Chapter 22

Gauss’s LawLecture 3

Dr. Armen Kocharian

Electric FluxElectric flux is the dot (scalar) product of the field vector E and area vector A=An (n-normal to surface, A magnitude)

Φ = E ·A= E ·n A=EA cos θ

If E||n then flux is product E and A, perpendicular to the field

ΦE = EA

Electric Flux, General AreaThe electric flux is proportional to the number of electric field lines penetrating some surfaceThe field lines may make some angle θ with the perpendicular to the surfaceThen ΦE = EA cos θ

Electric Flux, Interpreting the Equation

The flux is a maximum when the surface is perpendicular to the fieldThe flux is zero when the surface is parallel to the fieldIf the field varies over the surface, Φ = EA cos θ is valid for only a small element of the area

Electric Flux, General

In the more general case, look at a small area element

In general, this becomescosE i i i i iE A θΔΦ = Δ = ⋅ ΔE A

0surface

limi

E i iAE A d

Δ →Φ = ⋅ Δ = ⋅∑ ∫ E A

Electric Flux, finalThe surface integral means the integral must be evaluated over the surface in questionIn general, the value of the flux will depend both on the field pattern E and on the surface AThe units of electric flux will be N.m2/C

[ ] [ ][ ]

Φ =

Φ = = 2N mC

E

E

EA

E A

Electric Flux, Closed SurfaceAssume a closed surfaceThe vectors ΔAi point in different directions

At each point, they are perpendicular to the surfaceBy convention, they point outward from enclosed volume

Flux Through Closed Surface, cont.

At (1), the field lines are crossing the surface from the inside to the outside; θ < 90o, Φ is positiveAt (2), the field lines graze surface; θ = 90o, Φ = 0At (3), the field lines are crossing the surface from the outside to the inside;180o > θ > 90o, Φ is negative

Φ<0 Φ=0 Φ>0

Flux Through Closed Surface, final

The net flux through the surface is proportional to the net number of lines leaving the surface

This net number of lines is the number of lines leaving the surface minus the number entering the surface

If En is the component of Eperpendicular to the surface, then

E nd E dAΦ = ⋅ =∫ ∫E A

Gauss’s Law, IntroductionGauss’s law is an expression of the general relationship between the net electric flux through a closed surface and the charge enclosed by the surface

The closed surface is often called a gaussian surface

Gauss’s law is of fundamental importance in the study of electric fields

Gauss’s Law – General A positive point charge, q, is located at the center of a sphere of radius rThe magnitude of the electric field everywhere on the surface of the sphere of radius r is the same|E| = ke|q| / r2

E||dA

Gauss’s Law – General, cont.The field lines are directed radially outward and are perpendicular to the surface at every point

This will be the net flux through the gaussian surface, the sphere of radius rWe know E = keq/r2 and Asphere = 4πr2,

E d E dAΦ = ⋅ =∫ ∫E A

4E eo

qπk qε

Φ = =

Gauss’s Law – General, notesThe net flux through any closed surface surrounding a point charge, q, is given by q/εoand is independent of the shape of that surfaceThe net electric flux through a closed surface that surrounds no charge is zeroSince the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as

( )1 2d d⋅ = + ⋅∫ ∫E A E E A…

Gauss’s Law – FinalGauss’s law states

qin is the net charge inside the surfaceE represents the electric field at any point on the surface

E is the total electric field and may have contributions from charges both inside and outside of the surface

Although Gauss’s law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations

E A inE

o

qdε

Φ = ⋅ =∫

Applying Gauss’s LawTo use Gauss’s law, you want to choose a gaussian surface over which the surface integral can be simplified and the electric field determinedTake advantage of symmetryRemember, the gaussian surface is a surface you choose, it does not have to coincide with a real surface

Find electric flux through each surface, Problem

Flux through various surfaces is:

in

0E

qΦ =

∈Through

1S0 0

2E

Q Q Q− +Φ = = −

∈ ∈

2S0

0EQ Q+ −

Φ = =∈

3S0 0

2 2E

Q Q Q Q− + −Φ = = −

∈ ∈

4S 0EΦ =

Conditions for a Gaussian SurfaceTry to choose a surface that satisfies one or more of these conditions:

The value of the electric field can be argued from symmetry to be constant over the surfaceThe dot product of E.dA can be expressed as a simple algebraic product EdA because E and dAare parallelThe dot product is 0 because E and dA are perpendicularThe field can be argued to be zero over the surface

Uniform electric field, Exp

Find the flux through surface of the cube, edge length l

E is parallel to x direction

Φ = Φ +Φ +Φ +Φ +Φ +Φ =

⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫ ∫

1 2 3 4 5 6

1 2 3 4 5 6

E E E E E E E

d d d d d dE A E A E A E A E A E A

Uniform electric field, Exp, cont.

E is parallel to x directionE is perpendicular to y and z directions

( )

( )

( )

⋅ = = − = − = −

⋅ = = = =

⋅ = =

⋅ = ⋅ = ⋅ = ⋅ =

Φ +Φ = Φ = Φ = Φ = Φ = Φ =

∫ ∫ ∫

∫ ∫ ∫

∫ ∫

∫ ∫ ∫ ∫

0 2

1 1 1

0 2

2 2 1

0

3 3

3 4 5 61 2 3 4 5 6

cos180

cos0

cos90 0

0

0, 0 0E E E E E E E

d E dA EdA EA El

d E dA EdA EA El

d E dA

d d d d

E A

E A

E A

E A E A E A E A

Field Due to a Point Charge, Exp

Choose a sphere as the gaussian surface

E is parallel to dA at each point of the spherical surface

2

2 2

(4 )

4

Eo

eo

qd EdAε

E dA E πr

q qE kπε r r

Φ = ⋅ = =

= =

= =

∫ ∫

∫

E A

Field Due to a Spherically Symmetric Charge Distribution

Select a sphere as the gaussian surfaceFor r >a

in

2 24

Eo

eo

qd EdAε

Q QE kπε r r

Φ = ⋅ = =

= =

∫ ∫E A

Spherically Symmetric, Exp. 24.5.

Select a sphere as the gaussian surface, r < aqin < Q

( )Φ = ⋅ = = =

= =

⇒ = = =

∫ ∫ 2 in

3in

2 2

3 300

4

4 1 ,4 3 4

34 33

3

Eo

o o

e

qd EdA E πrε

q πrE ρπε r πε rρ Q QE r r k r

πa εε a

E A

= = =

= =

3 3

3

34 4

34

3in

Q Q QρπaV πa

πrq ρV ρ

Spherically Symmetric Distribution, final

Inside the sphere, Evaries linearly with r

E → 0 as r → 0The field outside the sphere is equivalent to that of a point charge located at the center of the sphere

Field Due to a Thin Spherical Shell

Use spheres as the gaussian surfacesWhen r > a, the charge inside the surface is Q and E = keQ / r2

When r < a, the charge inside the surface is 0 and E = 0

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 (bottom and top) of the cylinder is zero since the field is parallel to these surfaces

⊥ =Φ = Φ =

0, 900Bot Top

E A θ

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

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