1n̂

2n̂

3n̂ S1

S2S3

2 oE σ

ε=

(23-15)

Electric field generated by a thin, infinite, non-conducting uniformly charged sheet.We assume that the sheet has a positive charge of surface density . From symmetry, the electric field vector is perpendicular to the sheet and has a constant magnitude. Furthermore it points away from the sheet. We choose a cylindrical Gaussian surface Swith the caps of area A on either side of the sheet as shown in the figure.

σEr

S2

S1

iEr

2n̂

oEr

1̂n

(23-18)

Electric field generated by a uniformly charged sphere of radius R and charge q. Outside the sphere：Consider a Gaussian surfaceWhich is a sphere with radiusr < R and whose center coincides with that of the charge shell. The electric field flux

Thus

S1

24 / /r E q qo enc o oπ ε εΦ = = =

24

qEo roπε=

Inside the sphere：Consider a Gaussian surface which is a sphere with radius r < R and whose center coincides with that of the charge shell.

The electric field flux

Thus

S2

24qencr Eio

πε

Φ = =

3 3 34 / 243 3 34 /

r R R qq q q r Eenc iR r r o

π πεπ

= = → =

34

qE ri Roπε

⎛ ⎞⎜ ⎟=⎜ ⎟⎝ ⎠

(23-18)

24oo

qErπε

=

34io

qE rRπε

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

R

34 o

qRπε

r

E

O

S2

S1

iEr

2n̂

oEr

1n̂

(23-19)

Electric field generated by a uniformly charged sphere of radius R and charge q. Summary

Field at a Distance from a Line of ChargeSelect a cylindrical charge distribution

The cylinder has a radius of r and a length of ℓ

E is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface

Field Due to a Line of Charge, cont.

The end view confirms the field is perpendicular to the curved surfaceThe field through the ends of the cylinder is 0 since the field is parallel to these surfaces

Field Due to a Line of Charge, final

Use Gauss’s law to find the field

( )

in

2

22

Eo

o

eo

qd EdAε

λE πrε

λ λE kπε r r

Φ = ⋅ = =

=

= =

∫ ∫E A

ll

旄

Field Due to a Plane of Charge

E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the planeChoose a small cylinder whose axis is perpendicular to the plane for the gaussian surface

Field Due to a Plane of Charge, cont

E is parallel to the curved surface and there is no contribution to the surface area from this curved part of the cylinderThe flux through each end of the cylinder is EA and so the total flux is 2EA

Field Due to a Plane of Charge, final

The total charge in the surface is σAApplying Gauss’s law

Note, this does not depend on rTherefore, the field is uniform everywhere

22E

o o

σA σEA and Eε ε

Φ = = =

Electrostatic Equilibrium

When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium

Properties of a Conductor in Electrostatic Equilibrium

The electric field is zero everywhere inside the conductorIf an isolated conductor carries a charge, the charge resides on its surfaceThe electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/εoOn an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest

Property 1: Einside = 0Consider a conducting slab in an external field EIf the field inside the conductor were not zero, free electrons in the conductor would experience an electrical forceThese electrons would accelerateThese electrons would not be in equilibriumTherefore, there cannot be a field inside the conductor

Property 1: Einside = 0, cont.Before the external field is applied, free electrons are distributed throughout the conductorWhen the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external fieldThere is a net field of zero inside the conductorThis redistribution takes about 10-15s and can be considered instantaneous

Property 2: Charge Resides on the SurfaceChoose a gaussian surface inside but close to the actual surfaceThe electric field inside is zero (prop. 1)There is no net flux through the gaussian surfaceBecause the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface

Property 2: Charge Resides on the Surface, cont

Since no net charge can be inside the surface, any net charge must reside onthe surfaceGauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor

Property 3: Field’s Magnitude and DirectionChoose a cylinder as the gaussian surfaceThe field must be perpendicular to the surface

If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium

Property 3: Field’s Magnitude and Direction, cont.

The net flux through the gaussian surface is through only the flat face outside the conductor

The field here is perpendicular to the surface

Applying Gauss’s law

Eo o

σA σEA and Eε ε

Φ = = =

Conductors in Equilibrium, example

The field lines are perpendicular to both conductorsThere are no field lines inside the cylinder

Derivation of Gauss’s LawWe will use a solid angle, ΩA spherical surface of radius r contains an area element ΔAThe solid angle subtended at the center of the sphere is defined to be 2

ArΔ

Ω =

Some Notes About Solid Angles

A and r2 have the same units, so Ω is a dimensionless ratioWe give the name steradian to this dimensionless ratioThe total solid angle subtended by a sphere is 4π steradians

Derivation of Gauss’s Law, cont.

Consider a point charge, q, surrounded by a closed surface of arbitrary shapeThe total flux through this surface can be found by evaluating E.ΔA for each small area element and summing over all the elements

Derivation of Gauss’s Law, final

The flux through each element is

Relating to the solid angle

where this is the solid angle subtended by ΔA

The total flux is

( ) 2coscosE e

A θE θ A k qr

ΔΦ = ⋅ Δ = Δ =E A

2cosA θr

ΔΔΩ =

2cos

E e eo

dA θ qk q k q dr ε

Φ = = Ω =∫ ∫旄

Derivation of Gauss’s Law –Notes

This result matches the earlier result for the equation for Gauss’s lawThis derivation was independent of the shape of the closed surfaceThis derivation was independent of the position of the charge within the surface