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## STATIC ELECTRICITY

It can be shown that there are two kinds of charges by rubbing a glass rod with silk and hanging it from a long silk thread. If a second glass is rubbed with silk and held near the rubbed end of the first, the rod will repel each other.

On other hand, a hard rubber is rubbed with far, will attract the glass rod rubbed with silk. The modern view of bulk matter is that in its normal i.e neutral, it contains equal amount of positive and negative charges.

If two bodies like glass and silk are rubbed together, a small amount of charge is transferred from…… to the other upsetting the electric neutrality of each. In this case the glass would become positive and silk negative.

FORCE BETWEEN TWO CHARGES OR ANY TWO CHARGED BODIES

Columb found that a for force exist between two electrically charged bodies and that this force and the distance between the charged bodies obey the inccerse square law ie If is the distance between the charged bodies and F is the force of attraction between these charged bodies the

Fα……………………………….(i)

Equation (i) above is known as the inverse square law.

It has been found that if Q1 and Q2 are how charged then the force of attraction between the is given by

FαQ1Q2………………………………..(ii)

Combining eq (i) and (ii)

The permittimity of air at normal pressure is only about 1.005 times that of uacumum (Æ• 0) . For most purpose therefore we may assume the value of Æ• 0 for the permittinity of air.

- Calculate the value of two charges if they are one another with a force of 0.1 when situated 50cm apart in a vacuum
- What would be the size of the charges if they were situated in an insulating liquid whose permitting was 10 times that of vaccum.

Solution

F=0.1

R= 50cm=0.5m

Æ• â—¦=8.854×10-12

Q1=Q2 They repel=Q

Formula

Q2=Fxð• ›‗Æ• â—¦r2

Q=[0.1x4x3.16×8.8854×10-12x(0.5)2]1/2

Q=1.67×10-6C

(b) Given

F=0.1

Q1=Q2=Q

Æ• â—¦= 10

r=50c m=0.5m

Formula

Q2=Fx4xÆ• r2

Q= (0.1x4x3.14x10x8.854x (E-12)x(0.5×0.5)

Q=5.27x E-6 C

1998P2 Qns

The distance between the electrical proton in the hydrogen atom is about frictional force between those particles.

Me=9.11E-3 kg Mp=1.67E-27 e=1.6E-19c Æ• =6.67E-11Nm2kg-1 r=5.3E-11m

Fe=7.25E-29N, Fg=3.61E-19

- Gravitational force between particles

Fg α

=9.11E-31kgx1.67/(5.3E-11)2M2 Fg=3.7E-47

N

- Fe

Fe=8.1E-8 N or 8.1×10-8

ELECTRIC FIELD INTENSITY (E)

Electric field intensity is defined as the region which in electric force is experienced. So an electric field intensity E if an electrostatic field at any point is defined as the force per unit charges which it existence once positive charges.

· If Q is a small test charge placed on a point them

The SI unit of E is NC-1

QUESTION

Find the magnitude of an electric field strength such that an electron placed on the field would experience an electrical force equal to its weight mass of electron =9.1×10-31 kg, e=1.6×10-19C, g=8.9m/s2

Consider a test charge Q0 in vacuum which is placed a distance “r” from a point charge (isolated) charge Q0

By Coulombs Law the magnitude of a force acting on Qâ—¦ is given by

By putting eq (i) in eq(ii)

Note:

- E is the vector quantity
- From Æ•
- The direction of Æ• is radial line from Q point …… out towards. If Q is positive and inside if Q is negative

LINES OF FORCE

There (are) is a relationship between imaginary line of force and the electric field strength as follows.

(i)The tangent to a line of the force to any point gives the direction of Æ• at that point

The lines of force are drawn so that the member of force permit crossectional area is proportional to the magnitude of Æ• . Where the lines are close together Æ• is given as large and where they are far apart Æ• is small.

