Physics Impt Point

8/6/2019 Physics Impt Point

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Mechanics


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3/34

Graphical Representaion: Day 2

Constant Velocity. No change in velocity (either in magnitude or in direction). Thenvelocity = displacement / time

See the V-t and S-t graph

Slope of s-t graph gives velocity Area under V-t graph gives displacement

Constant Acceleration: Three equation of motion. See the S-t, V-t and V-s graph for various cases:

When initial velocity is zero

When initial velocity is not zero

Slope of S-t, V-t graph

Area under A-t and V-t graph

Solve cases where object undergoes motion in different segment with eachsegment having a different constant acceleration: We have to first identify suchsegments then calculate the individual segment separately.

Conversion of V-t graph into S-t graph and V-s graph

Calculation of distance and Displacement from area under V-t graph: For displacement: We take magnitude and sign

For distance: We take only magnitude

Analysis of impossible graphs: Time cannot be negative

Time cannot be decreasing

Distance cannot decrease

At a particular time, object cannot have multiple velocities

NOTE: Problem solving through graph is much easier and faster than theconventional method. So Practice solving problem through graph


4/34

Other Cases

Variable Acceleration:

The three equation of Motion is not applicablewhen acceleration varies.

The equation to be used for variable accelerationto be used are: V = ds /dt, a = dv/dt and a = v.dv/ds

Practice the concept of definite integralsincluding identifying the upper and lower limitsfor calculating the integral.

Relation between velocity (conditional cases):

Solve cases of Pulley Case of Ladder

Case of Wedge and block

Case of Wedge and Pulley


5/34

Relative Velocity

Measurement with respect to earth

Inertial frame of reference : references moving with uniform velocity

Relative Velocity Two dimension :Use vector for solving it.

Relative velocity Vs Difference of Velocity Relative velocity pertains to two objects. The order is also important

Difference of velocity can be for same or different object. Order is not important

Relative velocity Vs Resultant Velocity In relative velocity, two objects are separated

In resultant velocity, one object moves right within the second object. They are in contact with

each other. For eg boat moving in a river, wind flowing along with rain Relative velocity is calculated when the speed of both the objects are w.r.t same frame. The

resultant velocity is calculated for a particular object in a differenr frame (one w.r.t water andother w.r.t ground)

Relative velocity of an object w.r.t medium in which it is flowing = Velocity ofobject in still medium. The inherent ability of the object doesnt depend on whether medium is still or moving

Impt Points in River B

oat Problem: Shortest time is when the velocity of boat is perpendicular to the velocity of river Shortest path is when the boat reaches the opposite point Here the direction of boat is as

observed by the observer standing at the bank. This is not always possible For cases where velocity of boat w.r.t stream is greater than the velocity of stream, it is possible to

reach the directly opposite path

For cases where velocity of boat w.r.t stream is less than or equal to the velocity of stream, then theshortest path is when the drift is minimum. The conditions need to be calculated by differentiating.


6/34

Numericals on River Boat (http://cnx.org/content/m14034/latest/)

Example 1: A person can swim at 1m/s in still water. He swims to cross a

river of width 200m to a point exactly opposite to his initial position. If thewater stream in river flows at 2m/s in linear direction, then find the timetaken in secs to reach the opposite point

Example 2: A person can swim at a speed u in still water. He pointsacross the direction of water stream to cross a river. The water streamflows with a speed v in a linear direction. Find the direction in which heactually swims with respect to the direction of stream.

Example 3: A boat capable of sailing at 2m/s moves upstream in a river.The water stream flows at 1m/s. A person walks from the front end to therear end of the boat at a speed of 1m/s along the linear direction. What isthe speed of the person with respect to the ground.

Example 4: A boy swims to reach a point Q on the opposite bank, suchthat the line joining initial and final position makes and angle of 45 with

the direction perpendicular to the stream of water. If the velocity of waterstream is u, then find the minimum speed with which the boy shouldswim to reach his target.

