coulomb force
TRANSCRIPT
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COULOMBS LAW
INTRODUCTIONThe magnitude of the force of attraction or repulsion between two electriccharges 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 isdirectly proportional to the product of charges.
The electrostatic force of attraction or repulsion between two point charges isinversely proportional to the square of distance between them.
MATHEMATICALREPRESENTATION OF
COULOMB'S LAWConsider two point charges q1 and q2 placed at a distance ofr 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 ofK is equal to 1/4 0where ois permittivity of free space .Its volume is 8.85 x 10-12 c2/Nm2.Thus in S.I. system numerical value ofK is 8.98755 x 109 Nm2c-2.
OTHER FORMS OF
COULOMB'S LAW
Putting the value ofK = 1/4 0in equation (i)
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FORCE IN THE PRESENCEOF DIELECTRIC MEDIUM
If the space between the charges is filled with a non conducting medium or aninsulator called "dielectric", it is found that the dielectric reduces the electrostaticforce as compared to free space by a factor ( r) called DIELECTRICCONSTANT. It is denoted by r . This factor is also known as RELATIVEPERMITTIVITY. It has different values for different dielectric materials.In the presence of a dielectric between two charges the Coulomb's law isexpressed as:
VECTOR FORM OF
COULOMB'S LAW
The magnitude as well as the direction of electrostatic force can be expressed byusing Coulomb's law by vector equation:
Where is the force exerted by q1 on q2 and is the unit vector along theline joining the two charges from q1 to q2.
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ELECTRIC FIELD - ELECTRIC INTENSITY
ELECTRIC FIELD
When an electric charge is placed in space, the space around the charge is modifiedand if we place another test charge within this space, the test charge will experiencesome electrostatic force. The modified space around an electric charge iscalled '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
whichanother charge experiences some electrostatic force of attraction
or repulsion when placed at a point is called Electric Field.
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ELECTRIC INTENSITY
Electric intensity is the strength of electric field at a point.Electric intensity at a point is defined as the force experiencedper unit positive charge at a point placed in the electric field.
orIt 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 LINESOF FORCE
In order to point out the direction of an electric field we can draw a number of linescalled electric lines of force.
DEFINITION
An electric line of force is an imaginary continuous line or curve drawn in an electricfield such that tangent to it at any point gives the direction of the electric force atthat point.The direction of a line of force is the direction along which a small freepositive charge will move along the line. It is always directed from positive charge tonegative charge.www.citycollegiate.com
CHARACTERISTICS OF ELECTRICLINES OF FORCE
Lines of force originate from a positive charge and terminate to a negative charge.
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The tangent to the line of force indicates the direction of the electric field andelectric 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 atest charge 'q'.
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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 directlyproportional to the magnitude of charge q and inversely proportional to thesquare of distance.
EFFECT OFDIELECTRIC
MEDIUM
If there is a medium of dielectric constant ( r) between the source charge andthe field charge,intensity at a point will decrease r times i.e.
VECTORIAL
FORM
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ELECTRIC FLUX
GENERAL MEANING
OF ELECTRIC FLUXIn common language flux refers to the flow or stream of any thing from one point toanother point. In the similar way electric flux is the total number of lines of forcepassing through a surface.
PHYSICAL MEANING OFELECTRIC FLUX
In physical sense, electric flux is defined as:
"The total number of lines of force passing through the unit area of a surfaceheld 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
If the surface is placed perpendicular to the electric field then maximum electric linesof force will pass through the surface. Consequently maximum electric flux will passthrough the surface.
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ZERO FLUX
If the surface is placed parallel to the electric field then no electric lines of force willpass 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 directionof electric intensity varies point to point over the surface of sphere. In order toovercome this problem the sphere is divided into a number of small and equalpieces each of area A. The direction of electric field in each segment of sphereis the same i.e. outward normal.
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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 thealgebraic sum of all these flux i.e.
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This expression shows thatthe total flux through the sphere is 1/ Otimes thecharge enclosed (q) in the sphere.
The total flux through closed sphere is independent of the radius of sphere .
GAUSSS LAW
INTRODUCTION
Gausss law is a quantitative relation which applies to any closed hypotheticalsurface called Gaussian surface to determine the total flux () 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/ otimesthe total charge enclosed by the surface."
