Physics 12th Guass 's Law

Guass 's Law

1. Introduction

• Gauss's law was suggested by Kark Fredrich Gauss(1777-1855) who was german scientist and mathematecian.
• Gauss's law is basically the relation between the charge distribution producing the electrostatic field to the behaviour of electrostatic field in space.
• Gauss's law is based on the fact that flux through any closed surface is a measure of total amount of charge inside that surface and any charge outside that surface would not contribute anything to the total flux.
• Before we look further to study Gauss's law in detail let's study electric field due to continous charge distributions. 

2. Electric field due to continous charge distributions

• So far as in the previous chapter we have discussed force and field due to discrete charges.
• We now assume that charges on a surface are located very close togather so that such a system of charges can be assumed to have continous distribution of charges.
• In a system of closely spaced charges, total charge could be continously distributed along some line, over a surface or throughout a volume.
• First divide the continous charge distribution into small elements containing Δq amount of charge as shown in fig 1
  
  
  
• Electric field at point A due to element carrying charge Δq is
  
  
  where r is the distance of element under consideration from point A and rˆis the unit vector in the direction from charge element towards point A.
• Total electric field at point A due to all such charge elements in charge distribution is
  
  
  where index i refers to the i_th charge element in the entire charge distribution.
• Since the charge is distributed continously over some region , the sum becomes integral. Hence total field at A within the limit Δq→0 is,
  
  
  and integration is done over the entire charge distribution.
• If a charge q is uniformly distributed along a line of length L, the linear charge density λ is defined by
  
  and the unit of λ is Coulumb/meter(C/m).
• For charge distributed non-uniformly over a line, linear charge density is
  
  where dQ is the amount of charge in a small length element dL.
• For a charge Q uniformly distributed over a surfsce of area A, the surface charge density σ is
  
  and unit of surface charge density is C/m². For non uniform charge distributed over a surface charge density is
  
  where dA is a small area element of charge dQ.
• Similarly for uniform charge distributios volume charge density is
  
  and for non uniform distribution of charges
  
  and unit of volume charge distribution is C/m³.

3. Gauss's Law

• We already know about electric field lines and electric flux. Electric flux through a closed surface S is
  
  which is the number of field lines passing through surface S.
• Statement of Gauss's Law
  “ELectric flux through any surface enclosing charge is equal to q/ε₀ , where q is the net charge enclosed by the surface”
  mathematically,
  
  where q_enc is the net charge enclosed by the surface and E is the total electric field at each point on the surface under consideration.
• It is the net charge enclosed in the surface that matters in Gauss's law but the total flux of electric field Edepends also on the surface choosen not merely on the charge enclosed.
• So if you have information about distribution of electric charge inside the surface you can find electric flux through that surface using Gauss's Law.
• Again if you have information regarding electric fluxthrough any closed surface then total charge enclosed by that surface can also be easily calculated using Gauss's Law.
• Surface on which Gauss's Law is applied is known as Gaussian surface which need not be a real surface.
• Gaussian surface can be an imaginary geometrical surface which might be empty space or it could be partially or fully embedded in a solid body.
• Again consider equation 11
  
  In left hand side of above equation E·da is scalar product of two vectors namely electric field vector E and area vectorda. Area vector da is defined as the vector of magnitude |da| whose direction is that of outward normal to area element da. So,da=nˆda where nˆ is unit vector along outward normal to da.
   
  
  
  
  
  From above discussion we can conclude that,
  (1) If both E and surface area da at each points are perpandicular to each other and has same magnitude at all points of the surface then vector E has same direction as that of area vector as shown below in the figure.
  
  
  
  
  since E is perpandicular to the surface
   
  (2) If E is parallel to the surface as shown below in the figure
  
  
  
  
   
  E at all points on the surface.

4. Applications of Gauss's Law

(A) Derivation of Coulumb's Law

• Coulumb's law can be derived from Gauss's law.
• Consider electric field of a single isolated positive charge of magnitude q as shown below in the figure.
  
  
  
• Field of a positive charge is in radially outward direction everywhere and magnitude of electric field intensity is same for all points at a distance r from the charge.
• We can assume Gaussian surface to be a sphere of radius r enclosing the charge q.
• From Gauss's law
  
  since E is constant at all points on the surface therefore,
  
  surface area of the sphere is A=4πr²
  thus,
   
• Now force acting on point charge q' at distance r from point charge q is
  
  This is nothing but the mathematical statement of Coulumb's law.