Example 5: A boat crosses a river in minimum time, taking 10 mts duringwhich time it drifts by 120 m in the direction of stream. On the otherhand, boat takes 12.5 mts while moving across the river. Find (i) width ofriver (ii) velocity of boat in still water and (iii) speed of stream


7/34

Hints on Numericals on River Boat

Example 1: velocity of person relative to water = 1m/s. Theresultant velocity (velocity of person w.r.t ground) is along the yaxis. Ans: Not possible

Example 2: velocity of person relative to water = u . This directionis always perpendicular to direction of stream. Resultant velocity(velocity of person w.r.t ground) is at and angle theta with the x axis

where Tan(theta) = u/v Example 3: velocity of boat with respect to river = 2m/s (upstream).

Ans.. = 0m/s

Example 4: A boy swims to reach a point Q on the opposite bank,such that the line joining initial and final position makes and angleof 45 with the direction perpendicular to the stream of water. If thevelocity of water stream is u, then find the minimum speed withwhich the boy should swim to reach his target.

Example 5: Minimum time when velocity of person w.r.t river isperpendicular to velocity of river. Shortest path is when the startingand ending point are directly opposite. Ans..width = 200m, speedof stream = 12m/mt, velocity of boat w.r.t stream = 20 m /mt

Note: X axis is the flow of stream and y axis is perpendicular to X axis


8/34

Projectile

Following Cases to be seen:

Default case: Starting and ending point is at thesame vertical level

Important points

Relative change in height, range when one of the

variable say velocity or angle of projection isvaried. (Use concept of differentiation).

Inclined Plane:

Ending point is at the higher vertical level

Ending point is at the lower vertical level

Relative motion of two projectiles

Collision of projectiles


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Important Results (Starting and Ending Point at the same Level)

)/1(

)2/(

2//2

/2

222

22

2

RxxTany

vSecgxxTany

gSinvhgSinvR

gvSinT

!

!

!

!

!

U

UU

U

U

U

Here:

V is the velocity of object projected

is the angle of projection with horizontal

T is the time of flight

R is range (horizontal distance)

h is the height (vertical distance)

Other Results:

Motion along perpendicular direction

Are independent of each other.

Range is maxm when = 45

Range is same for and (90 )

If h1 and h2 is the height for the

same range R then:

If T1 and T2 is the time of flight for

the same range R, then

If and are the angle made to anypoint from the two extremities of a

range in a projectile path of an object

projected with an angle from the

horizontal then

2*14 hh!

U2../. TanEKEP !

At highest Point:

g/*22*1 !

))(/( xxyananan !! UFE


10/34

Relative Change in the T, R, h when one of the independent variable is changed

)2(*)2/(sin/

)2(*)/2(sin/

)2(*)/(sin/

2

vgdvdh

vgdvdR

gdvdT

U

U

U

!

!

!

When is constant. The v is varied

)cos*sin*2(*)2/(/

)2(*)2(cos*)/(/

)(cos*)/2(/

2

2

UUU

UU

UU

gvddh

gvddR

gvddT

!

!

!

When v is constant. The is varied

RdRhdh // !

TdThdh /*2/ !


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Important Results (Ending Point at the higher level)

)cos2/()(sin

)cos/()sin(*cos2

)cos/()sin(2

22

2

EEU

EEUU

EEU

gvh

gvR

gvT

!

!

!

Technique:

Resolve the motion along the incline and perpendicular to it

Resolve the component of g along the incline and perpendicular to it Solve independently for two directions.

Let:

V is the velocity of object projected

is the angle of projection with horizontal

is the angle of incline with the horizontalT is the time of flight

R is range (horizontal distance)

h is the height (vertical distance)

902 ! EURange is maximum when


12/34

Circular Motion

Horizontal: Here the weight (mg), which is pointing vertically downward is always

perpendicular to motion and as such doesnt contribute to centripetal

acceleration.