ELECTRIC FLUX (ØÆ• )

The electric flux ØÆ• through an area perpendicular total lines

of force is the product of. Ex area where E is the electric intensity at that place.

of force is the product of. Ex area where E is the electric intensity at that place.

Consider a sphere of radius drawn in a space concentric with a point charge.

Total flux through the sphere is given by

Ð¤Æ• =E x Area of the sphere

=Ex4ð• ›‗r²

=x 4ð• ›‗r²

i.e

If the charge is placed at any other medium apart from air or vacuum then

Ð¤ =

The above equation shows that the total flux crossing any point at drawn sphere concentrically outside the point charge is constant.

- (OUT SIDE THE CHARGED SPHERE)

ELECTRIC FIELD INTENSITY DUE TO A CHARGED SPHERE

The flux across a spherical surface of radius or concentric with a small sphere carrying charge Q is given

This result shows that the outside of a charged sphere the field behaves as if all charges on the sphere are concentrated at the centre.

- INSIDE A CHARGED EMPTY SPHERE.

Inside the empty charged sphere there are no charges so the electric field strength E = O therefore;

since Q=0,then the value of inside the sphere is also 0

##### 3. ELECTRIC INTENSITY OUTSIDE THE CHARGED PLATES

Consider a charged plane conductor S with a surface change density of 6cm-2 let the plane surface 1 as shown above be drawn outside the 5 which is parallel to 5 and has the area Am2

The intensity of in the field must be perpendicular to the surface and the charges will produce this field are those in projection of the area P on the surface S i.e those within the shaded region A.

Question.

A particle of mass M and charge of q is placed at rest in a uniform electric field see the fig below and released. Describe its motion.

The motion reassemble that of the falling body in the earth’s gravitational field. The constant to

acceleration is given by

The equation of uniform acceleration to apply therefore with

The vertical distance moved by a particle with initial velocity

By putting eqn(1) in (3) we get

From the third equation of motion we have

Putting eqn(1) into eqn(5)

The kinetic energy attached at the moving a distance y is formed from :-

substitute eqn(6) into (7) we get

ELECTRIC POTENTIAL (V)

The electric field around a charged and can be described not only by a vector electric field strength E but also by a scalar quantity i.e the electric potential,v.

To find the electric potential difference between two points A and B in an electric field we move a test charge q from A to B and we measure the work

WAB that must be done by agent moving the charge.

Electric potential difference ,v ca be expressed in the form of

……………………………………(i)

The unit of the potential difference is obtained for equation (i) that is JC. However volts is also used.

1JC1=1Volts

If point A is chosen to be at very far (say at infinity) then the electric potential at infinity distance is arbitrarily taken et zero.

Therefore then putting VA=0 1 in equation (i) and dropping the subscripts we get

…………………………………………………… (ii)

Definition:

The electric potential at the point is the work done by the force in taking the unit chart from infinity to that point.

CALCULATION OF WORK DONE

Consider a positive charge Q to be at Ra distance as indicate in the figure the work done in taking the charge from A to B is equal to the work done in taking the same distance from B to A. If Q0 is moved by the fo….. from A to B then the force acting on it is

If the charge has moved a distance ∂x the work done is

dw

Hence total workdone is taking Qo from A to B is

We get

WAB=+…………………………..(ii)

The potential difference between A and is given by

……………………………………….(iii)

Substitute eq-(ii) into (iii)

We get

Therefore

]

If rB is very large i.e B is at the infinity the potential at A is given by

So in general the potential v at a distance r from the point charge Q is given by:

POTENTIAL DUE TO SEVERAL CHARGES

The potential at any point due to a group is found by

- Calculating the potential Vn due to each charge as if other charges are not present.
- Adding the quantity so obtained or

1) Where question is the value of the charge and is the distance of this change from the point in question

##### Question

Calculate the potential at the centre of the squire shown below

2) Eight charges having the values. Shown in the figure below one arranged systematically on the circle of radius 0.4m in air Calculate the potential at the centre 0.

