Ch. 27 : GAUSS’ LAW

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# Ch. 27 : GAUSS’ LAW - PowerPoint PPT Presentation

Ch. 27 : GAUSS’ LAW. Electric Flux. Gauss’ Law. Applications of Gauss’ Law. Conductors. Flux of a Vector Field. rate of flow of a fluid through the surface A:  vA ,.  vA cosØ (when the rectangle is tilted at an angle Ø.). Flux of a Vector Field.

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Presentation Transcript
Ch. 27: GAUSS’ LAW
• Electric Flux
• Gauss’ Law
• Applications of Gauss’ Law
• Conductors
Flux of a Vector Field

rate of flow of a fluid through the surface A:  vA ,

vA cosØ

(when the rectangle is tilted at an angle Ø.)

Flux of a Vector Field

This rate is defined as the FLUXof the velocity vector v

=> A measure of number of field lines passing through area.

Electric flux

For a flat surface  to the field lines

the electric flux

ØE = EA

S.I. unit (Nm2/C)

Electric flux

If surface area is not perpendicular to the field,

A’ = A cos 

The # of lines through A’ = the # through A

=> the flux through A’ = the flux through A

ØE = EA cos 

Electric Flux for non uniform electric field
• divide the surface into large number of small elements, each of area DA.
• define a vector DAi ,

magnitude = the area of the ith surface element

direction = perpendicular to the surface element

Electric Flux for non uniform electric field

The flux through the element

DFE = EiDAi cos q = Ei .DAi

Electric Flux for non uniform electric field

the total flux through the surface

= sum of contributions of all elements.

Electric Flux through Closed Surface

Flux through area element

1 : positive

2 : zero

3: negative

Example: Electric Flux through a Cube

E along x-axis: To find the net electric flux.

1

2

3

4

5

6

=0

+0

+0

+0

+0

FE=

+El2

-El2

Electric Flux Through a Cone

A horizontal uniform field E penetrates the cone.

To find E that enters the left hand side of the cone.

E = - E (1/2  2R  h) = - E R h

Ex: 27-8:

To find the flux through the netting: relative to the outward normal.

Ans: E = -Ea2

Ex: 27-11:

A point charge q is placed at one corner of a cube of edge a .

To find the flux through each of the cube faces:

Ans: E = 0 (for each face touching q);

= 1/240 (for other faces)

GAUSS’ LAW

This law is useful to calculate E set up by a point charge/collection of charges.

Gauss’ Law relates the net flux through any closed surface to the net charge enclosed by the surface; as:

Gaussian surface

Spherical Gaussian surfaces

positive point charge

negative point charge.

+

Choosing the Gaussian surface

A spherical Gaussian surface of radius r centered on the charge +q.

S3

S2

S1

q

Flux is independent of the radius r of

the spherical Gaussian surface.

Flux is independent of the size of Gaussian surface

Flux through S1 = q / o

the number of field lines through S1

= number of field lines through S2 or S3.

=> the net flux through any closed surface is independent of the shape of thatsurface

Gauss’s Law – Charge outside closed surface

A point charge is located outside closed surface.

The number of lines entering the surface equals the number of lines leaving the surface.

The net flux within the surface is zero.

Quick Quiz

A spherical gaussian surface surrounds a point charge q. Describe what happens to the total flux through the surface if :

1.the radius of the sphere is doubled

2.the surface is changed to a cube

3. the charge is tripled

4.the charge is moved to another location inside the surface.

Gauss’ Law in Differential Form:

The integral form:

When the charge contained within the surface S is continuously distributed within a volume of charge density :

The divergence can be interpreted as the number of field lines starting/terminating at a given point:

=> The number of field lines starting or terminating at a given point is proportional to the charge density at that point.

Divergence of E

Electric field at any field point r:

THE CURL of E

Electric field at a field point r, due to a point charge at origin:

Recall: