COULOMB’S LAW   

 

INTRODUCTION www.citycollegiate.comThe magnitude of the force of attraction or repulsion between two electric charges at rest was studied by Charles Coulomb. He formulated a law ,known as "COULOMB'S LAW".

STATEMENT  According to Coulomb's law:

 The electrostatic force of attraction or repulsion between two point charges is directly proportional      to the product of charges.

 The electrostatic force of attraction or repulsion between two point charges is inversely proportional      to the square of distance between them.

MATHEMATICAL REPRESENTATION OF

COULOMB'S LAWwww.citycollegiate.com

Consider two point charges q1 and q2 placed at a distance of r from each other. Let the electrostatic force between them is F.

According to the first part of the law:

According to the second part of the law:

Combining above statements:

OR 

                                     ---------------------(I)Where k is the constant of proportionality.

VALUE OF K www.citycollegiate.comValue of K is equal to 1/40

where o is permittivity of free space .Its volume is 8.85 x 10-12 c2/Nm2.Thus in S.I. system numerical value of K is 8.98755 x 109 Nm2c-2.

OTHER FORMS OF COULOMB'S LAW  

Putting the value of K = 1/40 in equation (i)

FORCE IN THE PRESENCE OF DIELECTRIC MEDIUM www.citycollegiate.com

If the space between the charges is filled with a non conducting medium or an

insulator called "dielectric", it is found that the dielectric reduces the electrostatic force as compared to free space by a factor (r) called DIELECTRIC CONSTANT. It is denoted by r . This factor is also known asRELATIVE PERMITTIVITY. It has different values for different dielectric materials.In the presence of a dielectric between two charges the Coulomb's law is expressed as:

VECTOR FORM OF COULOMB'S LAW  

The magnitude as well as the direction of electrostatic force can be expressed by using Coulomb's law by vector equation:

Where   is the force exerted by q1 on q2 and   is the unit vector along the line joining the two charges from q1 to q2.

ELECTRIC FIELD - ELECTRIC INTENSITY   

 

ELECTRIC FIELD www.citycollegiate.comWhen an electric charge is placed in space, the space around the charge is modified and if we place another test charge within this space, the test charge will experience some electrostatic force. The modified space around an electric charge is called 'ELECTRIC FIELD'.For an exact definition we can describe an electric field as:Space or region surrounding an electric charge or a charged body within which 

another charge experiences some electrostatic force of attractionor repulsion when placed at a point is called Electric Field.

ELECTRIC INTENSITY  Electric intensity is the strength of electric field at a point.

Electric intensity at a point is defined as the force experienced per unit positive charge at a point placed in the electric field. 

or 

It may also be also defined as the electrostatic force per unit charge which the field exerts at a point.

Mathematically,E=F/q

UNIT www.citycollegiate.comN/C or Volt/m

The force experienced by a charge +q in an electric field depends upon. 1. magnitude of test charge (q)2. Intensity of electric field (E)It is a vector quantity. It has the same direction as that of force.

ELECTRIC LINES OF FORCE  

 In order to point out the direction of an electric field we can draw a number of lines called electric lines of force.

DEFINITION  

 

An electric line of force is an imaginary continuous line or curve drawn in an electric field such that tangent to it at any point gives the direction of the electric force at that point.The direction of a line of force is the direction along which a small free positive charge will move along the line. It is always directed from positive charge to negative charge. www.citycollegiate.com

 

  CHARACTERISTICS OFELECTRIC LINES OF FORCE www.citycollegiate.com

   Lines of force originate from a positive charge and terminate to a negative charge.

 

   The tangent to the line of force indicates the direction of the electric field and electric force.   Electric lines of force are always normal to the surface of charged body.

 

   Electric lines of force contract longitudinally.   They and expand laterally. www.citycollegiate.com   Two electric lines of force cannot intersect each other.   Two electric lines of force proceeding in the same direction repel each other.

   Two electric lines of force proceeding in the opposite direction attract each other. The line of force are imaginary but the field it represents as real.

   There are no lines of force inside the conductor.

ELECTRIC INTENSITY DUE TO A POINT CHARGE   

 

www.citycollegiate.comConsider a point charge q called SOURCE CHARGE placed at a point ‘O’ in space. To find its intensity at a point ‘p’ at a distance ‘r’ from the point charge we place a test charge 'q'.

The force experienced by the test charge q’ will be:

F = Eq’----(1)  According to coulomb's law the electrostatic force between them is given by:

  Putting the value of 'F' we get : www.citycollegiate.com

          

          

 

  This shows that the electric intensity due to a point charge is directly proportional to the magnitude of charge q and inversely proportional to the square of distance.

  EFFECT OF  

DIELECTRIC MEDIUM

  If there is a medium of dielectric constant (r) between the source charge and the field charge,intensity at a point will decrease r times i.e.