RELATIONSHIP BETWEEN ELECTRIC INTENSITY (E) AND ELECTRIC POTENTIAL V:

Consider two points A and B at a distance x and x +dx from 0 respectively V and B Vd are respectively potential A and B are very close. So that the electric intensity E is instant. Hence the potential between A and B is VAB = VA -VB

Qn 1985 p Qn 13

QN 1994. P, Qn B

An electron is liberated from the lower part two large parallel plates separated by a distance h. The upper plate has a potential of 2400V relative lower. How long does the electron take to reach

##### Solutions

Where m is the mass of an electron.

EQUIPOITENTIAL SURFACE

It is the evident from the equation V = Q/4ÐŸÆ• that all points which are at the same distance from point charge are at the same potential

Any surface over which the potential is constant is called equipotential surface

Equipotential surface has the surface property that along direction lying on the surface there is No electric field for there is no potential difference dv/dx =0 since v is constant.

figure (a) And (b) above shows equipotential surface dashes times while solid (continuously) lines duplicating lines force.

Graph- shown below shows the variation of electric field strength within an isolated sphere and that at external part of it.

Graph below shows variation of electric potential at an isolated sphere and at the external part of it.

TYPES OF CAPACITOR

(i)VARIABLE AIR CAPACITOR

Variable (capacitance) capacitor is the one in which the effective area of the plates can be adjusted. The capacitance of variable capacitor can be varied as you wish but at a certain limits.

These are widely used in the turning circuits of radio receivers. They are constructed of number affixed parallel metal plates. Plates. Connected together and constituting one plate of the capacitor. The second parts of movable plates also connected together and form the other plates.

By rotating plates on which the movable plates are mounted the second set may be caused to interleave the first to a lesser or greater extent.

The effective area of the capacitor is that of the interleaved portion of the plates only

The plats of the capacitor may be made of brass or Aluminum. The dielectric may be oil, air or mica.

(ii)A MULTIPLE CAPACITOR MICA DIELECTRIC

Tin foil

The plates of this type of capacitor are made of tin foil. The capacitance of the capacitor is ―n‖ times the capacitance (of the) between the two successive plates. When ―n‖ is the number of dielectric between the plates.

Qn 1999. P2 Qn Sc

Given that the distance of separation between the plates of a capacitor is 5mm and the plate has an area of 5m2 .A potential difference of 10V is supplied the capacitor which is parallel in vacuum. compute

- The capacitance.
- The electric intensity in a space between the plates
- The change in stored energy, if the separation of the plates is increased from 5m to 5.5mm. Æ• â—¦= 8.85 pF/m

(iii)PAPER CAPACITOR

The paper capacitor has a dielectric of paper impregnated with paraffin wax or oil unlike the mica capacitor the paper can be rolled and scatted into a cylinder relatively small volume. Now a days the paper has been replaced by thin layer of polystyrene.

- ELECTROLYTIC CAPACITOR

They are produced by passing a direct current between the two sheets of aluminum foil with a suitable electrolyte of lightly liquid conductor between them. A very thin film of aluminum oxide is then formed on the oxide plate which is of positive side of the d.c supply.

This film is an insulator. It forms dielectric in between two plates the electrolyte being of a good conductor.

Since the dielectric d is very small and C α 1/d the capacitance value can be very high.

ARRANGEMENT OF CAPACITOR

##### 1) Parallel arrangement of capacitor

All the left hand plates are connected together and all the right hand plates are connected together and in the case of parallel arrangement of capacitors (see figure above)

When a cell is connected across these capacitor is parallel they have the same potential difference (v)

So

- Q1 = C1V
- Q2 = C2V

(ii) Q3 = C3V

Let the total charge be Q then Q = Q1 + Q2 + Q3………..(4)

Put

(ii) SERIES

ARRANGEMENT OF CAPACITOR.

ARRANGEMENT OF CAPACITOR.

When the right hand plates of one capacitor is connected to the left hand of the next and so on then these capacitors are said to be connected in series.