The centripetal acceleration is contributed by the Tension.

The vertical component balances the weight

The horizontal component (radially towards the circle) providecentripetal acceleration

The speed remains constant of an object.

Vertical circle: In this we use the concept of law of conservation of energy.

The body undergoes change in K.E. and P.E. The speed keeps on

changing

The weight mg along with tension T contributes towards centripetal

acceleration.


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Types of Forces and its Direction

Gravitational Force:

This force always act vertically downward (along the ve y axis)

The force is assumed to be acting on the centre of Mass Reaction / Normal Force:

This force is exerted by the surface on which the body is in contact, away from the gravity

The direction is perpendicular to the tangent to the point of contact of the body with the surface.

Friction Force:

This force is perpendicular to the direction of the Normal force.

It is always away from the motion

Tension:

The force exerted by the end of a taut cord, string, or wire connected to a body is called the Tension

The direction of the tension-force is exerted along the direction of the cord

The magnitude of the tension is equal to the force that one would measure if the cord were cut and

a force spring-scale were inserted.

It is normally assumed that the cord is both massless and stretchless. When chord is massless the

tension is same across the chord. When the chord is inextensible, the acceleration is assumed to be

uniform across the string and the two bodies to which the string is connected.

When cord is both massless and stretchless, the forces at both ends of a cord have the same in

magnitude even if the cord changes direction through a pulley

At any moment the velocity and acceleration of two moving objects will be the same provided they

are connected by a massless, stretchless, taut-cord


14/34

Example on force and String

Different Forces:F1 is the Gravitational force = mg,

F2 is the reaction force and

F3 is the frictional force (if the body is sliding

downward) = F2 ( is the coefficient of friction)

Force resolution:The component of F1 along the y axis = -F1cos

(it is negative since the force is acting in the ve y axis)

The component of F1 along the x axis is = -F1sin

(It is negative since the force is along the ve x axis)

If this line

Doesnt cross

Through base,

The object

Will topple


15/34

Force Analysis in an accelerated frame of reference

Inertial frame (Ground) Accelerated frame (Lift, train)Case Description

A pendulum is suspended

from the roof of a train

compartment, which ismoving with a constant

acceleration a.

Find the deflection of the

pendulum bob from the

vertical as observed

from the ground and the

compartment

we consider a block of

mass 10 kg lying on the

floor of a lift, which itself is

moving up with an

acceleration of 2 m/s2.

Let g = 10 m/s2.

Find the weight measuredBy the spring balance .


16/34

Force Analysis in an accelerated frame of reference (Summary) Application of force analysis in accelerated frame of reference may have two approaches. We

can analyze using Newtons law from an inertial frame of reference. Alternatively, we can use

the technique of pseudo force and apply Newtons law right in the accelerated frame of

reference as described above. There is a school of thought that simply denies merit in pseudo force technique. The

argument is that pseudo force technique is arbitrary without any fundamental basis. Further,

this is like a short cut that conceals the true interaction of forces with body under

examination.

On the other hand, there are complicated situation where inertial frame approach may turn

out to be difficult to work with

Case Description: Here, a wedge isplaced on the

smooth surface of an

accelerated lift. We have to

study the motion of the block

on the smooth incline surface

of the wedge.Multiple accelerations here

complicates the situation.

The lift is accelerated with

respect to ground; wedge is

accelerated with respect to lift

(as the surface of the lift is smooth) ;

and finally block is accelerated with

respect to wedge (as wedge surface isalso smooth

Solution: In this case, it would bedifficult to assess or determine attributes

of motion by analyzing force in the

inertial ground reference. In situation

like this, analysis of forces in the non-

inertial frame of reference of the lift

eliminates one of the accelerations

involvedNote: when we analyze motion with

respect inertial frame of reference, then

all measurements are done with respect

to inertial frame. On the other hand, if

we analyze with respect to accelerated

frame of reference using concept of

pseudo force, then all measurements

are done with respect to acceleratedframe of reference


17/34

Important Cases to be seen

Accelerated frame of reference: For drawing F.B.D

When object is inside a frame which itself is accelerating: The force =ma in the opposite side of acceleration is added for newtons law to be

valid

An accelerated object moving in a circular path will have tangential

acceleration as well as centripetal acceleration, both at right angle to each

other. The resultant acceleration is the resultant of both.

