Electric Fields and Forces/E F q=

r r

F qE=r r

A region of space has an electric field of 104 N/C, pointing in the plus x direction. At t = 0, an object of mass 1 g carrying a charge of 1 µC is placed at rest at x = 0. Where is the object at t = 4 sec?

A) x = 0.2 m C) x = 20 mB) x = 0.8 m D) x = 80 m

F ma=r r

/a qE m=rr

Electric Flux•Electric Flux is the amount of electric field flowing through a surface•It is the electric field times the area, when the field is perpendicular to the surface•It is zero if the electric field is parallel to the surface•Normally denoted by symbol ΦE.•Units are N·m2/C

Electric Field

Electric Field

ΦE = EA ΦE = 0

Electric Flux•When electric field is at an angle, only the part perpendicular to the surface counts

•Multiply by cos θ

θ

E

ΦE = EnA= EA cos θ

En•For a non-constant electric field, or a curvy surface, you have to integrate over the surface

cosE E dA E dAθΦ = ⋅ =∫ ∫rr

R

Electric Flux•What is electric flux through surface surrounding a charge q?

24E R EπΦ =

4E ek qπΦ =

2E ek qπΦ = 2 ek qπ+4 ek qπ=

0+

4 ek qπ=224 ek q

RR

π=

Answer is always 4πkeq

charge q

Gauss’s Lawin

0E

qε

Φ =•Flux out of an enclosed region depends only on total charge inside

A positive charge q is set down outside a sphere. Qualitatively, what is the total electric flux out of the sphere as a consequence?

A) PositiveB) NegativeC) ZeroD) It is impossible to tell from the given information

charge q

Gauss’s Law

charge q

4E eE dA k qπΦ = ⋅ =∫rr

Ñ

charge q’charge q’’

4 'ek qπ+ ( )4 'ek q qπ= +

4 ''ek qπ+

in4E eE dA k qπΦ = ⋅ =∫rr

in

0

qε

=

Applying Gauss’s Law•Can be used to determine total flux through a surface in simple cases•Must have a great deal of symmetry to use easily

1 2 3 1 20

E S S S E Eqε

Φ = = Φ + Φ + Φ + Φ + Φ

charge q

Charge in a long triangular channelWhat is flux out of one side?

3 0S= Φ +

03S

qε

Φ =

Applying Gauss’s Law

•Infinite cylinder radius R charge density ρ•What is the electric field inside and outside the cylinder?

L

R

•Electric Field will point directly out from the axis•No flux through end surfaces

r

•Draw a cylinder with the desired radius inside the cylindrical charge

AEΦ = 2 rLEπ=0

V ρε

=2

0

r Lπ ρε

=0

inqε

=

02r

Eρε

=

Applying Gauss’s Law

•Infinite cylinder radius R charge density ρ•What is the electric field inside and outside the cylinder?

L

R

•Electric Field will point directly out from the center•No flux through endcaps

r

•Draw a cylinder with the desired radius outside the cylindrical charge

AEΦ = 2 rLEπ=0

V ρε

=2

0

R Lπ ρε

=0

inqε

=2

02R

Erρ

ε=

Applying Gauss’s Law

•Sphere radius R charge density ρ. What is E-field inside?

RSphere volume:

V = 4πa3/3

Sphere area: A = 4πa2

•Draw a Gaussian surface inside the sphere of radius r

r

What is the magnitude of the electric field inside the sphere at radius r?

A) ρR3/3ε0r2

B) ρr2/3ε0RC) ρR/3ε0 D) ρr/3ε0

AEΦ = 24 r Eπ=

0

V ρε

=3

0

43

rπ ρε

=

0

inqε

=

03r

Eρε

=

Conductors in Equilbirum•A conductor has charges that can move freely•In equilibrium the charges are not moving•Therefore, there are no electric fields in a conductor in equilibrium

F qE ma= =r r r

0 0inq E dAε ε= Φ = ⋅∫rr

= 0

•The interior of a conductor never has any charge in it•Charge on a conductor is always on the surface

= 0

Electric Fields near Conductors•No electric field inside the conductor•Electric field outside cannot be tangential – must be perpendicular

Surface charge σ

0 0 AEΦ = + +0

Aσε

=0

inqε

=

Area A

•Add a gaussian pillbox that penetrates the surface

0

ˆE nσε

=r

•Electric field points directly out from (or in to) conductor

Conductors shield chargesNo net charge

Charge q•What is electric field outside the spherical conductor?

•Draw a Gaussian surface•No electric field – no charge•Inner charge is hidden – except

Charge -q •Charge +q on outside to compensate•Charge distributed uniformlyCharge +q

20

ˆ4

qrE

rπε=

r