• (B) Electric field due to line charge

• Consider a long thin uniformly charged wire and we have to find the electric field intensity due to the wire at any point at perpandicular distance from the wire.
• If the wire is very long and we are at point far away from both its ends then field lines outside the wire are radial and would lie on a plane perpandicular to the wire.
• Electric field intensity have same magnitude at all points which are at same distance from the line charge.
• We can assume Gaussian surface to be a right circular cylinder of radius r and length l with its ends perpandicular to the wire as shown below in the figure.
  
  
  
• λ is the charge per unit length on the wire. Direction of E is perpandicular to the wire and components of E normal to end faces of cylinder makes no contribution to electric flux. Thus from Gauss's law
   
• Now consider left hand side of Gauss's law
  
  Since at all points on the curved surface E is constant. Surface area of cylinder of radius r and length l is A=2πrl therefore,
   
• Charge enclosed in cylinder is q=linear charge density x length l of cylinder,
  or, q=λl
  From Gauss's law
  
  Thus electric field intensity of a long positively charged wire does not depends on length of the wire but on the radial distance r of points from the wire. 

    (C) Electric field due to charged solid sphere

• We'll now apply Gauss's law to find the field outside uniformly charged solid sphere of radius R and total charge q.
• In this case Gaussian surface would be a sphere of radius r>R concentric with the charged solid sphere shown below in the figure.From Gauss's law
  
  where q is the charge enclosed.
• Charge is distributed uniformly over the surface of the sphere. Symmetry allows us to extract E out of the integral sign as magnitude of electric field intensity is same for all points at distance r>R.
• Since electric field points radially outwards we have
  
  
  
  
  
  also as discussed magnitude of E is constant over Gaussian surface so,
  
  where 4πr² is the surface area of the sphere.
  Again from Gauss's law we have
  
  THus we see that magnitude of field outside the sphere is exactly the same as it would have been as if all the charge were concentrated at its center. 

(D) Electric field due to an infinite plane sheet of charge

• Consider a thin infinite plane sheet of charge having surface charge density σ(charge per unit area).
• We have to find the electric field intensity due to this sheet at ant point which is distance r away from the sheet.
• We can draw a rectangular gaussian pillbox extending equal distance above and below the plane as shown below in the figure.
  
  
  
• By symmetry we find that Eon either side of sheet must be perpandicular to the plane of the sheet, having same magnitude at all points equidistant from the sheet.
• No field lines crosses the side walls of the Gaussian pillbox i.e., component of E normal to walls of pillbox is zero.
• We now apply Gauss's law to this surface
  
  in this case charge enclosed is
  q=σA
  where A is the area of end face of Gaussian pillbox.
• E points in the direction away from the plne i.e., E points upwards for points above the plane and downwards for points below the plane. Thus for top and bottom surfaces,
  
  thus
  2A|E|=σA/ε₀
  or,
  |E|=σ/2ε₀
  Here one important thing to note is that magnitude of electric fied at any point is independent of the sheet and does not decrease inversly with the square of the distance. Thus electric field due to an infinite plane sheet of charge does not falls of at all.

Physics 12th Electric Charge and Electric Field

1. Electric Charge

• Electric charge is a fundamental property like mass, length etc associated with elementary particles for example electron, proton and many more.
• Electric charge is the property responsible for electric forces which acts between nucleus and electron to bind the atom together.
• Charges are of two kinds
  (i) negative charge
  (ii) positive charge
• Electrons are negatively charged particles and protons, of which nucleus is made of, are positively charged particles. Actually nucleus is made of protons and neutrons but neutrons are uncharged particles.
• electric force between two electrons is same as electric force between two protons kept at same distance apart i. e., both set repel each other but electric force between an electron and proton placed at same distance apart is not repulsive but attractive in nature.
  
  Conclusion
  (a) Like charges repel each other
  
  
  
  (b) Unlike charges attract each other
  
  
• Assignment of negative charge on electron and positive charge on proton is purely conventional , it does not mean that charge on electron is less than that on proton.
• Importance of electric forces is that it encompasses almost each and every field associated with our life; being it matter made up of atoms or molecules in which electric charges are exactly balanced or adhesive forces of glue associated with surface tension, all are electric in nature.

