GAUSS’(S) LAW Lecture 4 Dr. Lobna Mohamed Abou El-magd

Electric Flux   The concept of electric flux involves a surface and the (vector) values of the electric field at all points of the surface.  consider a flat surface of area A and an electric field which is constant over the surface.  This is a vector which points perpendicularly (normal) to the surface and has magnitude E

example  A flat disk with radius r = 0.1m, is oriented with its normal unit vector relative the a constant E field with magnitude 2X 10 3 N/c as shown above. 1. What is the electric flux through the disk area? 2. if it turned so that its normal is perpendicular to E? 3. If its normal is parallel to E? 1- A= π (0.1m) 2 = m 2  =E.A.cosө = ( 2X 10 3 N/C) ( m 2 )(cos 30 0 ) =54 N.m 2 /c 2- ө = 90, cosө = 0,  =0 3- ө =0, cosө = 1  = E.A.cosө =( 2X 10 3 N/C) ( m 2 ) = 63 N.m 2 /c

 a surface is not flat, and the electric field will not be uniform over the surface. In practice one must use the machinery of advanced calculus to find the flux for the general case

 idea of the process: 1. divide up the surface into little sections (squares) 2. Suppose the i th little square has area vector  Ai,the value of the electric field on that square is close to Ei 3. Then the electric flux for the little square is found as before the electric flux over the surface S is

Gaussian Surfaces  a special class of surfaces, ones which we call Gaussian surfaces  this surface is a closed surface, it encloses a particular volume of space and doesn’t have any holes in it  Thus for a Gaussian surface S, the electric flux is written

Gauss’(s) Law  Suppose we choose a closed surface S in some environment where there are charges and electric fields.  We can (in principle, at least) compute the electric flux  on S  We can also find the total electric charge enclosed by the surface S, q enc  Gauss’(s) Law  The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

Point Charge

Spherically–Symmetric Distributions of Charge  Suppose we have a ball with total charge q, where the charge density only depends on the distance from the center of the ball.  We can draw a Gaussian surface of radius r and use the same arguments as for the point charge to find the electric field.

Electric Fields and Conductors  The electric field is zero inside any conductor. Using Gauss’(s) law it follows that if a conductor carries any net charge, the charge will reside on the surface(s) of the conductor.  Also using Gauss’(s) law one can show that the electric field just outside a conducting surface is perpendicular to the surface and is given by  + ve  if  is positive the E field points outward and if it is negative the E field points toward the surface. -ve