When the cell is connected across of the end of the system a charge is transferred from the plates H to A, A charge –Q being left on it. This charge induce a charge +Q on plate Q. This process is repeated with other plate.

Question

Two capacitors of capacitance C1=2μF and C2= 8μF are connected in series and the resulting combination is connected across 300volt. Calculate the charge and potential difference

Solution

Qn. 1998 Qn 13

Capacitor of 5μF and 25μF are connected in series and the combination is connected to the battery of 90 volt. Calculate

- Charge on each capacitor
- The p.d across each capacitor

ENERGY STORED IN A CAPACITOR

Consider a capacitor of capacitance

C to have been charged to a potential difference V and let a small charge dQ be transferred from the negative plate to positive plate. Then the work don‘t in moving a charge dQ will be

C to have been charged to a potential difference V and let a small charge dQ be transferred from the negative plate to positive plate. Then the work don‘t in moving a charge dQ will be

dw = VdQ but V=Q/C

Suppose a capacitor is at first discharge d and then charged until the final charge on the plate is Q The work done in charging it is given by

W=

In general if C is capacitance of a capacitor carrying charge Q at potential difference then,

W=…………………(i)

But Q is equal to CV then (i) becomes

W=CV²…………………….(ii)

Also C=…………………………(iii) put into (i) we get

W=QV…………………………..(iv)

Equation (i), (ii) and (iv) give the energy stored in a capacitor.

Question

1998 P2B Qn

A capacitor of a capacitance 3ð• œ‡F is charged until a potential difference of 200v is developed across. Its plat…..another capacitor of capacitance 2ð• œ‡F developed apd of 100v across its plates on being charged.

- What is the energy stored on each capacitor?
- The capacitors…them connected by a wire of negligible resistance so that the plates carrying like charges are connected together. What is the total energy stored in the combined capacitors?

Solution

C1=3×10-6F

V1= 200v

C2=2×10-6F

V2=100v

Formula

E1=C1V²

=1/2x3x10-6x(200)

=6×10-2 Joules

E1=0.06 Joules

E2=C2V2

=1/2x2x10-6x104 =10-2Joules

(ii) =

C=

=

=6/5

=1.2

C=1.2×10-6F V=V1+V2, 100+200=300V

E=CV²

=1/2×1.2×10-6x(300)²

=0.6×10-6x 9×104

=5.4×10-2 Joules.

The total energy stored in the combined capacitor 5.4×10-2

Question

The capacitance of a parallel plate capacitor is 400 pic…Farad and its plates are separated by 2m of air

i. What will be the energy when it is charged to 500 volts.

Æ• â—¦=8.854×10-12C2NM

How much work (energy) is needed to double the distance between the plates?

Solution

C=400×10-12 F

D=2×10-3m

Æ• â—¦=8.854×10-12C2N-1M-2

Energy stored= C V or QV or QV

=5X10-5

=2.5X10-5

Energy needed to separate /double the space twice is 2.5×10-5 J

DISCHARGE IN C-R CIRCUIT

Consider a capacitor initially charged to a p.d V0 so that its charge is that Q=CV0

At a time t, after the discharge through R has begun the current, I flowing=V/R………………(i) where V is the potential difference across C

Now V=………………….(ii) and I=……………………………(iii)

The minus sign shows that Q decreases with increasing time from equation (i), (ii) and (iii) we have

…………………………….(iv)

=mQ-mQ0=t

=ln

==Q=Q0……………………(v)

From equation (v) Q decreases experimentally with time

Since the P.d, v across C is proportional to Q then V=V0e-t/C also since is the circuit is proportional to v

then, I=I0e-t/CR where I0 is the initial current value from the equation(i) Q decreases from Q0 to half of its value

in time t given by

e-t/C=1/2=2-1

Taking logarithm to base e both sides we get

=-ln2

Therefore the time for a charge to diminish to half its initial value no matter what the initial value may be is always the same.