PROOF
Consider a Gaussian surface as shown below which encloses a number of pointcharges q1,q2,q3.qnDraw imaginary spheres around each charge.Now we make use of the fact that
the electric flux through a sphere is q/ oFlux due to q1 will be1 = q1/ o
Gaussian Surface
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Flux due to q2 will be2 = q2/ o
Flux due to q3 will be
3 = q3/ o
Flux due to qn will be
n = qn/ o
Hence the total flux e will be the sum of all flux i.e
= 1 + 2+ 3 +4 .+ n
=q1/ o+ q2/ o+ q3/ o+.. +qn/ =1/ o(q1+ q2+ q3+.. +qn)
= 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 to1 / o times the total charge enclosed bythe 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 = qTotal area of sheet =ACharge density ( )= q/A (charge per unit area )Take two points p and p near the sheet. Draw a cylinder from P to P'. Take thiscylinder as a Gaussian surface. Consider a closed surface in front of cylinder
such that p lies at one of its end faces.
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The angle between E and normal n to the cylindrical surface is 90 .So the flux
through the cylindrical surface:www.citycollegiate.com
= E A cos = E A cos90 = E A(0) = 0
The angle between E and normal n at the end of the surface P and P' is 0.hencethe flux through one end surface P:
1 = E A cos 1 = E A cos 01 = E A (1)1 = E A
Similarly the flux through other end face P':
2 = E A
Since electric flux is a scalar quantity, therefore, total flux through both surfacesis:
t = 1 + 2t = E A + E A
t = 2EA ...........(i)
According to gausss law : www.citycollegiate.com
Total flux through a closed surface is 1/ x (charge enclosed) i.e. t = 1/ x q.............(ii) Comparing equations (i) and (ii)
2EA = 1/ x qE = q/ x 1/2A
or
E = q/ 2AE = (q/A) X 1/2
But q/A = , therefore,
In vector formate: www.citycollegiate.com
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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 theseplates are separated by a small distance as compared to their size.
Surface density of charge on each plate is ' ' .Since the electric lines of forceare parallel except near the edges, each plate may be regarded as a sheet ofcharges.
Electric intensityat a point between the plates due to positiveplate: www.citycollegiate.com
Electric intensity at a point between the platesdue to negative plate:
Since both intensities are directed from +ve to ve plate hence total intensitybetween the plates will be equal to the sum of E1 and E2
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CAPACITOR
CAPACITOR
Capacitor is an electronic device, which is used to store electric charge or electricalenergyA system of two conductors separated by air or any insulating material forms acapacitor asshown below:
WE CANNOT STORE ELECTRIC CHARGE ON A SINGLE CONDUCTOR
The size of a conductor required to store large amount of electric charge becomesvery 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
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certain limit as its potential rises to breakingvalue 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 isgreatly reducedand its capacity is increased without affecting the electric charge in it by placinganother earth connected conductor or an oppositely charged conductor in itsneighborhood. This arrangement is therefore able to store electric charge.Capacitor are designed to have large capacity of storing electric charge withouthaving large dimensions.For latest information , free computer courses and high impact notesvisit :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 anyinsulating material (dielectric). One plate is connected to positive terminal andother is connected to negative term-inal of power supply. The plate connected to positive terminal acquires positivecharge and theother plate connected to negative terminal acquires equal negative charge .Thecharges are stored between the plates of capacitor due to attraction.
DEPENDENCE OF CHARGE STORED IN ACAPACITOR
Electric charge stored on any one of the plates of a capacitor is directly proportionalto the potential difference between the plates . i.e.,
Q V OR
Q = (constant) VQ = CV
Where C = capacitance of capacitor
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CAPACITANCE OF CAPACITOR
The ability of a capacitor to store electric charge between its plates is its capacityor 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 differencebetween the plates."
Mathematically
C = Q / V
UNIT OF CAPACITANCE
In S.I. system unit of capacitance isCoulomb / volt
ORFarad
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 PLATECAPACITOR
Consider a parallel plate capacitor as shown below:
LetThe area of each plate = AThe separation between plates = dMedium = air
Surface 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 thefollowing relation
V = Ed
Putting the value of "E"
V = ( / o) x dBut = Q/AV = Qd/A. o
A oV = Qd(a)
We know thatThe electric charge stored on any one of plate of capacitor is
Q = CV(b)
Putting the value of Q in (a)
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A oV = CVdCd = A o
C = A o/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 thecapacitance increases by r times.
C'= C x r
1-When dielectric is partially filled between the plates
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CONCLUSION
1- Capacitance can be increased by increasing the dimensions of plates.
C A2- Capacitance can be increased by decreasing the separation between the plates.
C 1/d3- Capacitance of a capacitor increases by the presence of dielectric medium betweenthe plates.
C r