JJlecture++Notes+5+PDF

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1/13/2012 1 Announcements 1. Lecture Notes available at: https://sakai.iitd.ac.in/portal login yourself; Fields and Waves, Resources 2. Time slot to meet the teacher: Wednesdays 5.30 to 6.30 pm in respective office rooms 1 Brief Summary For a single charge at origin, we have the electric field Flux of Electric Field through a surface S For any closed surface, Gauss’s Law Gauss’s Law in differential form Applications of Gauss’s Law

Transcript of JJlecture++Notes+5+PDF

Page 1: JJlecture++Notes+5+PDF

1/13/2012

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Announcements

1. Lecture Notes available at:https://sakai.iitd.ac.in/portallogin yourself; Fields and Waves, Resources

2. Time slot to meet the teacher:Wednesdays 5.30 to 6.30 pm inrespective office rooms

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Brief SummaryFor a single charge at origin,we have the electric field

Flux of Electric Field through a surface S

For any closed surface, Gauss’s Law

Gauss’s Law in differential form

Applications of Gauss’s Law

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Some important results from H.W. onapplications of Gauss’s law

• E field outside a uniformly charged sphere (solid / shell) is exactlythe same as if all charges were at center.

• E field inside a uniformly charged hollow sphere is zero.• E field from an infinite plane: , independent of distance.• E field from two infinite planes of opposite charges:

zero outside, inside.

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Dirac Delta Function [Sect. 1.5 of Griffiths]We now have the concept of how a point charge is to be described.

Consider the function

We know that this is a diverging function.

So, let us take the divergence of this function,

We thus have the paradoxical situation that a point charge at theorigin produces a Coulomb field, but the lines do not start at theorigin!

0)1(11122

22

rrr

rrr

rr

ˆ12

???

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Dirac Delta Function ……

rrv

0

4ˆsinˆ1 2

0

2

02

dv

rddRrR

adv

dvadv

Apply divergence theorem by considering a sphere of radius R centeredat the origin

Does this mean that divergence theorem is false?

If the divergence theorem is right, we should get 4 dv

Dirac Delta Function …

1)(

000

)(

dxx

andxifxif

x

To account for this paradox, we define a new kind of function.

In the first instance in only one space variable x (One dimension).

This function is known as the Dirac Delta function and is definedbelow.

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The Dirac delta function is not really a function but whatmathematicians call a distribution, but we will treat it just like afunction most of the time.

δ(x) is limit of a sequence of functions , e.g. a RectangleRn(x) of height ‘n’ and width ‘1/n’ in the limit or anIsosceles triangles Tn(x) of height ‘n’ and base 2/n in the samelimit. There are many other examples like this.

n

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Some properties of delta function

Delta function is always intended for use under an integral sign

Example 1.5

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Three dimensional delta function

We can also write these results in spherical polar coordinates as

)()()( 3 afdarrfallspace

1)()()()(

)()()()(

3

3

dzdydxzyxdr

zyxr

allspace

Paradox solved using delta function

• We can now explain the paradox related with the Coulombfield

• Everywhere except at the origin the divergence vanishes, at theorigin we have a delta function sitting there to give us thecorrect form for Gauss’s law or equivalently Coulombs law.

• So, we define

Also,

Now, Paradox solved!

2ˆrr

)(4ˆ 32 rrr

4 dv

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H.W.

• Problems 1.43, 1.44, 1.47

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12

Divergence of E

Using divergence theorem,Gauss’s law

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Curl of E

Line integral is zero over closed path,

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means

Stokes’ theorem

For many charges using principle of superposition

So, we have, in electrostatics,