Week #3 Gauss’ Law September 7, 2005

What’s up Doc?? At this moment I do not have quiz grades unless I get them at the last minute. There was a short WebAssign due and there is a rather long set of problems also due next Monday. First EXAM is on Friday the 16 th.  May be pushed out a week because I will not be here on the 23 rd so it would be a good test date.  Material will be the same.  Will decide this week.

Gauss’s Law

Recall what we done so far? We were given Coulomb’s Law We defined the electric field. Calculated the Electric Field given a distribution of charges using Coulomb’s Law. (Units: N / C)

A Question: Given the magnitude and direction of the Electric Field at a point, can we determine the charge distribution that created the field? Is it Unique? Question … given the Electric Field at a number of points, can we determine the charge distribution that caused it?  How many points must we know??

Another QUESTION: Solid Surface Given the electric field at EVERY point on a closed surface, can we determine the charges that caused it??

The answer to the question is buried in something called GAUSS’ LAW or Gauss’s Law or The Law Of Gauss or The Chapter that Everybody hates! The Answer …

The “Area Vector” Consider a small area. It’s orientation can be described by a vector NORMAL to the surface.  We usually define the unit normal vector n.  If the area is FLAT, the area vector is given by A n, where A is the area.  A is usually a differential area of a small part of a general surface that is small enough to be considered flat.

Area Vector Area

The normal component of a vector  The normal vector to a closed surface is DEFINED as positive if it points OUT of the surface. Remember this definition!

ANOTHER DEFINITION: Element of Flux through a surface E NORMAL AA E  =|E NORMAL| x |  A| (a scalar)

“Element” of Flux of a vector E leaving a surface n is a unit OUTWARD pointing vector.

This flux was LEAVING the closed surface 

Definition of TOTAL FLUX through a surface

Total Flux  =Flux leaving the surface It is a SCALAR You can only do this integration explicitly for “well behaved” surfaces and fields.

Visualizing Flux n is the OUTWARD pointing unit normal.

Definition: A Gaussian Surface Any closed surface that is near some distribution of charge

Remember n E  A  Component of E perpendicular to surface. This is the flux passing through the surface and n is the OUTWARD pointing unit normal vector!

Example Cube in a UNIFORM Electric Field L E E is parallel to four of the surfaces of the cube so the flux is zero across these because E is perpendicular to A and the dot product is zero. Flux is EL 2 Total Flux leaving the cube is zero Flux is -EL 2 Note sign area

Simple Example r q

Gauss’ Law n is the OUTWARD pointing unit normal. q is the total charge ENCLOSED by the Gaussian Surface. Flux is total EXITING the Surface.

Simple Example UNIFORM FIELD LIKE BEFORE E AA E E No Enclosed Charge

Line of Charge L Q

From SYMMETRY E is Radial and Outward

What is a Cylindrical Surface?? Ponder

Looking at A Cylinder from its END Circular Rectangular Drunk

Infinite Sheet of Charge  cylinder E h We got this same result from that ugly integration!

Materials Conductors –Electrons are free to move. –In equilibrium, all charges are a rest. –If they are at rest, they aren’t moving! –If they aren’t moving, there is no net force on them. –If there is no net force on them, the electric field must be zero. THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO!

More on Conductors Charge cannot reside in the volume of a conductor because it would repel other charges in the volume which would move and constitute a current. This is not allowed. Charge can’t “fall out” of a conductor.

Isolated Conductor Electric Field is ZERO in the interior of a conductor. Gauss’ law on surface shown Also says that the enclosed Charge must be ZERO. Again, all charge on a Conductor must reside on The SURFACE.

Charged Conductors E=0 E Charge Must reside on the SURFACE  Very SMALL Gaussian Surface

Charged Isolated Conductor The ELECTRIC FIELD is normal to the surface outside of the conductor. The field is given by: Inside of the isolated conductor, the Electric field is ZERO. If the electric field had a component parallel to the surface, there would be a current flow!

Isolated (Charged) Conductor with a HOLE in it. Because E=0 everywhere inside the surface. So Q (total) =0 inside the hole Including the surface.

A Spherical Conducting Shell with A Charge Inside.

Insulators In an insulator all of the charge is bound. None of the charge can move. We can therefore have charge anywhere in the volume and it can’t “flow” anywhere so it stays there. You can therefore have a charge density inside an insulator. You can also have an ELECTRIC FIELD in an insulator as well.

Example – A Spatial Distribution of charge. Uniform charge density =  charge per unit volume (Vectors) r E O A Solid SPHERE

Outside The Charge r E O R Old Coulomb Law!

Graph R E r

Charged Metal Plate E is the same in magnitude EVERYWHERE. The direction is different on each side. E  E A A

Apply Gauss’ Law  E A A Same result!

Negatively Charged ISOLATED Metal Plate  E is in opposite direction but Same absolute value as before

Bring the two plates together A     e e B As the plates come together, all charge on B is attracted To the inside surface while the negative charge pushes the Electrons in A to the outside surface. This leaves each inner surface charged and the outer surface Uncharged. The charge density is DOUBLED.

Result is ….. A     e e B  E E=0

VERY POWERFULL IDEA Superposition –The field obtained at a point is equal to the superposition of the fields caused by each of the charged objects creating the field INDEPENDENTLY.

Problem #1 Trick Question Consider a cube with each edge = 55cm. There is a 1.8  C charge In the center of the cube. Calculate the total flux exiting the cube. NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE! Easy, yes??

Problem #2 (15 from text) Note: the problem is poorly stated in the text. Consider an isolated conductor with an initial charge of 10  C on the Exterior. A charge of +3mC is then added to the center of a cavity. Inside the conductor. (a) What is the charge on the inside surface of the cavity? (b) What is the final charge on the exterior of the cavity? +3  C added +10  C initial

Another Problem from the book   m,q both given as is  Gaussian Surface Charged Sheet

-2   m,q both given as is  mg qE T Free body diagram 

-3 (all given)

A VERY Hard One! a A uniformly charged VOLUME With volume charge density=  What is E at the end of the vector? VECTOR!!! (We already did this)

-2- a c A cavity is now carved out in The volume. Find the electric field at the point indicated in the Diagram at the end of the vector c.Call the point b and let it be Located at the end of the vector b Solution: We use the SUPERPOSITION of the original volume and a new negatively charged sphere (small) as shown of same charge density but opposite sign Vector sum: b=a+c

-3 And the answer is….. a c E(b)=E first sphere – E second sphere E is constant in The cavity!

A Last Problem A uniformly charged cylinder.   R