This box contains

1. a net positive charge.2. no net charge. 3. a net negative charge.4. a positive charge. 5. a negative charge.

More Quiz Fun!

This box contains

1. a net positive charge.2. no net charge. 3. a net negative charge.4. a positive charge. 5. a negative charge.

More Quiz Fun!

FLUX

d Ecos dAθΦ =

The amount of FLOW perpendicular to a surface!

dΦ = ⋅∫vv

E A

Remember that dot products ‘select’ mutually perpendicular components of vectors. For flux, a vector normal to the surface area determines the direction. On a closed surface, positive flux always points OUTWARDS.(-) (()) (+)

ˆd dAΦ = ⋅vE n

Electric FLUX

Gauss’s Law

E dε

Φ = ⋅∫vv

Ñ in

0

qE A =

The net electric flux through ANY closed surface is due to the net charge contained inside that surface (divided by ε0)!

Maxwell’s EquationsGauss is #1!

Spherical Shell

Spherical Shell0inE =

# 57. A solid, insulating sphere of radius a has a uniform charge density ρ and a total charge Q. Concentric with this sphere is an uncharged, conducting hollow sphere whose inner and outer radii are b and c, as shown. (a) Find the magnitude of the electric field in the regions r < a,  a < r < b, b < r < c, and r > c. (b) Determine the induced charge per unit area on the inner and outer surfaces of the hollow sphere.

A conductor in an external Electric Field

Faraday Cage

•Mobile phones and radios may have no reception inside elevators or similar structures. Some traditional architectural materials act as Faraday shields in practice.•Coaxial cables are in fact data cables wrapped by a hollow, flexible conductor, effectively a Faraday cage.•Cars and aircraft. When lightning strikes an aircraft or a car the electric currents induced on it are forced to travel on the outer skin of the vehicle's body.

A Faraday cage or Faraday shield is an enclosure formed by conducting material, or by a mesh of such material. Such an enclosure blocks out external static electrical fields. Faraday cages are named after physicist Michael Faraday, who built one in 1836. Externally applied electric fields produce forces that rearranges the charges in the conductor so as to cancel the applied external field inside.

Quickie Review QuestionA point charge q is located at the center of a uniform ring having linear chargedensity λ and radius a, as shown. Determine the total electric flux through asphere centered at the point charge and having radius R, where R < a.

0

inqε

Φ =

Only the charge INSIDE contributes to the Flux!

Why/How Use Gauss’s Law?•Electric fields of some continuous distributions of charge can be found much more easily than using Coulomb’s Law.•Gauss’s Law is valid for moving charges, but Coulomb’s law is not. Thus, Gauss’s Law is more fundamental.•Use Gauss’s Law to find the Electric field when you have symmetry and already know what the field should look like.•The Gaussian surface must have the same symmetry and every part of it must either be tangent to or perpendicular to the electric field.•You need not enclose all the charge within the Gaussian surface.

A Line of ChargeCoulomb or Gauss?

Solid Sphere

E dε

Φ = ⋅∫vv

Ñ in

0

qE A =

Inside the Earth the Gravitational Force is Linear.

Acceleration decreases as you fall to the center (where your speed is the greatest) and then the acceleration increases

but in the opposite direction, slowing you down to a stop at the other end…but then you would fall back in again,

bouncing back and forth forever!

Gravitational Force INSIDE the Earth

Cylinder

ADD PROBLEM 70!!!

29. Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ. Find the electric field at distance r from the axis where r < R

Infinite Plane

02E σ

ε=

Parallel Plates

0

E σε

=

The total electric flux through this box is1. 0 Nm2/C.2. 1 Nm2/C.3. 2 Nm2/C.4. 4 Nm2/C.5. 6 Nm2/C.

The total electric flux through this box is1. 0 Nm2/C.2. 1 Nm2/C.3. 2 Nm2/C.4. 4 Nm2/C.5. 6 Nm2/C.

Conductors in Electrostatic Equilibrium

0 0

insurface

Q AAE σε ε

Φ = = =

0surfaceE σ

ε=

At Electrostatic Equilibrium:•the field inside is zero.•External field lines are perpendicular to the surface.•External field lines end on charges on the surface.•Any excess charges lies entirely on the surface. •At Electrostatic Equilibrium, the conducting surface shields the interior of external fields. •The magnitude of the Electric field near a conducting surface is: σ/ε0.•On asymmetric objects, charges collect at pointy edges, such as on lightening rods.

Free electrons move to the surface until the interior is neutral. They move until the parallel component of the field is zero and the conductor is at Electrostatic Equilibrium.

Fig 24-CO, p.737