An inclined object will topple before it slide if:

Angle is such that the centre of gravity at that instant no longer crosses the base of the

object

The angle is less than the limiting angle required as per friction.

Free Body Diagram: Practice for various cases

Sliding and Toppling One object over another:

Force on one body

Force on other body: When the object will move together. When they will move relative

to each other.


18/34

Work, Power and Energy

Work is done when Force acts over a displacement (The component of displacement is along

the direction of force, in same or opposite direction)

Work = F.d. It is a scalar quantity. Here F is the force, d is the displacement Work = F.d.Cos

Work is zero when: F = 0, d = 0 or Force is perpendicular to d

Gravitational Potential Energy: An object is said to have a potential energy if it has

the ability to do work by way of its position or state.

If an object of mass m is lifted to a height h, the change in P.E is given by:Ep = mgh As the object moves upward there is a increase in P.E due to work done by the external

force against gravity

The change in gravitational P.E depends upon change in vertical height. In an

inclined plane, change in P.E. depends upon change in vertical height.

Hookes Law: The stretch produced by a force applied to a string is directly

proportional to force applied, within elastic limit

Elastic Potential energy:

Work Energy Theorem:

In all cases, work done by a non-zero net force results in a change in K.E. but the

applied forces on an object may cause change in P.E. or K.E. or both.

2*2

1E kxp !

Wpk !(( EE


19/34

Isolated and Non-Isolated System

Isolated System: within an isolated system, energy may be transferred from one object to

another or transformed from one form to another, but it cannot be increased or decreased.

This is the law of conservation of energy. So Total Mechanical energy remains constant

Alternatively:

Conservative Forces: Forces that act within systems but do not change their

mechanical energy .Eg: Force of a gravity, elastic force,

Since the conservative force does not affect the mechanical energy of a system , the work done by a

conservative force to move an object from one point to another within the system is independent of

the path the object follows.

Non Conservative Forces: It causes mechanical energy of the system to change.

Eg friction force

Since the Work done by the friction force is negative, the final mechanical

energy is less than the initial mechanical energy. So, frictional force

reduces the mechanical energy of the system.

Comparision of Energy Position graph for Isolated and Non Isolated System: The component of the force of gravity parallel to the motion can be determined by calculating the

slope of the gravitational P.E. position graph

The net force can be determined by calculating the slope of the K.E. position graph

In the non-isolated system, the slope of the total energy curve gives the force of friction.

Power = Rate of doing work =

fm

fm

W

W

!

!

m12

m12

-

pk (!(

aveFVt

dF

t

WP !

(

(!

(!


20/34

More on Work and Collision

Work done when force is varying with distance

Area of F-X curve

Here F is the force, X is the displacement

Potential Energy: The notion of P.E. is applicable only to the class of forces where

work done against the force gets stored up as energy. When external constraints

are removed , it manifests itself as kinetic energy

The P.E. V(x) is defined if F(x) can be written as

This implies that

T

Collisions: Elastic: Both Kinetic Energy and the Linear momentum is conserved

Inelastic : Only Linear Momentum is conserved

Completely inelastic: After collision, both objects move together.

fi

f

VVdVdxx

V

Vi

xf

xi!! )(

!

f

i

x

x

fdxW

dx

dVx !)(


21/34

System of Particles and Rotational MotionParticle (point mass, no size) body of finite size (system of particles) Centre of Mass

Rigid body:

Definition: Perfectly definite and unchanging shape: No real body is truly rigid since real bodies deform under the influence of force.

