# Lecture Set 3 Gauss’s Law

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Lecture Set 3 Gauss’s Law

Calendar for the Weekj Today (Monday) Wednesday, Friday
One or two problems Introduction to the concept of FLUX Wednesday, Friday Gauss’s Law & some problems EXAM DATE WILL BE NEXT WEDNESDAY through Gauss

Protons are projected with an initial speed vi = 9
Protons are projected with an initial speed vi = m/s into a region where a uniform electric field E = (-720 j) N/C is present, as shown in Figure P The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons cross the plane and enter the electric field in Figure

Summary from last week VECTOR VECTOR
(Note: I left off the unit vectors in the last equation set, but be aware that they should be there.)

Electric Field We will now introduce a convenient way to represent the overall electric field in a region of space. It is kinda sorta a map of the field strength. We will then introduce a new concept FLUX We will use this new concept to introduce Gauss’s Law

Look at the “Field Lines” of an infinite sheet of charge …

How do you do that?

Ignore the Dashed Line … Remember last time .. the big plane?
s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 E= E=s/e E=0

NEW RULES (Bill Maher) Imagine a region of space where the ELECTRIC FIELD LINED 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.

Point Charges

They don’t like each other …

Mr. Gauss …

Recall… 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??

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.

The normal component of a vector
q 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
ENORMAL DA E DF=|ENORMAL| | |DA| (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
q

Definition of TOTAL FLUX through a surface

Visualizing Flux 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

Example Cube in a UNIFORM Electric Field
Flux is EL2 Flux is -EL2 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 n is the OUTWARD pointing unit normal.
Flux is total EXITING the Surface. n is the OUTWARD pointing unit normal. q is the total charge ENCLOSED by the Gaussian Surface.

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

Line of Charge L Q

Line of Charge From SYMMETRY E is Radial and Outward

What is a Cylindrical Surface??
Ponder

Looking at A Cylinder from its END
Drunk Circular Rectangular

Infinite Sheet of Charge
cylinder We got this same result from that ugly integration!

Materials Conductors THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO!
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 s 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.
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.

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

Outside The Charge R r O E Old Coulomb Law!

Graph E r R

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

Apply Gauss’ Law s s + + + + + + + + + + + + + + + + A A E
Same result!

Negatively Charged ISOLATED Metal Plate
-s s - E is in opposite direction but Same absolute value as before

Bring the two plates together
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 ….. +2s1 -2s1 A +s s E=0 E=0 E B e e

VERY POWERFULL IDEA Remember It!! 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. Remember It!!

Problem #1 Trick Question
Consider a cube with each edge = 55cm. There is a 1.8 mC 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) +10 mC initial +3 mC added
Note: the problem is poorly stated in the text. Consider an isolated conductor with an initial charge of 10 mC 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? +10 mC initial +3 mC added

Another Problem from the book
Charged Sheet +s a m,q both given as is a Gaussian Surface

-2 a m,q both given as is a a T qE +s mg Free body diagram

-3 (all given)

A Last Problem A uniformly charged cylinder.