cilindro ejercicio

8/17/2019 cilindro ejercicio

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Electron and proton and other particles ( chargeparticle ) exert a long – range force on one

another , like gravitation , this force is inversely

proportional to the square of the distancebetween the particle ; but unlike gravitation ,

this force is attractive or repulsive , depending

on what particle are involved .

Coulos law :

The electric force that a point charge exerts on a

point charge q at a distance r has magnitude


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The force is directed along the line joining thecharges . the force is repulsive if the charge have thesame sign , and it is attractive if the charge have

opposite sign .

If the charge at the origin of our coordinate system ,

thenThe potential energy associated with the coulombforce is

The eleti field ; Gauss la :

For a charge placed at the origin of coordinates , wedefine the electric field that this charge produces atsome distance r as


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According to eq.(2) , we can then express the force

on a second charge as

• The net electric field of several charges if thecharges are with position , then

the net electric field

• In the calculation of electric field , to describe thedistribution of charges by a continuous chargedensity . Since is mount of charge in the

volume near the point .


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Gausss law says that the total normal outward

flux over a closed surface in an electric field is

equal to times the charge enclosed by thesurface. returning to the div. theorem we can

write,

From div. theorem we haveThereforeLet us now suppose that charge uniformly

distribution in space and charge density then

Gauss's equation become this is differential

euatio of Gauss la .


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Application :

1-We wont to find out the electric field at a point pdue to a point charge q placed at o as shown in the

diagram .Sol.

Consider a sphere of radius r passing through thepoint p . let the electric field at p be . then by

Gauss law

Or

.


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2-To find out the electric field at a point p due to

an infinite line of charge :

Consider a cylindrical surface through the point pas shown in the diagram and let be the electric

field at the point p . by Gauss law


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Ex. Find the electric field to the electric dipole

consisting of two point like charge at z ?

Sol.

For 2 then


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at θ=0 p location on the z direction (r)

For θ=90 ,ф=0


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The electrostatic potentialPoissos euatio :We know that a vector field of zero curl is conservative andcan be express as the gradient of scalar function . theelectrostatic field satisfied this condition , it is thereforeconservative , for any arbitrary closed path

And it can be expressed as the gradient of scalar function

From Gauss law , we have that

OrThis equation is known as Poisson equation and solution ofthis equation can be used for solving any electrostaticproblem .


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For electrostatic problem in which the charge

density at most of point is zero , we can write the

Poisson equation as (charge free region)This equation is known as Laplaes equation in

conductors , the charge is on the surface and

therefore for any point other than surface thecharge density , hence Laplaes equation can

be applied

rectangular coordinates

cylindrical coordinate

Spherical coordinates


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Theorem No.1:

State that if are different solutions

of Laplaes equation , thenIs also a solution of Laplaes equation. Where

c1,c2, ……cn are different constants

Theorem No.2:

States that if and are the two solution of

Laplaes equation with same conditions, then

either or .


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2-4 Solutios of Laplaes euatio :

1-when the potential is dependent on only one

variable:

1-a:when is a function of x only in coordinates.

The Laplaes euatio the eoe

Integral eq.(25) we get

, where a and b are constant.

1-b:when is a function of r only in coordinates

the Laplaes equation in this case becomes

, where a and b are constant.


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1-c: when is a function of r only in

oodiates the Laplaes euatio, theefoe

becomes

,

, where a and b are constant.Ex.1: find out the capacity of a parallel plate condenser

using Laplaes equation :

Let A and B be two plates of the condenser and let theirdistance be d and area of one plate be A as shown in the

diagram. from Laplaes equation we have in this have

u=ax+b in this case when x=0 , u=u1 and when x=d ,

u=u2 .


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Fo Laplaes euatio e hae u=ax+b (1)in this case when x=0 , u=u1 and when x=d and u=u2

u1=a*0+b or u1=b u1=a*d+b

u1-u2= -a.d (2)

From eq. (1)

Now we know that E between the two plates isgiven by

Where is charge per unit area therefore from

eq. (3) and (4),

Hence


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Ex. 2: find out the capacity of spherical condenser

usig Laplaes euatio

From eq.(5) and eq.(6)


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Ex. 3 find out the capacity of cylindrical condenser

usig Laplaes euatio

Fo Laplaes euatio e hae Let ,


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2- solution of Laplaes equation in sphericalcoordinate In coordinates Laplaes equation is:

To make it simple let us assume that u is a function ofr and θ only and is independent of ф . ThereforeLaplaes equation become:

Let u = z pθ hee ) is pue futio of ad pis a pue futio of θ .