Types of Motion:

Only Translational: All particles have same liner velocity at any instant of time

Only Rotational: All particles have same angular velocity at any instant of time

Translational + Rotational: Eg Rolling cylinder

Rotational Motion: Axis of Rotation: about which the body rotates

In rotation of rigid body about a fixed axis, every particle of the body moves in a circle

which lie in a plane perpendicular to axis of rotation and has its centre on the axis of

rotation.

For any particle on axis, the particle remains stationary while the body rotates

There are cases where axis may not be fixed only one of the point is fixed. Eg spinning

top whose nail is at a particular point (Here we assumed that the top is not slipping)

while its axis rotates to form a cone.

The movement of axis of the top around vertical is termed precession

Summary:

Motion of rigid body which is not pivoted or fixed may have either pure translational or

a combination of translation and rotation

Motion of a rigid body which is pivoted or fixed in some way is rotational


22/34

Some Important Formulae

)...mm)/(mxm...xmx(mX n21nn2211 !The Centre of Mass of a linear system:

The Centre of Mass in a 3 dimensional system:

)...mm)/(mrm...rmr(mR n21nn2211 !Here r is the position vector of the system of particles

R and r are both vector quantity.

Note: If the origin of the frame of reference is chosen to be centre of mass, then R = 0

For cases: where rigid body is considered as a continous distribution of mass (when

spacing between particle is small), we subdivide body into n small elements of mass

)m...,m,m n21 (((

If we make n bigger and bigger and each mass element

Smaller and smaller, we denote the sums by integrals

! xdmM *1

X ! ydmM *1

Y ! zdmM *1

Z

! rdm

M *

1

R

The Centre of Mass

Calculation

For a continous

Distribution of mass


23/34

Motion of Centre of Mass

!

!

!

!

!

!

!

!!

ext

ext

ii

FA

FFA

A

A

V

rmR

M

M

F...FFM

am...amamMdt

dvm...

dt

dvm

dt

dvm

dt

dVM

vm...vmvmM

dtdrm...

dtdrm

dtdrm

dtdRM

)rm...rmr(mM

int

n21

nn2211

n

n

2

2

1

1

nn2211

n

n

2

2

1

1

nn2211

The total internal force

Cancel each other due toNetwtons third law of motion

The Centre of Mass of a system of particles moves as if all the mass of a system

Was concentrated at the centre of mass and all the external forces were applied

At that point.

Further, to determine the motion of Centre of Mass, no knowledge of internal

force is required. Internal force do not contribute to motion of C.O.M.


24/34

ElectroStatics It deals with the study of forces, fields and potential energy arising out of static charge

State of Charge:

Static: Equilibrium state. No flow of charge - current Motion: Lead to flow of Current

Basic Properties of Charge:

Additivity

Charge is conserved

Charge is quantized

Coulombs Law: Fundamental law. Proved by experiment. For Point Charge: When linear size of charged body is much smaller than the distance

between them then the body can be considered as point charge.

It indicates the force between two point charges

The force is directly proportional to product of two charges and inversely proportional to

square of distance between them

It is a conservative force. Force is a vector qunatity. Its direction is along the line joining the two charged body

Like charges repel each other and unlike charges attract each other.

Force due to Multiple Charge:

Calculate the force due to each combination of charge using Couloms law. Magnitude

and direction

Calculate the resultant force using law of addition of vector


25/34

Electric Field Electric Field: due to a point charge at any point is defined as a force

experienced by a unit charge at that point.

It is a vector quantity. It is emnated in all directions (3 Dimensional) For +charge, it is directed radially outward

For ve charge, it is directed radially inward.

Electric field is independent of test charge.

Here test charge is taken to be near to zero to reduce the interference effect that the

test charge may have on the existing field of Source Charge.

It can be concluded that Electric field is same for points which are at a same distancefrom the Source Charge.

