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I-2 Gauss’ Law

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Main Topics

• The Electric Flux.• The Gauss’ Law.• The Charge Density.• Use the G. L. to calculate the field of a

• A Point Charge• An Infinite Uniformly Charged Wire• An Infinite Uniformly Charged Plane• Two Infinite Charged Planes

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The Electric Flux

• The electric flux is defined as

It represents amount of electric intensity which flows perpendicularly through a surface, characterized by its outer normal vector . The surface must be so small that can be considered constant there.

• Let’s revisit the scalar product.

AdEd e

E

Ad

E

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The Gauss’ Law I

• Total electric flux through a closed surface is equal to the net charge contained in the volume surrounded by the surface divided by the permitivity of vacuum .

• It is equivalent to the statement that field lines begin in positive charges and end in negative charges.

0

QAdEd e

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The Gauss’ Law II

• Field lines can both begin or end in the infinity.

• G. L. is roughly because the decrease of intensity the with r2 in the flux is compensated by the increase with r2 of surface of the sphere.

• The scalar product takes care of the mutual orientation of the surface and the intensity.

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The Gauss’ Law III• If there is no charge in the volume each field line

which enters it must also leave it. • If there is a positive charge in the volume then

more lines leave it than enter it. • If there is a negative charge in the volume then

more lines enter it than leave it.• Positive charges are sources and negative are sinks

of the field.• Infinity can be either source or sink of the field.

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The Gauss’ Law IV• Gauss’ law can be taken as the basis of

electrostatics as well as Coulomb’s law. It is actually more general!

• Gauss’ law is useful: • for theoretical purposes

• in cases of a special symmetry

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The Charge Density

• In real situations we often do not deal with point charges but rather with charged bodies with macroscopic dimensions.

• Then it is usually convenient to define the charge density i.e. charge per unit volume or surface or length, according to the symmetry of the problem.

• Since charge density may depend on the position, its use makes sense mainly if the bodies are uniformly charged e.g. conductors in equilibrium.

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A Point Charge I

• As a Gaussian surface we choose a spherical surface centered on the charge.

• Intensity is perpendicular to the spherical surface in every point and so parallel (or antiparallel) to its normal.

• At the same time E is constant on the surface, so :

E

0

24

qrEdAEAdE

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A Point Charge II

• So we get the same expression for the intensity as from the Coulomb’s Law :

• Here we also see where from the “strange term” appears in the Coulomb’s Law!

204

)(r

qrE

041

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An Infinite Uniformly Charged Wire I

• Conductive wire (in equilibrium) must be charged uniformly so we can define the length charge density as charge per unit length:

• Both Q and L can be infinite, yet have a finite ratio.

• The wire is axis of the symmetry of the problem.

][ 1 CmLQ

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An Infinite Uniformly Charged Wire II

• Intensity lies in planes perpendicular to the wire and it is radial.

• As a Gaussian surface we choose a cylindrical surface (of some length L) centered on the wire.

• Intensity is perpendicular to the surface in every point and so parallel to its normal.

• At the same time E is constant everywhere on this surface.

E

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Infinite Wire III

• Flux through the flat caps is zero since here the intensity is perpendicular to the normal.

• So :

0

2 L

rLEdAEAdE

02)(

r

rE

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Infinite Wire IV

• By making one dimension infinite the intensity decreases ~ 1/r instead of 1/r2 which was the case of a point charge!

• Again, we can obtain the same result using the Coulomb’s law and the superposition principle but it is “a little” more difficult!

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An Infinite Charged Conductive Plane I

• If the charging is uniform, we can define the surface charge density :

• Again both Q and A can be infinite yet reach a finite ratio, which is the charge per unit surface.

• From the symmetry the intensity must be everywhere perpendicular to the surface.

][ 2 CmA

Q

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Infinite Plane II

• As a Gaussian surface we can take e.g. a cylinder whose axis is perpendicular to the plane. It should be cut in halves by the plane.

• Nonzero flux will flow only through both flat cups (with some magnitude A) since is perpendicular to them.

0

2A

AEdAEAdE

ErE 02

)(

E

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Infinite Plane III

• This time doesn’t change with the distance from the plane. Such a field is called homogeneous or uniform!

• Note that both magnitude and direction of the vectors must be the same if the vector field should be uniform.

E

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Quiz: Two Parallel Planes

• Two large parallel planes are d apart. One is charged with a charge density , the other with -. Let Eb be the intensity between and Eo outside of the planes. What is true?

• A) Eb= 0, Eo=/0

• B) Eb= /0, Eo=0

• C) Eb= /0, Eo=/20

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Homework

• The one from yesterday is due tomorrow!

• The next one will be assigned tomorrow.

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Things to read

• This lecture covers:

Giancoli: Chapter 22

• Advance reading :

Giancoli : Chapter 23-1, 23-2

The scalar or dot product Let

Definition I. (components)

Definition II. (projection)

3

1iiibac

cosbac

Can you proof their equivalence?

^

bac

Gauss’ Law• The exact definition:

0q

AdEd e

• In cases of a special symmetry we can find Gaussian surface on which the magnitude E is constant and is everywhere parallel to the surface normal. Then simply:

0

qAEd e

^

E

Infinite Wire by C.L.– die hard!Only radial component Er of is non-zero

2

sin

sin

;;sin

dxkdE

arctgrxEE

r

rr

• We have to substitute all variables using and integrate from 0 to :

rr

k

r

dkEz

00 2

2sin2

2

• “Quiz”: What was easier?

^

E