Unit

• Charge on a system can be measured by comparing it with the charge on a standard body.
• SI unit of charge is Coulomb written as C.
• 1 Coulomb is the charge flowing through the wire in 1 second if the electric current in it is 1A.
• Charge on electron is -1.602 × 10 ^-19 C and charge on proton is positive of this value.

2. Basic properties of electric charge

(i) Additivity of charges

• Charges adds up like real numbers i. e., they are Scalars more clearly if any system has n number of charges q₁, q₂, q₃, q_n then total charge of the system is
  q = q₁ + q₂ + q₃ + ................ q_n
• Proper sign have to be used while adding the charges for example if
  q₁ = +1C
  q₂ = -2C
  q₃ = +4C
  then total charge of the system is
  q = q₁ + q₂ + q₃
  q = (+1) + (-2) + (+4) C
  q = (+3) C
(ii) Charge is conserved


• Charge of an isolated system is conserved.
• Chage can not be created or destroyed but charged particles can be created or destroyed.

(iii) Quantization of charge


• All free charges are integral multiples of a unit of charge e, where e = -1.602 × 10 ^-19 C i. e., charge on an electron or proton.
• Thus charge q on a body is always denoted by
  q = ne
  where n = any integer positive or negative
• 3. Frictional Electricity
  
  
  • If we pass a comb through hairs, comb becomes electrically charged and can attract small pieces of paper.
  • Many such solid materials are known which on rubbing attract light objects like light feather, bits of papers, straw etc.
  • Explaination of appearance of electric charge on rubbing is simple.
  • Material bodies consists of large number of electrons and protons in equal number and hence is in neutral in their normal state. But when the body is rubbed for example when a glass rod is rubbed with silk cloth, electrons are transferred from glass rod to silk cloth. The glass rod becomes positively charged and the silk cloth becomes negatively charged as it recieves extra electrons from the glass rod.
  • In this case rod after rubbing, comb after passing through dry hairs becomes electrified and these are the example of frictional electricity.
  
  4. Coulumb's law
  
  
  • Coulomb's law is the law of forces between electric charges.
    Statement
    " It states that two stationary point charges q₁ and q₂ repel or attract each other with a force F which is directly proportional to the product of charges and inversly proportional to the square of distance between them."
    
    This dependence can be expressed by writing
    

    F∝	q1q2
    	r2

                                             (1)
  • These forces are attractive for unlike charges and repulsive for like charges .
  • We now try to express Coulomb's law in vector form for more clearity of magnitude and direction of forces.
  • Consider two point charges q₁ and q₂ at points with position vector r₁ and r₂ with respect to the origin
    
    
    
    vector r₂₁= r₂ - r₁ is the difference between r₂ and r₁ and the distance of separation r is the magnitude of vector r₂₁.
    pointwise it can be written as
    r₁ = position vector of charge q₁ with respect to origin
    r₂ = position vector of charge q₂ with respect to origin
    r₂₁ = vector from 1 to 2 (r₂ - r₁)
    r₁₂ = -r₂₁ = vector from 2 to 1 (r₁ - r₂)
    r = r₁₂ = r₂₁ = distance between 1 and 2.
    
    
    
    
    Coulomb's law can then be expressed as
    F₂₁ = force on q₂ due to q₁
    

    F21=	kq1q2r21
    	r3

                                            (2a)
    and, F₁₂ = force on q₁ due to q₂
    

    F12= -F21	kq1q2r12
    	r3

                                             (2b)
  
  Special Case
for simplicity we can choose q₁ being placed at origin
r₁ = 0
and if we write r₂ = r the position vector of q₂ then
F₂₁ = force on q₂ due to q₁


F21=	kq1q2r
	r3

                                         (3a)
F₁₂ = force on q₁ due to q₂


F12=	- kq1q2r
	r3

                                         (3b)
unit vector r^ˆ₂₁ and r^ˆ ₁₂ can be defined as
r^ˆ₂₁ = r₂₁/r directed from q₁ to q₂
r^ˆ₁₂ = r₁₂/r directed from q₂ to q₁ (4)
       = -r₂₁/r
force can now be written in terms of unit vector given as follows