Thus electric field due to a point charge at the centre of sphere is same across all the

points of sphere. It has a spherical symmetry

Electric field due to a system of Charges:

Calculate the electric field for each point charge at a particular point

It will have both magnitude and direction

The resultant electric field can be calculated using vector addition.

Force experienced by a Charge q in Electric Field

The magnitude is given by F = qE

The direction is given by: Along the Electric field for positive charge

Opposite to electric field for negative charge

)(0

Eq

F

q

Lim

p

!


26/34

Electric Field Lines and Electric Flux Electric Field Lines:

For representing Electric Field

Its is denser where the electric field is strong. It is the relative density across differentpoints that matter.

The tangent to the electric field lines gives the direction of electric field at that point

Two Electric Field lines never intersect each other

Electrostatic field lines do not form a closed loop. This follows from the conservative

nature of electric field.

A field line is a space curve: That is curve in three dimensions In a charge free region, electric field lines can be taken to be continous curves without

any break.

Electric Flux:

The number of field lines crossing an area placed normal to Electric field

If is the angle between the Electric field and the normal to area dA and assuming that

electric field is constant in that area then , then Electric flux

= E.dA = EdA Cos

The direction of dA is a normal vector from area element dA and pointing outward of a

closed surface,

The electric flux is zero if Electric Field is zero

Electric field is parallel to the normal of the area


27/34

Electric Dipole Pair of equal and opposite point charge separated by a distance 2a.

Dipole Moment : Denoted by p.

It is vector quantity. Direction is along the line joining two point charges from negativeto positive

Magnitude is given by: 2aq

Electric field at a point along axis:

At a distance r from the centre of dipole

Here p is a unit vector

For distance (r>>a)

Electric field at a point along the equatorial plane:

At a distance r from the centre of dipole

For distance (r >>a)

Impt Point:

Dipole field at a large distance falls of not as 1/r2 but as 1/r3

The eqn (for r>>a) is exact for any r for point dipole (size 2a approaches zero the charge

q approaches infinity in such a way that 2aq is finite)

par

arq222 )(4

4*E

!

TI

pr

aq3

4

4*

TI!

par

aq2/322 )(4

2*E

!

TI

pr

aq3

4

2*E

TI!


28/34

Electric Dipole Contd Physical Significance of Electric Dipole:

In most molecules, the centre of positive and negative charge lie at the same place. So

their dipole moment is Zero. They are called non-polar molecule. Eg CO2, CH4. Theydevelop a dipole moment when electric field is applied.

In some molecules, centre of positive and negative charge do not coincide. Thus they

have a permanent dipole moment even in the absence of electric field. Eg water

Dipole in a Uniform External Electric Field:

The net force is Zero

The torque (Rotational force) is given by = PXE = 2aqSinp The direction of torque will be such that it will tend to align the dipole in the direction of

electric field.

Dipole in a Non-Uniform External Electric Field:

The net force is not zero

Torue will also be there

So dipole will be subjected to both linear and rotational motion


29/34

Continous Charge Distribution Continous charge Distribution: Charge is distributed uniformly across the object

Ring: Eg bangle. (negligible thickness). Circumference = 2r. Linear Charge density

Disc: We use Area: We use Surface Charge density = Q / Area of object Sphere: We use Volume charge density = Q / Volume of Object

For Calculating Electric Field in a Continous Charge:

Take a small area where charge is constant

Calculate the Electric field due to that small segment of Charge

The total electric field is sum of electric field due to such small segment across the source body

For linear Object: Wire or Ring

Linear charge density = Q/l or Q/2r.

For Surface charge density: Eg disk

A disk is considered to be made up of muliple ring of varying radii of negligible thickness joined

together

We take a ring at a radius r of thickness dr Its Area is given by 2r*dr

For Volume Charge density: Eg Solid Sphere

Made up of multiple disk of varying radii of negligible thickness and joined together

We take a test sphere of radius r and thickness dr.

Its Volume is given by = 4r2 * dr.