TIME CONSTANT

The time constant (T) of the discharge circuit is defined as CR seconds where C is the in Farad and R is in Ohms.

A resistor of the resistance R=10â„¦ is connected in series with a capacitor of capacitance 1F. Find the time constant and half life.

Solution

C=1×10-6F

R=10×106â„¦

Time constant CR=10-6x107

=10 Seconds

T1/2=10……..=10×0.693

The time constant =10 second

Half life=0.693

CHARGING OF CAPACITOR THROUGH A RESISTOR

Consider a charging of a capacitor with capacitance C through a resistor R in series

.

.

Suppose the supplied battery has an e.m.f E and negligible internal resistance.

Initially there is no charge on care so no p.d across after connecting the battery the P.d across R=E the

applied circuit P.d the initial current flow I0=. Suppose I is the current flowing after a time t, the if VC is the P.d across C

CR=CE-Q……………………………(iv)

CE is the final charge on C when no further current flows through Q0.

Therefore equation (iv) becomes

CR=CE-Q

Integrating the above expression gives

=

Let U=Q0-Q du=-dθ

–-ln u=-ln(Q0-Q)from 0-Q

-ln[ln(Q0-Q)-L=lnQ]= +ln(

ln(

Q0-Q=Q0

To show that CR i.e the time constant takes the unit of time i.e seconds

CR can be expressed as

C(Farad)=

R=Ohms= =

CR==Seconds

The plates have the electric flux

Ð¤=E x A…………………………..(i)

But also Ð¤=…………………………….(ii)

From equation (i) and (ii) we have

EA=……………………………………(iii)

The work required to take a test charge Q0 from the plate to the other is

W=Force x Distance

W=EQ0d…………………………………………….(iv)

But also work done

W=Q0V………………………………………(v)

Q0V= Æ• Q0d

V=Æ• d……………………………..(vi)

Put equation(iii) into

equation (vi

equation (vi

V=……………………………………(vii)

But capacitance=C=…………………..(viii)

Substitute equation (vii) into equation (viii) we get

C=

If the space between plates is filled with air or vacuum, the equation above becomes

C=

QUESTION

1997P2 Qn 6

What is the capacitance of a parallel plates capacitor if it consist of a mica 0.1mm thick and 1.5cm2 silvered o both sides the permittivity of mica plate is 6.0 (Æ• â—¦=8.854×10-12C2N-1M-2)

QUESTION

What is the electric field strength at the surface of a…………..metal sphere of radius 100mm if it carries the charge 2×10-7 in vacuum Æ• â—¦=8.854×10-12C2N-1M-2)

Solution

=1.8×10 NC-1

CAPACITANCE

It can be shown by sending a positive or negative charge close to a charged body lower s or va….the potential difference between the system of the charged bodies. This shows that the charge is proportional to the potential difference.

If Q is the charge and v is the p.d then

Q αV i.e

Q=CV……………………..(i)

Where the constant C is known as Capacitance.

Definition

The capacitance of a system of bodies is the charge necessary to raise potential by a unit i.e

C=

Capacitance is measured in Farad but other units like ð• œ‡F and Ηf are also used.

The symbol of capacitor is The two parallel…………found in the symbol above can also be horizontal i.e depending on the circuit.

THE DERIVATION OF CAPACITANCE IN PARALLEL PLATES CAPACITOR Consider the two parallel plates capacitor with plates ……………….

