Why is the Electric Field in a Conductor Zero After Removing Negative Charge?

  • Thread starter Thread starter goodboy
  • Start date Start date
  • Tags Tags
    Field
goodboy
Messages
11
Reaction score
0
a problem.put a conductor in a electric field .then we konw that in the conductor ,the positive charge will move along with the direction of the electric field,and the negative charge will move along against the direction of the field.then at all point in the conductor ,the external field+the internal field is 0.there is no problem.
but,now if we move the negative charge away(the external field do not change all the time),in this case,the field in the conductor is still 0(i can get this conclusion with Gauss law ).but why?
 
Physics news on Phys.org
goodboy said:
a problem.put a conductor in a electric field .then we konw that in the conductor ,the positive charge will move along with the direction of the electric field,and the negative charge will move along against the direction of the field.then at all point in the conductor ,the external field+the internal field is 0.there is no problem.
but,now if we move the negative charge away(the external field do not change all the time),in this case,the field in the conductor is still 0(i can get this conclusion with Gauss law ).but why?
If you remove a negative charge from a neutral conductor it becomes a charged conductor.

The field in the conductor is 0 because if it was not, the free charges in the conductor would move (opposite to the direction of the field) until the field became 0.

AM
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top