 

  VECTORIAL FORM www.citycollegiate.com

 

ELECTRIC FLUX   

 

GENERAL MEANING OFELECTRIC FLUX www.citycollegiate.com

In common language flux refers to the flow or stream of any thing from one point to another point. In the similar way electric flux is the total number of lines of force passing through a surface.

PHYSICAL MEANING OF ELECTRIC FLUX  

In physical sense, electric flux is defined as:

"The total number of lines of force passing through the unit area of a surface held perpendicularly."

  MATHEMATICAL MEANING OF ELECTRIC FLUX  

Mathematically the electric flux is defined as: "The dot product of electric field intensity (E) and the vector area (A) is called electric flux."

 

Where is the angle between E and A  MAXIMUM FLUX www.citycollegiate.com If the surface is placed perpendicular to the electric field then maximum electric lines of force will pass through the surface. Consequently maximum electric flux will pass through the surface.

 

  ZERO FLUX   If the surface is placed parallel to the electric field then no electric lines of force will pass through the surface. Consequently no electric flux will pass through the surface.

 

 Flux is a scalar quantity .  UNIT OF FLUX www.citycollegiate.com

 

ELECTRIC FLUX THROUGH A SPHERE     www.citycollegiate.com

Consider a small positive point charge +q placed at the centre of a closed sphere of radius "r". The relation  is not applicable in this situation because the direction of electric intensity varies point to point over the surface of sphere. In order to overcome this problem the sphere is divided into a number of small and equal pieces each of area A. The direction of electric field in each segment of sphere is the same i.e. outward normal.

Now we will determine the flux through each segment. www.citycollegiate.com   For latest information , free computer courses and high impact notes visit : www.citycollegiate.com Electric flux through the first segment:

Electric flux through the second segment:

 

Similarly,  Electric flux through other segments: www.citycollegiate.com

 

  Being a scalar quantity, the total flux through the sphere will be equal to the algebraic sum of all these flux i.e.

 

             

    

 

  

           

  This expression shows that the total flux through the sphere is 1/O times the charge enclosed (q) in the sphere.

  The total flux through closed sphere is independent of the radius of sphere .GAUSS’S LAW  

 

 

INTRODUCTION  Gauss’s law is a quantitative relation which applies to any closed hypothetical surface called Gaussian surface to determine the total flux (Ø) through the surface and the net charge(q) enclosed by the surface.

STATEMENT  "The total electric flux through a closed surface is equal to 1/o 

times the total charge enclosed by the surface."

PROOF  

 

Consider a Gaussian surface as shown below which encloses a number of point charges q1,q2,q3…….qn

Draw imaginary spheres around each charge. Now we make use of the fact that the electric flux through a sphere is q/o

Flux due to q1 will be Ø1 = q1/o Gaussian Surface

Flux due to q2 will be Ø2 = q2/o

Flux due to q3 will be Ø3 = q3/o

Flux due to qn will be Øn = qn/o

  Hence the total flux Øe will be the sum of all flux i.e  Ø = Ø1 + Ø2+ Ø3 +Ø4 ……….+ Øn                       

  Ø = q1/o+ q2/o+ q3/o+……….. +qn/  Ø = 1/o(q1+ q2+ q3+……….. +qn)                

  Ø = 1 /o  x ( total charge)                                

 This shows that the total electric flux through a closed surface regardless of its shape or size is numerically equal to 1 /o times the total charge enclosed by the surface.

ELECTRIC INTENSITY DUE TO A SHEET OF CHARGES       www.citycollegiate.com

Consider a plane infinite sheet on which positive charges are uniformly spread.Let ,The total charges on sheet = q Total area of sheet =A Charge density ()= q/A (charge per unit area )Take two points p and p’ near the sheet. Draw a cylinder from P to P'. Take this cylinder as a Gaussian surface. Consider a closed surface in front of cylinder such that p lies at one of its end faces.

The angle between E and normal n to the cylindrical surface is 90 .So the flux through the cylindrical surface: www.citycollegiate.com

 Ø = E A cos  Ø = E A cosØ = E A(0)     Ø = 0             

  The angle between E and normal n at the end of the surface P and P' is 0.hence the flux through one end surface P:

 Ø1 = E A cos Ø1 = E A cos Ø1 = E A (1)     Ø1 = E A            

  Similarly the flux through other end face P':  Ø2 = E A               Since electric flux is a scalar quantity, therefore, total flux through both surfaces is:

 Øt =  Ø1 + Ø2Øt = E A +E A     

      Øt = 2EA ...........(i)  According to gauss’s law : www.citycollegiate.com  Total flux through a closed surface is 1/  x (charge enclosed) i.e.              Øt = 1/  x q.............(ii)  Comparing equations (i) and (ii)

 

2EA = 1/  x q       E = q/  x 1/2A    

orE = q/  2A            E = (q/A) X 1/2 

  But q/A =  therefore,

 

  In vector formate: www.citycollegiate.com

 

  Which is the expression for electric intensity due to a infinite sheet of charge.

