640 likes | 653 Views
Lecture Set 3 Gauss’s Law. Spring 2007. Calendar for the Week. Today (Wednesday) One or two problems on E Introduction to the concept of FLUX Friday 7:30 session if wanted Quiz – Chapter 23 – Electric Field Gauss’s Law & some problems EXAM APPROACHING. Note.
E N D
Lecture Set 3Gauss’s Law Spring 2007
Calendar for the Week • Today (Wednesday) • One or two problems on E • Introduction to the concept of FLUX • Friday • 7:30 session if wanted • Quiz – Chapter 23 – Electric Field • Gauss’s Law & some problems • EXAM APPROACHING
Note • Pleas read the material on Electric Dipoles • We will NOT cover it in class but it is part of the course. • We will use it later in the semester. • It could show up on the exam.
The figure shows two concentric rings, of radii R and R ' = 2.95R, that lie on the same plane. Point P lies on the central z axis, at distance D = 2.20R from the center of the rings. The smaller ring has uniformly distributed charge +Q. What must be the uniformly distributed charge on the larger ring if the net electric field at point P due to the two rings is to be zero?[-3.53] Q
Ignore the Dashed Line … Remember last time .. the big plane? s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 E=0 E=s/e0 E=0 We will use this a lot!
NEW RULES (Bill Maher) • Imagine a region of space where the ELECTRIC FIELD LINES HAVE BEEN DRAWN. • The electric field at a point in this region is TANGENT to the Electric Field lines that have been drawn. • If you construct a small rectangle normal to the field lines, the Electric Field is proportional to the number of field lines that cross the small area. • The DENSITY of the lines. • We won’t use this much
How about this?? • Positive point charge • Negative point charge • Large Sheet of charge • No charge • You can’t tell from this
All of the E vectors in the bottom box are twice as large as those coming from the top box. The top box contains a charge Q. How much charge do you think is in the bottom box? • Q • 2Q • You can’t tell • Leave me alone.
So far … • The electric field exiting a closed surface seems to be related to the charge inside. • But … what does “exiting a closed surface mean”? • How do we really talk about “the electric field exiting” a surface? • How do we define such a concept? • CAN we define such a concept?
Mr. Gauss answered the question with.. Yup .. Gauss's Law
Another QUESTION: Not Quite Solid Surface Given the electric field at EVERY point on a closed surface, can we determine the charges that caused it??
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??
Still another question Given a small area, how can you describe both the area itself and its orientation with a single stroke!
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 An, 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.
E n En The “normal component” of the ELECTRIC FIELD
E n En DEFINITION FLUX
q We will be considering CLOSED surfaces The normal vector to a closed surface is DEFINED as positive if it points OUT of the surface. Remember this definition!
“Element” of Flux of a vector E leaving a surface For a CLOSED surface: n is a unit OUTWARD pointing vector.
q This flux is LEAVING the closed surface.
Flux is • A vector • A scaler • A triangle
VisualizingFlux n is the OUTWARD pointing unit normal.
Definition: A Gaussian Surface Any closed surface that is near some distribution of charge
Remember Component of E perpendicular to surface. This is the flux passing through the surface and n is the OUTWARD pointing unit normal vector! n E q q A
Flux is -EL2 ExampleCube in a UNIFORM Electric Field Flux is EL2 E area L Note sign 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. Total Flux leaving the cube is zero
Simple Example r q
Gauss’ Law Flux is total EXITING the Surface. n is the OUTWARD pointing unit normal. q is the total charge ENCLOSED by the Gaussian Surface.
Simple ExampleUNIFORM FIELD LIKE BEFORE E No Enclosed Charge A A E E
Q L Line of Charge
Line of Charge From SYMMETRY E is Radial and Outward
Drunk Looking at A Cylinder from its END Circular Rectangular
Infinite Sheet of Charge +s h E cylinder 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 Charge Must reside on the SURFACE - - E=0 - - E - s 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 conductor. So Q (total) =0 inside the hole Including the surface.
A Spherical Conducting Shell withA Charge Inside. A Thinker!
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.
E O r Example – A Spatial Distribution of charge. Uniform charge density = r = charge per unit volume (Vectors) A Solid SPHERE