F21=	kq1q2rˆ21
	r2

                                         (5a) 

F12=	kq1q2rˆ12
	r2

                                         (5b)
from this we can immidiately find factors giving magnitude and the directions


• in equation (2) we find a positive constant K and experimentally found value of k is
  K = 8.98755 × 10 ⁹ Nm²/C²
  K ≅ 9 × 10 ⁹ Nm²/C²
  sometimes K is written as 1/4π ε₀ where ε₀ is the permittivity of the vaccum whose value is
  K = 1/4πε₀
  (ε₀ = 9 × 10 ^-12 C²/Nm²)
• 5. Principle Of Superposition
  
  
  • Coulomb's law gives the electric force acting between two electric charges.
  • Principle of superposition gives the method to find force on a charge when system consists of large number of charges.
  • According to this principle when a number of charges are interacting the total force on a given charge is vector sum of forces exerted on it by all other charges.
  • This principle makes use of the fact that the forces with which two charges attract or repel one another are not affected by the presence of other charges.
  • If a system of charges has n number of charges say q₁, q₂, ...................., q_n, then total force on charge q₁ according to principle of superposition is
    F = F₁₂ + F₁₃ + .................................. F_1n
    Where F₁₂ is force on q₁ due to q₂ and F₁₃ is force on q₁ due to q₃ and so on.
    
    
    
  • F₁₂, F₁₃, .................. F_1n can be calculated from Coulomb's law i. e.
    

    F12=	kq1q2rˆ12
    	4πε0(r12)2

    to,
    

    F1n=	kq1q2rˆ1n
    	4πε0(r1n)2

  • The total force F₁ on the charge q₁ due to all other charges is the vector sum of the forces F₁₂, F₁₃, ................................. F_1n.
    F₁ = F₁₂ + F₁₃ + ..................................
     
  • The vector sum is obtained by parellogram law of addition of vector.
  • Similarly force on any other charge due to remaining charges say on q₂, q₃ etc. can be found by adopting this method.
  
  6. Electric Field
  
  
  • Electrical interaction between charged particles can be reformulated using the concept of electric field.
  • To understand the concept consider the mutual repulsion of two positive charged bodies as shown in fig (a)
    
    
  • Now if remove the body B and label its position as point P as shown in fig (b), the charged body A is said to produce an electric field at that point (and at all other points in its vicinity)
  • When a body B is placed at point P and experiences force F, we explain it by a point of view that force is exerted on B by the field not by body A itself.
  • The body A sets up an electric field and the force on body B is exerted by the field due to A.
  • An electric field is said to exists at a point if a force of electric origin is exerted on a stationary charged (test charge) placed at that point.
  • If F is the force acting on test charge q placed at a point in an electric field then electric field at that point is
    E = F/q
    or F = qE
  • Electric field is a vector quantity and since F = qE the direction of E is the direction of F.
  • Unit of electric field is (N.C^-1)
  Q. Find the dimensions of electric field
  Ans. [MLT^-3A^-1]
  
  7. Calculation of Electric Field
• In previous section we studied a method of measuring electric field in which we place a small test charge at the point, measure a force on it and take the ratio of force to the test charge.
• Electric field at any point can be calculated using Coulomb's law if both magnitude and positions of all charges contributing to the field are known.
• To find the magnitude of electric field at a point P, at a distance r from the point charge q, we imagine a test charge q'to be placed at P. Now we find force on charge q' due to q through Coulomb's law.
  

  F=	qq'
  	4πε0r2

  electric field at P is 

  E=	q
  	4πε0r2

  The direction of the field is away from the charge q if it is positive
  
  
• Electric field for either a positive or negative charge in terms of unit vector r directed along line from charge q to point P is 

  E=	qrˆ
  	4πε0r2

  r = distance from charge q to point P.
• When q is negative , direction of E is towards q, opposite to r.
  
  Electric Field Due To Multiple Charges
• Consider the number of point charges q₁, q₂,........... which are at distance r_1P, r_2P,................... from point P as shown in fig
  
  
• The resultant electric field is the vector sum of individual electric fields as
  E = E_1P + E_2P + .....................
  
  
  This is also a direct result of principle of superposition discussed earlier in case of electric force on a single charge due to system of multiple charges.
• E is a vector quantity that varies from one point in space to another point and is determined from the position of square charges.

