COULOMB’S LAW
INTRODUCTION
|
|
| The 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 LAW
|
|
| 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
|
|
| Value of K is equal to 1/4pe0 |
| where eo 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 = 1/4pe0 in equation (i) |
|
|
FORCE IN THE PRESENCE OF DIELECTRIC MEDIUM
|
|
| 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 (er) called DIELECTRIC CONSTANT. It is denoted by er . This factor is also known as RELATIVE 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
|
|
| When 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 attraction
or 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
|
|
N/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. |
|
|
|
CHARACTERISTICS OF ELECTRIC LINES OF FORCE
|
|
|
 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. |
| 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
| Consider 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 : |
|
|
|
|
|
|
| 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 (er) between the source charge and the field charge,intensity at a point will decrease er times i.e. |
|
|
|
VECTORIAL FORM
|
|
|
|
|
ELECTRIC FLUX
GENERAL MEANING OF ELECTRIC FLUX
|
|
| 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 (DA) is called electric flux." |
|
Where q is the angle between E and DA
|
|
MAXIMUM FLUX
|
|
|
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
|
|
|
|
ELECTRIC FLUX THROUGH A SPHERE
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 DA. 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.
Electric flux through the first segment:
|
|
|
|
| Electric flux through the second segment: |
|
|
| Similarly, |
| Electric flux through other segments: |
|
|
| 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/eO 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/eo 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/eo |
| Flux due to q1 will be Ø1 = q1/eo
Flux due to q2 will be Ø2 = q2/eo
Flux due to q3 will be Ø3 = q3/eo
Flux due to qn will be Øn = qn/eo |
|
| Hence the total flux Øe will be the sum of all flux i.e |
|
Ø = Ø1 + Ø2+ Ø3 +Ø4 ……….+ Øn
|
|
Ø = q1/eo+ q2/eo+ q3/eo+……….. +qn/e
|
|
Ø = 1/eo(q1+ q2+ q3+……….. +qn)
|
|
Ø = 1 /eo 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 /eo times the total charge enclosed by the surface.
ELECTRIC INTENSITY DUE TO A SHEET OF CHARGES
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 (s)= 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: |
|
|
Ø = E A cos q
Ø = E A cos90Ø = 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 q Ø1 = E A cos 0 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 = D Ø1 + D Ø2
Øt = E A + E A
Øt = 2EA ...........(i)
|
| According to gauss’s law : |
| Total flux through a closed surface is 1/ eO x (charge enclosed) i.e. |
|
Øt = 1/ eO x q.............(ii)
|
| Comparing equations (i) and (ii) |
|
2EA = 1/ eO x q
E = q/ eO x 1/2A
or
E = q/ eO 2A
E = (q/A) X 1/2 eO
|
| But q/A = s, therefore, |
|
|
| In vector formate: |
|
|
| Which is the expression for electric intensity due to a infinite sheet of charge.
ELECTRIC INTENSITY BETWEEN TWO OPPOSITELY CHARGED PLATES
| Consider 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 's' .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: |
|
|
| 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 energy
A 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 CONDUCTOR
|
| The 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.
|
|
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 a 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 / V
|
UNIT OF CAPACITANCE
| |
| In S.I. system unit of capacitance is
Coulomb / volt
OR
Farad
Farad 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 F
|
|
CAPACITANCE OF A PARALLEL PLATE CAPACITOR
| |
| Consider a parallel plate capacitor as shown below: |
| Let
The area of each plate = A
The separation between plates = d
Medium = air
Surface density of charge on each plate = s
The electric field intensity between the plates of capacitor is given by |
|
E = s / eo
|
| Potential difference between the plates of capacitor can be calculated by the following relation |
|
V = Ed
Putting the value of "E"
V = (s/eo) x d
But s = Q/A
V = Qd/A.eo
AeoV = 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)
AeoV = CVd
Cd = Aeo
|
|
C = Aeo/d
|
|
CAPACITANCE IN THE PRESENCE OF DIELECTRIC
|
|
| | |
| 1-When dielectric is completely filled between the plates |
Let the space between the plates of capacitor is filled with a dielectric of relative permittivity er.
The presence of dielectric reduces the electric intensity by er times and thus the capacitance increases by er times.
C'= C x er
|
|
|
|
|
| |
|
| 1-When dielectric is partially filled between the plates |
| |
|
|
|
|
CONCLUSION
| |
| 1- Capacitance can be increased by increasing the dimensions of plates. |
|
C a A
|
| 2- Capacitance can be increased by decreasing the separation between the plates. |
|
C a 1/d
|
| 3- Capacitance of a capacitor increases by the presence of dielectric medium between the plates. |
|
C a er
|
|
|
|
|
