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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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, ,
The solution of eq. (5) is
From 2nd part of eq. (4) we have,
Where m is the second separation constant.
The solution of eq. (7) is
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 .
ل
ذ
د
ما
تا
دبصف
را
امتغ
•ا
اد
ق
د
إ
اث
تا
د
ما
ل
امط
من
د
ا
م
د
اد
ط
اشر
من
•مدعy=0•مدع•مدع
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تح إن ع z دا اuمن اشرط ادد ال كذاكم بتz فن بمك إن ختر ت اد
بن
نتتج
ف
ع
كم خ ترض ت من اشرط ال أ تب ن إ
كذا فن
اد
ا
ال
إن
f 1
اد ا لا نإ f 2
ن
اد ا لاf 3
اد
ا
ال
د
كذا
ات
اشكل
خذ
:
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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
•عدم
y=0ف
•مدع ف
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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
• أن
تح
كذا
ككم
ب
خ
كم
عدم
.أ
تا ددا طرشا ع دمت خ آ مك اختر خ ث د أن ددصب ن تا ما مت
ن
ف
ع
رى
كم
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م
د
اد
ط
اشر
من
ن
ف
كذا
when m=0,1,2,3,…
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تم دمتا صخب ن ت عد ق د رفدف
رة
اخ
د
ما
طرف
ر
)*(ب
مل
اتك
ر
ع لص
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