Gauss's Law and Divergence: Understanding Electric Flux and Potential, Study notes of Electronics

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Module 2

Gauss’s Law and Divergence

Energy, Potential and Conductors

Syllabus

• Gauss’s law and Divergence : Gauss ‘law, Application of Gauss’ law to point charge, line charge, Surface charge and volume charge, Point (differential) form of Gauss law, Divergence. Maxwell‘s First equation (Electrostatics), Vector Operator ▼ and divergence theorem, Numerical Problems ( Text: Chapter 3. 2 to 3. 7 ).
• Energy, Potential and Conductors : Energy expended or work done in moving a point charge in an electric field, The line integral, Definition of potential difference and potential, The potential field of point charge, Potential gradient, Numerical Problems (Text: Chapter 4. 1 to 4. 4 and 4. 6 ). Current and Current density, Continuity of current. (Text: Chapter 5. 1 , 5. 2 )

Gauss’s Law

◼ The results of Faraday’s experiments with the concentric spheres could be summed up as an experimental law by stating that the electric flux passing through any imaginary spherical surface lying between the two conducting spheres is equal to the charge enclosed within that imaginary surface. ◼ Faraday’s experiment can be generalized to the following statement, which is known as Gauss’s Law: “The electric flux passing through any closed surface is equal to the total charge enclosed by that surface.”

ψ = Q

Gauss’s Law

◼ Imagine a distribution of charge, shown as a cloud of point charges, surrounded by a closed surface of any shape. ◼ If the total charge is Q , the Q coulombs of electric flux will pass through the enclosing surface. ◼ At every point on the surface the electric-flux-density vector D will have some value D S (subscript S means that D must be evaluated at the surface).

Gauss’s Law

◼ The resultant integral is a closed surface integral, with d S always involves the differentials of two coordinates ► The integral is a double integral. ◼ We can formulate the Gauss’s law mathematically as: ψ charge enclosed S S =  d = = Q

D S

Q =  Qn L

Q =  dL

 S S

Q =  dS

 vol v

Q =  dv

◼ The charge enclosed meant by the formula above might be several point charges, a line charge, a surface charge, or a volume charge distribution.

Gauss’s Law

◼ We now take the last form, written in terms of the charge distribution, to represent the other forms: S vol

 D^ S^ ^ d^ S =^  vdv

◼ Illustration. Let a point charge Q be placed at the origin of a spherical coordinate system, and choose a closed surface as a sphere of radius a. ◼ The electric field intensity due to the point charge has been found to be: 2 (^4 ) r Q  r E = a D =  0 E 2 4 r Q  r  D = a

Application of Gauss’s Law: Some Symmetrical Charge Distributions ◼ Let us now consider how to use the Gauss’s law to calculate the electric field intensity D S : S S Q =  d  D S ◼ The solution will be easy if we are able to choose a closed surface which satisfies two conditions:

1. D S is everywhere either normal or tangential to the closed surface, so that D S  d S becomes either D S dS or zero, respectively.

2. On that portion of the closed surface for which D S  d S is not zero, D S is constant. ◼ For point charge ► The surface of a sphere. ◼ For line charge ► The surface of a cylinder.

Application of Gauss’s Law: Some Symmetrical Charge Distributions D   D = a ◼ From the previous discussion of the uniform line charge, only the radial component of D is present: ◼ The choice of a surface that fulfill the requirement is simple: a cylindrical surface. ◼ D ρ is every normal to the surface of a cylinder. It may then be closed by two plane surfaces normal to the z axis.

Application of Gauss’s Law: Some Symmetrical Charge Distributions ◼ The problem of a coaxial cable is almost identical with that of the line charge. ◼ Suppose that we have two coaxial cylindrical conductors, the inner of radius a and the outer of radius b , both with infinite length. ◼ We shall assume a charge distribution of ρ S on the outer surface of the inner conductor. ◼ Choosing a circular cylinder of length L and radius ρ , a < ρ < b , as the gaussian surface, we find: 2 S Q = D  L ◼ The total charge on a length L of the inner conductor is: 2 0 0 2 L S S z Q ad dz aL       = = = =   S S a D    =

Application of Gauss’s Law: Some Symmetrical Charge Distributions ◼ For one meter length, the inner conductor has 2 πaρ S coulombs, hence ρ L = 2 πaρ S

2 L    D = a ◼ Everly line of electrix flux starting from the inner cylinder must terminate on the inner surface of the outer cylinder: outer cyl ,inner cyl 2 S Q = −  aL  2  bL  S ,outer cyl = − 2  aL  S ,inner cyl S ,outer cyl S ,inner cyl a b  = −  ◼ If we use a cylinder of radius ρ > b , then the total charge enclosed will be zero. ► There is no external field, 0 S D =

• Due to simplicity, noise immunity and broad bandwidth, coaxial cable is still the most common means of data transmission over short distances.

Application of Gauss’s Law: Differential Volume

Element

◼ Consider any point P , located by a rectangular coordinate system. ◼ The value of D at the point P may be expressed in rectangular components: 0 x 0 x y 0 y z 0 z D = D a + D a + D a ◼ We now choose as our closed surface, the small rectangular box, centered at P , having sides of lengths Δ x , Δ y , and Δ z , and apply Gauss’s law: S  d = Q  D S S front back left right top bottom  d = + + + + +        D S

Application of Gauss’s Law: Differential Volume

Element

◼ We will now consider the front surface in detail. ◼ The surface element is very small, thus D is essentially constant over this surface (a portion of the entire closed surface): front front front    D S front x D    y z a x ,front D   y z ◼ The front face is at a distance of Δ x /2 from P , and therefore: ,front 0 rate of change of with 2 x x x x D D D x 

•  0 2 x x x D D x  



◼ If we combine the two integrals over the front and back surface, we have:

Application of Gauss’s Law: Differential Volume

Element

front back x D x y z x 

•       right left D y y x z y 
•       top bottom D z z x y z 
•       ◼ Repeating the same process to the remaining surfaces, we find: ◼ These results may be collected to yield: S Dx D y Dz d x y z x y z  (^)      (^)  + + (^)     ^ ^    D S S x y z D D D d Q v x y z  (^)      = (^)  + + (^)   ^ ^    D S

Charge enclosed in volume Dx Dy Dz v v x y z  (^)      (^)  + + (^)    ^ ^  

Application of Gauss’s Law: Differential Volume

Element

◼ The previous equation is an approximation, which becomes better as Δ v becomes smaller. ◼ For the moment, we have applied Gauss’s law to the closed surface surrounding the volume element Δ v , with the result: