محاضرة 4 فيزياء عامة (2) كهربية ساكنة

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درسنا سابقا كيفية حساب المجال لتوزيع معين من الشحنات باستخدام قانون كولوم.  وهنا سنقدم طريقة أخرى لحساب المجال الكهربي باستخدام “قانون جاوس” الذي يسهل حساب المجال الكهربي لتوزيع متصل من الشحنة على شكل توزيع طولي أو سطحي أو حجمي.  يعتمد قانون جاوس أساساً على مفهوم التدفق الكهربي الناتج من المجال الكهربي أو الشحنة الكهربائية، ولهذا سنقوم أولاً بحساب التدفق الكهربي الناتج عن المجال الكهربي، وثانياً سنقوم بحساب التدفق الكهربي الناتج عن شحنة كهربية، ومن ثم سنقوم بإيجاد قانون جاوس واستخدامه في بعض التطبيقات الهامة في مجال الكهربية الساكنة.

 4.1 The Electric Flux due to an Electric Field

We have already shown how electric field can be described by lines of force.  A line of force is an imaginary line drawn in such a way that its direction at any point is the same as the direction of the field at that point.  Field lines never intersect, since only one line can pass through a single point.

The Electric flux (F) is a measure of the number of electric field lines penetrating some surface of area A.

Case one:

The electric flux for a plan surface perpendicular to a uniform electric field

To calculate the electric flux we recall that the number of lines per unit area is proportional to the magnitude of the electric field.  Therefore, the number of lines penetrating the surface of area Ais proportional to the product EA.  The product of the electric filed E and the surface area Aperpendicular to the field is called the electric flux F.

 The electric flux F has a unit of N.m²/C.

Case Two

   The electric flux for a plan surface make an angle q to a uniform electric field

Note that the number of lines that cross-area is equal to the number that cross the projected area A`, which is perpendicular to the field.  From the figure we see that the two area are related by A`=Acosθ.  The flux is given by:

   = E A cosθ   →  

Where θ is the angle between the electric field E and the normal to the surface 

إذاً يكون الفيض ذا قيمة عظمى عندما يكون السطح عمودياً على المجال أي θ = 0 ويكون ذا قيمة صغرى عندما يكون السطح موازياً للمجال أي عندما θ = 90.  لاحظ هنا أن المتجه  هو متجه المساحة  وهو عمودي دائما على المساحة وطوله يعبر عن مقدار المساحة.

Case Three

In general the electric field is nonuniform over the surface

The flux is calculated by integrating the normal component of the field over the surface in question.

The net flux through the surface is proportional to the net number of lines penetrating the surface

والمقصود بـ  net number of lines أي عدد الخطوط الخارجة من السطح (إذا كانت الشحنة موجبة) – عدد الخطوط الداخلة إلى السطح (إذا كانت الشحنة سالبة).

Example 4.1

What is electric flux F for closed cylinder of radius R immersed in a uniform electric field as shown in figure?

Solution

نطبق قانون جاوس على الأسطح الثلاثة الموضحة في الشكل أعلاه

Since E is constant then

            F = – EA + 0 + EA = zero

Exercise

Calculate the total flux for a cube immersed in uniform electric field .

4.2 The Electric Flux due to a point charge

To calculate the electric flux due to a point charge we consider an imaginary closed spherical surface with the point charge in the center figure 4.5, this surface is called gaussian surface. Then the flux is given by

        (θ = 0)     

                                                 (4.3)

Note that the net flux through a spherical gaussian surface is proportional to the charge q inside the surface.

4.3 Gaussian surface

Consider several closed surfaces as shown in figure 4.6 surrounding a charge Q as in the figure below.  The flux that passes through surfaces S₁, S₂ and S₃ all has a value q/ε_o.  Therefore we conclude that the net flux through any closed surface is independent of the shape of the surface.

Consider a point charge located outside a closed surface as shown in figure 4.7.  We can see that the number of electric field lines entering the surface equal the number leaving the surface.  Therefore the net electric flux in this case is zero, because the surface surrounds no electric charge.

Example 4.2

In figure 4.8 two equal and opposite charges of 2Q and -2Q what is the flux F for the surfaces S₁, S₂, S₃ and S₄.

Solution

For S₁ the flux F = zero

For S₂ the flux F = zero

For S₃ the flux F = +2Q/ ε_o

For S₄ the flux F = -2Q/ ε_o

4.4 Gauss’s Law

Gauss law is a very powerful theorem, which relates any charge distribution to the resulting electric field at any point in the vicinity of the charge.  As we saw the electric field lines means that each charge q must have q/e_o flux lines coming from it.  This is the basis for an important equation referred to as Gauss’s law.  Note the following facts:

1.  If there are charges q₁, q₂, q₃, ……q_n inside a closed (gaussian) surface, the total number of flux lines coming from these charges will be

(q₁ + q₂ + q₃ + ……. +q_n)/ε_o                                            (4.4)

2. The number of flux lines coming out of a closed surface is the integral of  over the surface, 

We can equate both equations to get Gauss law which state that the net electric flux through a closed gaussian surface is equal to the net charge inside the surface divided by e_o

                 Gauss’s law                 (4.5)

where q_in is the total charge inside the gaussian surface.

Gauss’s law states that the net electric flux through any closed gaussian surface is equal to the net electric charge inside the surface divided by the permittivity.

4.5 Gauss’s law and Coulomb’s law

We can deduce Coulomb’s law from Gauss’s law by assuming a point charge q, to find the electric field at point or points a distance r from the charge we imagine a spherical gaussian surface of radius r and the charge q at its center as shown in figure 4.10.

 Because E is constant for all points on the sphere, it can be factored from the inside of the integral sign, then

                  →         →        

                                                                       (4.6)

Now put a second point charge q_o at the point, which E is calculated.  The magnitude of the electric force that acts on it F = Eq_o

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