Monday, November 26, 2012

Relation Between Field And Potential


Consider two closely spaced equipotential surfaces A and B Fig. with potential values V and V + δV, where δV is the change in V in the direction of the electric field E.

Let P be a point on the surface B. δl 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|δl.

This work equals the potential difference VA –VB. Thus,

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.

1 comment:

  1. let a point P at which a test charge is situated with same fig A and B having potential of V nd V+ dv respectively which are at r and r+dr from the test charge then
    v + dv = v + dw/q.
    this implie
    dv = dw/q. ......... (1)
    now word done in movinge test chare with a infinitesimally small distance dr dr
    then
    dw = fdrcos180 ( moving in opp dir of E)
    =-q.Edr
    dw/q = -Edr .........(2)

    now comparing both we get
    dv = -Edr
    E = -dv/dr
    therefore electric field is eqal to the neg of potential gradient

    ReplyDelete