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Electrostatic Potential and Capacitance 2.6.1 Relation between field and potential Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values V and V + V, where V is the change in V in the direction of the electric field E. Let P be a point on the surface B. l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E| l. This work equals the potential difference V –V . A B Thus, |E| l = V – (V + V)= – V δ V i.e., |E|=− (2.20) δl Since V is negative, V = – V|. we can rewrite FIGURE 2.12 From the Eq (2.20) as potential to the field. δV δV E = − = + (2.21) δl δl We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point. 2.7 P OTENTIAL E NERGY OF A S YSTEM OF C HARGES Consider first the simple case of two charges q1and q 2ith position vector r and r relative to some origin. Let us calculate the work done 1 2 (externally) in building up this configuration. This means that we consider the charges q 1nd q i2itially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q is brought from infinity to the point r . There is no 1 1 external field against which work needs to be done, so work done in bringing q 1rom infinity to r1is zero. This charge produces a potential in space given by 1 q1 V1 = 4π ε0r1P where r 1P is the distance of a point P in space from the location of q . 1 From the definition of potential, work done in bringing charge q from 2 infinity to the point 2 is 2 times the potential at r2due to q1: 1 q 1 2 work done on q =2 4π ε0 r12 55 Reprint 2026-27