8. Electric Field Lines

• For a single positive point charge q, electric field is
  

  E=	qrˆ
  	4πε0r2

  
  now to get feel of this field one can sketch a few representative vectors as shown in fig below
  
  
• Since electric field varies as inverse of square of the distance that points from the charge the vector gets shorter as you go away from the origin and they always points radially outwards.
• Connecting up these vectors to form a line is a nice way to represent this field .
• The magnitude of the field is indicated by the density of the field lines.
• Magnitude is strong near the center where the field lines are close togather, and weak farther out, where they are relatively apart.
• So, electric field line is an imaginary line drawn in such a way that it's direction at any point is same as the direction of field at that point.
• An electric field line is, in general a curve drawn in such a way that the tangent to it ateach point is the direction of net field at that point.
• Field lines of a single position charge points radially outwards while that of a negative charge are radially inwards as shown below in the figure.
  
  
• Field lines around the system of two positive charges gives a different picture and describe the mutual repulsion between them.
  
  
• Field lines around a system of a positive and negative charge clearly shows the mutual attraction between them as shown below in the figure.
  
  
• Some important general properties of field lines are
  1.Field lines start from positive charge and end on a negative charge.
  2.Field lines never cross each other if they do so then at the point of intersection there will be two direction of electric field.
  3.Electric field lines do not pass through a conductor , this shows that electric field inside a conductor is always zero.
  4.Electric field lines are continuous curves in a charge free region.

9. Electric Flux

• Consider a plane surface of area ΔS in a uniform electric field E in the space.
• Draw a positive normal to the surface and θ be the angle between electric field E and the normal to the plane.
  
  
• Electric flux of the electric field through the choosen surface is then
  Δφ = E ΔS cosθ
• Corresponding to area ΔS we can define an area vector ΔS of magnitude ΔS along the positive normal. With this definition one can write electric flux as
  Δφ = E . ΔS
• direction of area vector is always along normal to the surface being choosen.
• Thus electric flux is a measure of lines of forces passing through the surface held in the electric field.
  
  Special Cases
• If E is perpendicular to the surface i. e., parallel to the area vector then θ = 0 and
  Δφ = E ΔS cos0
• If θ = π i. e., electric field vector is in the direction opposite to area vector then
  Δφ = - E ΔS
• If electric field and area vector are perpendicular to each other then θ = π/2 and Δφ = 0
• Flux is an scaler quantity and it can be added using rules of scaler addition.
• For calculating total flux through any given surface , divide the surface into small area elements. Calculate the flux at each area element and add them up.
• Thus total flux φ through a surface S is
  φ ≅ ΣE.ΔS
• This quantity is mathematically exact only when you take the limit ΔS→0 and the sum in equation 3 is written as integral
  φ = ∫ΣE.dS

9. Electric Dipole

• Electric dipole is a pair of equal and opposite charges, +q and -q, separated by some distance 2a.
• Total charge of the dipole is zero but electric field of the dipole is not zero as charges q and -q are separated by some distance and electric field due to them when added is not zero.
  
  (A)Field of an electric dipole at points in equitorial plane
• We now find the magnitude and direction of electric field due to dipole.
  
  
  
• P point in the equitorial plane of the dipole at a distance r from the centre of the dipole. Then electric field due to +q and -q are 

  E-q=	-qPˆ
  	4πε0(r2 + a2)

                                           (1a) 

  E+q=	qPˆ
  	4πε0(r2 + a2)

                                           (1b)
  and they are equal
  Pˆ = unit vector along the dipole axis (from -q to +q)
• From fig we can see the direction of E_+q and E_-q. Their components normal to dipole cancel away and components along the dipole add up. 
• Dipole moment vector points from negative charge to positive charge so in vector form.
  E = -(E_+q + E_-q) cos θ
  
  
  
               
                                          (2)
  At large distances (r>>a), above equation becomes
  

  E=	-2qaPˆ
  	4πε0(r3)

                                          (3)
  and they are equal
  
  (B) Field of an electric dipole for points on the axis 
  
  
• Let P be the point at a distance r from the centre of the dipole on side of charge q. as shown in the fig.
  

