unit positive charge against field from r to r . The 2 1 displacement parallel to the force is 2a cos Thus, [V(r 1–V (r )2 = –E × 2a cos . We thus obtain, q2 q2 U′ ( )= −pE cosθ − = −p.E − (2.34) 4 πε0× 2a 4 πε0× 2a WenotethatU ( )differsfromU( )byaquantitywhichisjustaconstant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32). We can now understand why we took In this case, the work 0 done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V (1 ) – V 2r )]=0. Example 2.6 A molecule of a subs–29ce has a permanent electric dipole moment of magnitude 10 C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 10 V m . The direction of the field is suddenly changed by an angle of 60º. Estimate the heat released by the substance in aligning 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 –29C m 23 As 1 mole of the substance contains 6 × 10 molecule23 –29 total dipole moment of all the molecules, p = 6 × 10 –6× 10 C m = 6 × 10 C m E Initial potential energy,iU = –pE cos = –6×10 ×10 cos 0° = –6 J A Final potential energy (when = 60°), U = –6 × 10 × 10 cos 60° = –3 J P f E Change in potential energy = –3 J – (–6J) = 3 J 2 So, there is loss in potential energy. This must be the energy released 6 by the substance in the form of heat in aligning its dipoles. 2.9 E LECTROSTATICS O FC ONDUCTORS Conductors and insulators were described briefly in Chapter 1. Conductors contain mobile charge carriers. In metallic conductors, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of ‘gas’; they collide with each other and with the ions, and move randomly in different directions. In an external electric field, they drift against the direction of the field. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. In electrolytic 61 conductors, the charge carriers are both positive and negative ions; but Reprint 2026-27