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Electrostatic:Electric Field

Chapter 22 Halliday-Resnick 9th Ed.

Tuesday, 31 January 2012

0. Create a sketch (diagram)1. Define the charge distribution, and the corresponding

spatial element of distribution ( , , , etc.)2. Relate the charge with the spatial element1. = for 1D2. = for 2D

3. Describe the expression for on the point of interest,substitute any possible variables with regards to point 1 &2.

4. Complete the sketch by drawing the direction of , checkfor symmetry and any possible cancelling components(that needs no further attention).

5. Solve the integral, if the result is to be stated in totalcharge Q of the distribution, replace od the with thecorresponding charge density distribution.

Hint:

=== /22 +

= /22 + ⁄

= 2 + ⁄

≫ → 2 + ≈ = = − ̂

( ) ( )

( ) + ( )

==== == / sin= / cos= / tan= tan= cos= cos cos sin

= sin/ = − cos | /20= −

=

=

= cos cos cos= cos/ = sin | /2− /2= =

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Electrostatic:Gauss’ Law

Chapter 23 Halliday-Resnick 9th Ed.

Tuesday, 30 January 2012

Physics: Solving seemingly complex problem →using symmetry

Gauss’ Law: Imaginary surface enclosing charge distribution Relates the electric fields at points an a closed

Gaussian surface to the net charge enclosed by thesurface

Flux of Electric Field:The electric flux through a Gaussian surface isproportional to the net number of electric field linespassing through that surface.

Flux of Electric Field:

For a uniform electric fieldΦ = ∙ Non-uniform electric fieldΦ = ∙ Δ element of area Over a closed surfaceΦ = ∮ ∙

Number of electric fieldThrough a surface

?

Flux of Electric Field:

For a uniform electric fieldΦ = ∙ Non-uniform electric fieldΦ = ∙ Δ element of area Over a closed surfaceΦ = ∮ ∙

Number of electric fieldThrough a surface

The figure here shows a Gaussiancube of face area immersed in auniform electric field that hasthe positive direction of theaxis. In terms of E and A, what isthe flux through(a) the front face (which is in the

plane),(b) The rear face,(c) the top face, and(d) the whole cube?

(a) Φ = ∙ = ∙ cos = cos 0° =(b) Φ = ∙ = ∙ cos = cos 180° = −(c) Φ = ∙ = ∙ cos = cos 90° = 0(d) 0 (?)

Cylindrical surface

Flux through a closed cylinder, uniform field ?

Flux through a closed cube,nonuniform field

What is the total flux through the cubicalsurface if the Electrical Field passing throughthe surface is described as = ̂ + 4 ̂ ?

Relates the electric fields at points an a closedGaussian surface to the net charge enclosed bythe surface Φ =∙ =

Gauss’ Law:

= = 1 ?Gauss' Law and Coulomb's Law

=

= cos cos cos= cos/ = sin | /2− /2= =

Coulomb'sLaw Approach

= = ?

Gauss’Law Approach

0. Draw the diagram1. Select a closed surface2. Solve ∮ ∙3. Solve ∮ ∙ =