  E-q=	-qPˆ
  	4πε0(r+a)2

                                           (4a)
  Pˆ = unit vector along the dipole axis (from -q to +q)
  also
  

  E+q=	qPˆ
  	4πε0(r - a) 2

  
                                           (4b)
• Total field at P is
  E = E_+q + E_-q
  
  
                                           (5)
  for r>>a
  

  E=	4qaPˆ
  	4πε0r3

  
                                           (6)
• For equation (3) and (6) charge q and dipole separation 2a appear in combination qa. This leads us to define dipole moment vector P of electric dipole. Thus, electric dipole moment P = q × 2a Pˆ                                         (7) 
• Unit of dipole moment is Coulomb's meter (Cm).
• In terms of electric dipole moment, field of a dipole at large distances becomes
  
  (i) At point on equitorial plane (r>>a)
  E = -P/4πε₀r³
  

  E=	-P
  	4πε0r3

  
  (ii) At point on dipole axis (r>>a)
  

  E=	2P
  	4πε0r3

  Note:-
  (i) Dipole field at large distances falls off as 1/r³
  (ii) Both the direction and magnitude of dipole an angle between dipole moment vector P and position vector r
  
  (C) Dipole in a uniform external field
• Consider a dipole in a uniform electric field E whose direction makes an angle θ with dipole axis (line joining two charges)
  
  
• Force F₁ of magnitude qE, acts on positive charge in direction of electric field and a force F₂ of same magnitude acts on negative charge but it acts in direction opposite to F₁.
• Resultant force on dipole is zero, but since two forces do not have same line of action they constitute a couple.
• We now calculate torque (r × F) of these forces about zero.
  
  Torque of F₁ about O is
  τ₁= OB × F₁
       = q (OB × E)
  
  Torque of F₂ about O is
  τ₂ = OA × F₂
       = -q(OA × E)
       = q(AO × E)
  
  net torque acting on dipole is
  τ = τ₁ + τ₂
       = q(OB+ AO) × E 
       = q (AB × E)
  
  AB = 2a and p = 2qa (dipole moment)
  τ =p × E
• Direction of torque is perpendicular to the plane containing dipole axis and electric field.
• Effect of torque is to rotate the dipole to a position in which dipole moment p is parallel to E the electric field vector is shown above in figure b and for uniform electric field dipole is in equilibrium in this position.
• magnitude of this torque is
  τ = |τ| = pE sinθ

Examples

Question 1
Two point charges q₁ and q₂are located with points having position vectors r₁ and r₂ 
(1) Find the position vector r₃ where the third charge q₃ should be placed so that force acting on each of the three charges would be equal to zero.
(2) Find the amount of charge q₃

Question 2
Consider a thin wire ring of radius R and carrying uniform charge density λ per unit length.
(1) Find the magnitude of electric field strength on the axis of the ring as a function of distance x from its centre.
(2) What would be the form of electric field function for x>>R.
(3) Find the magnitude of maximum strength of electric field.

Question 3
Two equally charged metal balls each of mass m Kg are suspended from the same point by two insulated threads of length l m long. At equilibrium, as a result of mutual separation between balls, balls are separated by x m. Determine the charge on each ball.

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Tips And Tricks And Formulaes For Physics

Tips and Tricks on Alternating Current

Tip #1

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Tip #2

Formulae on Alternating Current

Important Formula #1

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Important Formula #2

Tips and Tricks on Capacitance

Tip #1

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Tip #2

In series ,charge remains the same and potential across the capacitors may be different .In parallel ,potential difference across each capacitor on the capacitors may be different

Formulae on Capacitance

Important Formula #1

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Important Formula #2

Tips And Tricks on Center of Mass, Impulse and Momentum

Tip #1

Linear momentum is conserved whenever external force is zero or in a two body system net external force is zero.

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Tip #2

Elastic collision occurs only in atomic or subatomic particles. In real world, collision is inelastic .Only collision between two ping-pong balls may be considered elastic. During elastic collision kinetic energy and linear momentum are conserved.

Formulae on Center of Mass, Impulse and Momentum

Important Formula #1

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Important Formula #2

Formulae on Heat and Thermodynamics

Important Formula #1

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Important Formula #2

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Important Formula #3

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Important Formula #4

Important Formula #5

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Important Formula #6

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Important Formula #7