Each plate is carrying charge equal to Q. If the electric field strength between the plates

The plates have electric flux Ð¤ = E x A ……………(1)

But also Ð¤ = Q/ Æ ……………………………….(2)

From equation (1 ) and (2) we have

EA = Q/ Æ• â—¦ or E = Q/A ………………………..(3)

The work required to take a test charge from one plate to another is

W = Force x distance

W =E Qâ—¦d……………………………(4)

But also work done= w Qâ—¦V………….(5)

Qâ—¦V = Æ• â—¦ Qâ—¦d

V = Æ d ……………………….(6)

Put equation (3) into (6) we have

V = Qd/A Æ• â—¦ …………………..(7)

But capacitance ,C = Q/V …………………………(8)

Substitute eqn(7) into (8) we get

C = A Æ• â—¦ /d

If the distance between plates is field with air or vacuum the equation above becomes C = A Æ• â—¦ /d

NECTA 1997 P2 Qn 6

What is the capacitance if parallel plates capacitor it is consist of a sheet of mica 0.1mm thick and 1.5cm square silvered on both sides? The permittivity of mica is 6.0 times that of vacuum (Æ• â—¦ = 8.854 x 10-12 JN-1m2)

Solution C=AÆ•/d

But

A=22.5×10-4m2

D=1.0×10-4m

Æ•=6×8.854x10CN-1m-2

C=1.2×10-10F

FACTORS WHICH DETERMINE CAPACITANCE

i.Distance between the plates

From the equation V = Cd/AÆ• It follows that if more far apart the p.d will increase and hence C=Q/v

The capacitance decrease. Therefore the capacitance decreases when the separation of the plates increase.

ii.Dielectric

When the dielectric e.g sheet of glass or ebonite between the potential difference between the plates decreases i.e V=Qd/AÆ•

Hence from C=Q/v

The capacitance has increased. Therefore so when Æ• r increases the capacitance will also increase.

iii.Area of plates

From V = Qd/Æ• A

It follows that the p.d decreases area of the plates.Increases, so the capacitance C = Q/V must increase.

DIELECTRIC CONSTANT(Æ• r)

Dielectric constant is also known relative permittivity

The ratio of the capacitance with dielectric to the one without the dielectric between the plates is called dielectric constant or relative permittivity of the material used.

Consider the case of parallel plates capacitor capacitance with dielectric (C)=AÆ• /d and capacitance without dielectric (Câ—¦)=AÆ• â—¦/d

Hence;

Æ• r=

Æ• r= or Æ• r.Æ• 0…………………..(i)

From equation (i) i

t follows that the capacitance of a capacitor ca also be given as

t follows that the capacitance of a capacitor ca also be given as

…………………………..(ii)

A capacitor which leads to capacitance is a device for storing charges. Essentially all capacitors five metal plates separated by an insulator. The insulator are called dielectric in some capacitors dielectric used are oil, air, polyethylene, etc.

1997P2 Qn 6

Determine the capacitance of a parallel plates capacitor if it consist of a sheet of mica 0.1mm thick and surface area 15cm2, silvered on both sides and that mica has dielectric constant of 6.0 Solution

Given A=1.5×10-4m2

d= 1×10-4=10-4

Æ• 0=8.854×10-12

Æ• r=6

Formula

C=7.96×10-11 Farad.

Question

A paper capacitor consists of a sheet of paper 35 width, 10mm long and 2.0×102mm thick between sheets metal foil. If the relative permittivity of paper is 2.7, what is the capacitance? Æ• 0=8.85×10-12 Solution

Æ• r=2.7

Area=35x10x10-3m2=3.5×10-1m2

d=2.0 x10-2 x 10-3=2 x 10-5 m

Æ• 0= 8.85 x 10-12

VAN DE GRAAFF GENERATOR

If a positively charged body makes extrexual contact with an uncharged body the first will loose some of its charge and its potential will decrease while the sea will gain charge and its potential will increase, the flow of charge will cease when both bodies are at the same potential but there will be still remain some charges from the first body.

Consider the figure shown below of Van De Graaff generator

It consists of a hallow metal conductor A approximately spherical is supported and an insulating tube B mounted on a metal base C which is normally grounded. A non conducting ended belt D runs over two non conducting pulley E and F. Pulley F may be driven by hand or by small electric motor pulley and F are covered by different materials chosen so that when the belt D makes contact with F acquires positive charges while on contact with E it acquires negative charges

The charges developed on the belt as it makes contact with the pulleys, stick to it and are carried along by it. As the belt passes through it induces a charge on it , this conductor…..which because of the sharp point result in sufficiently high field intensity to ionize the air between the point and the belt. As the belt leaves the pulley, E it becomes negatively charged and the right hand side of the belt carries negative charge out of the upper terminal. Removal of negative is equivalent to addition of positive charge . So sides of the belt act to increase the net positive charge of the terminal A.