ELECTRIC INTENSITY BETWEEN TWO OPPOSITELY CHARGED PLATES

 

 

 

www.citycollegiate.comConsider two oppositely charged plates placed parallel to each other. Let these plates are separated by a small distance as compared to their size.

Surface density of charge on each plate is '' .Since the electric lines of force are parallel except near the edges, each plate may be regarded as a sheet of charges.

Electric intensity at a point between the plates due to positive plate: www.citycollegiate.com

 

Electric intensity at a point between the plates due to negative plate:

 

  Since both intensities are directed from +ve to –ve plate hence total intensity between the plates will be equal to the sum of E1 and E2

 

 

CAPACITOR  CAPACITOR

 

Capacitor is an electronic device, which is used to store electric charge or electrical energyA system of two conductors separated by air or any insulating material forms a capacitor as shown below:

  WE CANNOT STORE ELECTRIC CHARGE ON A SINGLE CONDUCTORThe size of a conductor required to store large amount of electric charge becomes very inconvenient because as the charge increases potential of plate also increases. Electric charge generated by a machine can not be stored on a conductor beyond a certain limit as its potential rises to breaking value and the charge starts leaking to atmosphere.

PRINCIPLE OF CAPACITOR  The principle of capacitor is based on the fact that the potential of a conductor is greatly

reduced

and its capacity is increased without affecting the electric charge in it by placing another earth connected conductor or an oppositely charged conductor in its neighborhood. This arrangement is therefore able to store electric charge.Capacitor are designed to have large capacity of storing electric charge without having large dimensions.For latest information , free computer courses and high impact notes visit : www.citycollegiate.com

PARALLEL PLATE CAPACITOR

 

A parallel plate capacitor consists of two conducting plates of same dimensions. These plates are placed parallel to each other. Space between the plates is filled with air or any insulating material (dielectric). One plate is connected to positive terminal and other is connected to negative term-inal of power supply. The plate connected to positive terminal acquires positive charge and the other plate connected to negative terminal acquires equal negative charge .The charges are stored between the plates of capacitor due to attraction.

 

  DEPENDENCE OF CHARGE STORED IN A CAPACITOR  

  Electric charge stored on any one of the plates of a capacitor is directly proportional to the potential difference between the plates . i.e.,

 

                                                                                 Q  V                                                                                     OR                                                                                 Q = (constant) V                                                                                  Q = CVWhere C = capacitance of capacitor

      

CAPACITANCE OF CAPACITOR     The ability of a capacitor to store electric charge between its plates is its capacity or

capacitance.  DEFINITION  

The capacitance of a capacitor is defined as: 

                                         "The ratio of electric charge stored on any one of the plates of capacitor to potential difference between the plates."Mathematically

  C = Q / VUNIT OF CAPACITANCE  

In S.I. system unit of capacitance is                                  Coulomb / volt                                              OR                                          FaradFarad is a large unit therefore in general practice we use small units(1) uF (microfarad) 1 uF = 1x10-6 F(2) uuF (Pico farad) 1 uuF = 1x10-12 FFor latest information , free computer courses and high impact notes visit : www.citycollegiate.com

  CAPACITANCE OF A PARALLEL PLATE CAPACITOR 

  Consider a parallel plate capacitor as shown below:

 

Let The area of each plate = AThe separation between plates = dMedium = airSurface density of charge on each plate =  The electric field intensity between the plates of capacitor is given by

  E = / o

  Potential difference between the plates of capacitor can be calculated by the following relation

 

V = Ed

Putting the value of "E"

V = /o) x d

But  = Q/A

V = Qd/A.o

AoV = Qd…………(a)  We know that 

The electric charge stored on any one of plate of capacitor is

 

Q = CV……………(b)

Putting the value of Q in (a)

AoV = CVd

Cd = Ao  C = Ao/d

CAPACITANCE IN THE PRESENCE OF DIELECTRIC

   

 

1-When dielectric is completely filled between the platesLet the space between the plates of capacitor is filled with a dielectric of relative permittivity r.The presence of dielectric reduces the electric intensity by r times and thus the capacitance increases byr times. 

C'= C x r

 

  

1-When dielectric is partially filled between the plates  

  CONCLUSION   1- Capacitance can be increased by increasing the dimensions of plates.  C  A 2- Capacitance can be increased by decreasing the separation between the plates.  C  1/d 3- Capacitance of a capacitor increases by the presence of dielectric medium between the plates.

  C  r