1.4 CHARGING BY CONTACT/ INDUCTION

When we touch a pith ball with an electrified plastic rod, some of the negative charges on the rod are transferred to the pith ball and it also gets charged. Thus the pith ball is charged by contact.
Let take  two metal spheres A and B in contact with each other and standing on an insulating stand.
When we bring a positively charged rod near one of the spheres{A}, the free electrons in the spheres are attracted towards the rod. This leaves an excess of positive charge on the rear surface of sphere B. Both kinds of charges are bound in the metal spheres and cannot escape. They, therefore, reside on the surfaces,  The left surface of sphere A, has an excess of negative charge and the right surface of sphere B, has an excess of positive charge. However, not all of the electrons in the spheres have accumulated on the left surface of A. As the negative charge starts building up at the left surface of A, other electrons are repelled by these. In a short time, equilibrium is reached under the action of force of attraction of the rod and the force of repulsion due to the accumulated charges. The process is called induction of charge  The accumulated charges remain on the surface,till the glass rod is held near the sphere. If the rod is removed, the charges are not acted by any outside force and they redistribute to their original neutral state.In this process, the metal spheres will each be equal and oppositely charged. This is charging by induction. The positively charged glass rod does not lose any of its charge, contrary to the process of charging by contact.

CHARGING BY INDUCTION



Example 1  How can you charge a metal sphere positively without touching it?
Ans=use charging by induction law.

Example 2 If 109 electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?
Ans= use [q=]n.e=i.t
              n=i.t/e

Example 3 How much positive and negative charge is there in a cup of water?
ans=use q=ne and assume mass of one cup water='m' gram
     no of molecules of water in a cup=m/18 *6.02*10P[-23]
   here P denotes power