Question

A 100 volts battery is connected across two and 3ð• œ‡F capacitor in series. Calculate the potential difference between each capacitor and the energy/.

Solution

V=100u

From the series capacitor formula

+…………………..

C=

C=1.2×10-6

Potential difference across each capacitor is given by

but Q=θ=CxV=100×1.2×10-6

Q=1.2×10-4

=0.6x 102=60 volts

V2==0.4x 102=40 volts

Energy=CV2=1/2×1.2×10-6x104=6×10-3J

Energy=6×10-3J

The potential difference at each V1 = 60v and V2 = 40v and the total energy = 6×10-3J

SURFACE TENSION

INTERMOLECULAR FORCE

Force which exist between the molecules of solid, liquid and gaseous are known as intermolecular forces. These forces arise from the two main causes (i) The potential energy of the molecules (ii) The thermal energy of the molecules this is K.E of the molecules and it depends on the temperature of the substance concerned

The figure below shown a force separation graph from inter molecular forces between two molecules.

ð•š›o is the (separation) Equilibrium separation of the to molecules. The net force here is zero. When the separation of the two molecules is less than ro the net force is repulsive and when the separation is greater than ro the net force is attractive.

N:B the force between two molecules changes as they gradually brought together from the separation greater than ro to one less than ro. At large distance the force is negligible. As the molecules are brought closer there is a net attractive force which increases to a maximum value before diminishing to zero ro. Further bringing the molecules together results cute a net repulsive force.

The figure below shows that the potential energy separation graph for intermolecular forces between two molecules.

From the figure above than be observed that the separation of the molecules for the minimum potential energy is ro and the net force on the molecules at the minimum potential energy position is zero. The potential energy have positive values at small separations but negative values at large separations. Positive value means that the work has to be done molecules.

Molecules in solids are said to be in a condensed phase or state. They have relatively low thermal energy compared with their R.E U and vibrating about which the minimum of the curve

If the thermal Energy is increase by amount corresponding to cc the molecules oscillate between the limits X AND Y the molecules on the left of C experienced as greater force to wards it than when on the right. Thus the molecule return (quickly) e quicker to CI therefore the means position is on the right of CI this corresponds to a mean separation of molecules which is greater than ro . Thus the sold expands when its. Thermal energy is in creased.

##### SURFACE TENSION

If the clean glass rod is dipped into the water and then removed some water aling to the glass rod. We say that the water wet the glass. The adhesion of molecules of water to glass must therefore be greater than the cohesion to glass.

If you carefully place the razor blade on the surface of water it remains in the surface even though it has more than seven times as dense as water

The water acts as through it has a thin elastic film

These as observations shows that surface of a liquid acts like an elastic skin covering the liquid or in state of tension. All liquids shows surface tension.

In many liquids the surface tension is not as strong as that of water or mercury. The cleaning action as that of water or mercury. The cleaning action of the water or mercury. The cleaning action of the water or mercury. The cleaning action detergents is due to their ability to lower the surface tension of the water making it possible for the water and detergents more readily into the pores of the substances being cleaned.

Molecules at A attracted in all directions by cohesion force of the surrounding molecules. A molecules at B attracted equally on all side but move strongly downwards. The molecules at c is not attracted in upward direction at all. There is unbalanced force tendency to pull such surface molecules towards the enter of the liquid and keep the free surface of the liquid as small as possible.

##### CAPILLARITY TUBE É¤ ANGLE OF CONTACT

When the thread is tied cross the ring and dipper the soap of the soap film and the thread is loose.

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