Sunday, June 24, 2012

1.8 Electric Field

Consider a point charge Q placed in vacuum at the point O. If we place another point charge q at a point P, where OP=r, then the charge Q will exert a force on q according to Coulomb's law. We may ask the question: If the charge q is removed, then what is left at P? If there is nothing at the point P, then how does a force act when we place a charge q at P?
         
                To overcome this troublesome idea of non contact force, Michael Faraday introduced the concept of electric field.  The idea is that, surrounding every charge, there is its electric field like the light around a burning lamp. The modification made on the surrounding space by a charge or charge distribution is called it's electric field.

              When two charges q and Q are placed near to each other, each of them is situated in the field of the other. In other words, we can consider the interaction between two electric charges as the interaction between their electric fields. So if we know the electric field at a point we can calculate the electric force acting on a charge kept at that point.
         
              One of the important problems in electrostatics is to find the electric field due to different charge distributions

 Electric field Intensity
  
                   Electric field intensity at a point is defined as the force experienced by a a unit positive charge kept at that point. Electric field is a vector quantity.

           Let 'P' be a point in an electric field. To find the electric field intensity 'E' at 'P', imagine a small test charge q kept at  'P' so that it experiences a force 'F'. Then force acting on unit charge = F/q.
         By definition,  this is the electric field at P
                                E = limit q to zero (F/q)
( The test charge must be very small; otherwise it may disturb the electric field).
  Unit of electric field is newton/coulomb (N/C) or volt/meter(V/m). It's dimentional formula is MLT^-3A^-1

No comments:

Post a Comment