University:
Manchester University
Course:
PHYS 111 | College Physics I
Academic year:

2020
Views:

244

Pages:

4
Author:

StellarStrategist

   B  dA  0 From lab … Faraday’s Law of Induction Vs  EMF  N s BP   dB d   d     BP  As   N s As cos  P   B dA N P s    dt  dt  dt o iP N P 2rP  oVP N P 2rP RP  o N P 2rP RP VP  peak sin(t   ) Things you can prove with Gauss’ Law: The two shell theorems  For a uniform, spherical charge distribution …  For a uniform spherical charge distribution … Properties of conductors:  Any excess charge on a conductor …  The electric field inside the bulk of a conductor is always …. A perfect conductor … is an equipotential surface & throughout! has all excess charge on its outer surface. has no internal E field (does this make sense based on first statement, above?). does not necessarily have a uniform charge distribution on its surface. has a field just outside its surface that is perpendicular to the surface. Some of these ideas are very important in circuit analysis! Electric Field due to a point charge within a conducting spherical shell Spherical shell is electrically neutral Charges that make up conducting sphere are free to move about in response to field created by point charge at center Those charges arrange themselves in such a way that the field they create cancels out the field due to the point charge (within the conducting material) So, the field in the conductor is zero!!!! For Example: A small charged ball lies within the hollow of a metallic spherical shell of radius R. Here, for three situations, are the net charges on the ball and the shell, respectively:  (1) + 4q, 0  (2) – 6q, +10q  (3) +16q, -12q Rank situations according to the charge on the (A) inner surface of the shell and (B) the outer surface of the shell, most positive first.