Form eq. (3)


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Multiplying eq.(6) by we get

If eq. (8) the left hand side is purely dependent on rand the right hand side is pure function of ,

therefore both side be equal to a constant

Eq. (10) can be written as

Eq. (11) is the equation known as Legendre equationand it acceptable solution are obtain when k=n(n+1) where n= 0,1,2,3,….the different value of givesdifferent solutions and these different solutions are

known as Legendre polynomial .


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these polynomials are as below:

n Pn θ

0 1*constant

1 cosθ

2 1/2(3cos2θ-1)

3 1/2(5cos3θ-3cosθ

4 1/8(35cos4θ-30cos2θ+3)

5 1/8(63cos5θ-70cos3θ+15cosθ

Solution of eq.(13) is z = r n or z= r -(n+1)

Prove , ,

, , ,

Therefore


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eq. (14) will give different solution for different values of n.these solution are called (Zonal Harmonics) or sphericalharmonics. Now if we give different value of n we get

when n=0when n=1

when n=2

from theorem 1, know the complete solution is given by:

the complete solution in this case is

In eq. (15), the term is similar to the potential due to a pointcharge and the term is similar to potential due to an electricdipole . Therefore eq. (15) can be used to solving differentelectric problems.


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Ex.1: To study the behavior of a spherical conductor place inuniform electric field.

Sol.

Let us suppose that there is a uniform electric field E0 asshown in fig. , let a spherical conductor of radius a be placedin this field. According to Laplace's equation, the potential isgiven by

In this problem, the conductor has no charge and therefore ineq. (1) the term c1r

-1 must be zero c1 = 0 with this condition

eq. (1) becomes

At infinity,

But


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The polar direction (z- direction), and if we make the origincoincide with center of sphere.

Then we apply this condition to eq.(2), we find that A2 = -E0 and A3 , A4 , A5,……. should become zero.

Therefore eq. (2) becomes

Now, at r=a, u is constant = u0

in eq. (3) when r=a (at thesurface of sphere the potential must become independentof a) then we apply this condition, we get A1=u0 and at thesurface of sphere the potential must become independentof angle θ, the two term involving cosθ must be cancel

each other, but the term with higher inverse power of rcannot be cancelled one against the other because thecontain different Legendre function, the only possibility isto set all the ci s with i ≥ 3 equal to zero. equalzero.c3,c4,c5,.. equal zero.


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So eq.(2) becomes

Where r=a, the field is Er

the charge density


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The total charge on the conductor

x=sinθ

Q:Show that the charge on spherical conductor

in uniform field is zero.


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2- Solutio of Laplaes euatio i ylidial oodiates:

I this oodiates Laplaes euatio is

For problems in which is independent of eq.(1) becomes

Therefore where k is the separation constant

And

Let where Y is pure function of r and S is pure

function of θ

OR


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We know that solution of eq. (6) is

If this solution is correct then

Placing this value of k in eq. (4) we get,

The solution of eq. (9) is

solution eq. (2)

And

When n=0

The solution are 1 and lnr

When n=1,2,3,4

The solution are

These solutions are known as cylindrical harmonics.


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Ex. 2 A long cylindrical conductor of radius a bearing no net chargeis placed in an infinity uniform electric field E0 .The direction of E0 isperpendicular to the cylinder axis. Find the potential and theelectric field at point exterior to the cylinder.

Sol.

Let the a be small, so the cylinder can only change the potentialnearby at large distance, the potential will remain unchanged. Thisgives us the boundary condition

Another boundary condition is that the electric field at thecylindrical must be normal to the surface

According to Laplaes equation, the potential given by

According to the boundary condition any where in space consist of-E0ρcosθ plus extra terms that vanish as


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1- at ρ→∞

2- at

Therefore


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3- solution of Laplaes equation in(x,y,z) coordinates:

The Laplaes equation in(x,y,z) coordinates iswritten as

Let

Where and are independent function ofx,y,z respectively.

Finding out and from eq. (2) andsubstituting in eq. (1) we get

, dividing by , we get

Let

Where k is the separation constant .


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, ,


From 2nd part of eq. (4) we have,

Where m is the second separation constant.


ThereforeFrom (2)

or

eq. (8) give different solution .


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A very interesting case is when k=0 and m=0 we

get

Where are constant


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Ex.1 If the electric potential in the y=0 plane is given by

find the potential and the electric field component at any pointlike, p

Sol.The Laplaes equation in coordinate is written as

We will look for special solution of the type .

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تح إن ع z دا اuمن اشرط ادد ال كذاكم بتz فن بمك إن ختر ت اد

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ب ن أن• A=0اشرط ادد ال

•ا ددا طرشا نأ طت C=0

for

D=0تط أن اشرط ادد اث•

for

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Ex. 2 Find the electric potential distribution functionsinside rectangular as shown in diagram. Note the potential

is constant in the z – direction.

In this problem is independent of z,

Laplaes euatio eoes

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