30/34

Gausss Law Used for Calculating Electric Field

Definition: Total Electric Flux through a closed surface = Net Charge Enclosed / Epsilon

In the situation, when the surface is so chosen that are some charges inside and some outside, the

electric field (whose flux appears on the L.H.S.,) is due to all charges, both inside and outside. The

term q on the R.H.S. is however the net charge enclosed by the surface.

If the total electric flux through a closed surface is zero then the net charge enclosed by the surface is

zero.

Gausss Law is true for any closed surface no matter what is its shape or size.

Gaussian Surface: A Closed surface . Ensure that Gaussian surface doesnt cross through any discrete

charge as electric field is not well defined at the location of the charge.

For continous volume charge distribution, it is defined at any point in the distribution

For surface charge distribution, electric field is discontinous across the surface.

With Gaussian symmetry such that Electric field is constant in magnitude at that point. All points are

equivalent to given charge distribution. Typical Gaussian Surface is Sphere, Cylinder

TheGausss Law is based on the inverse square dependence on distance contained in the Coulombs

law. Any violation ofGauss Law will indicate departure from the inverse square law.

Note:

The electric field due to a charge configuration with total charge zero is not zero; but for

distances large compared to the size of the configuration, its field falls of faster than 1/r2, typical

of field due to a point charge. An electric dipole is the simplest example of this.


31/34

Application ofGausss Law (Do Yourself)

Electric field due to a infinite long straight uniformly charged wire

Electric field due to a uniformly charged infinite plane Sheet

Electric Field due to a uniformly charged thin spherical shell

Electric field due to a uniformly charged insulating sphere


32/34

Conductors

All the Charge in the conductors lies on the external surface

The electric field is zero inside the conductor

If the conductor is enclosing a insulated cavity with charge q then

Charge q will be induced on the inside of the conductor. This is because for electric field

inside the surface of conductor to be zero, the net charge shall be zero. This will be zero

only when the charge q which is in the cavity is nullified by the equal and opposite

charge. This charge has to be at the inside of the surface of the conductor

Chare +q will be there on the outside of the conductor to compensate for the charge q

on inside of the surface.

Q1

Q2

Q3

Initial State

Q1

Final State

-Q1


33/34

Electrostatic Potential Energy

Potential Energy: When an external force does work in moving a body from one point to

another against a force such as gravitational force or spring force of electrostatic force, that

force gets stored as potential energy in the object. When the external force is removed, the body gains kinetic energy and loses potential energy

The sum of kinetic and potential energy is thus conserved.

Such forces are called conservative forces

Electrostatic Potential Energy: Work done by an external force in moving a test charge q from

point R to point P, against that force. Two points to be taken care are:

The test charge q is so small that it has no effect on the source Charge The body is moved from point R to P with zero acceleration i.e at any instant the net

force acting on the charged body is zero. This is possible when

Thus Work done by external force

This work is stored as Potential Energy = U = Up Ur = Wrp Impt. Points regarding Electrostatic Potential Energy:

Depends only on the initial and final position and not on the path followed. This is a

fundamental characteristic of conservative force

It is the difference of potential energy that is significant and not the absolute potential

energy. So there is freedom in choosing a point where potential energy is zero.

A convenient choice is to take potential energy as zero at infinity

Eext F!F

drFdrFW

P

R

P

R

extERP .. !!


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Electrostatic Potential

Electrostatic Potential: due to a Source Charge Q At a point is defined as work done in

bringing a unit positive charge q from infinity to that point. It is denoted by V.

It is a scalar quantity. Potential due to a point Source charge Qat a distance r

Thus E at a point is inversely proportional to square of distance while V is inversely

proportional to distance. Thus E falls off faster than potential

Potential Due to Electric dipole

For distance r >>a then

Here is the angle between p and radius vector joining the mid point of dipole to point The above eqn is approx. equal

For point dipole: The said equation is exact

For point along the axis, = 0 or 180, so

For point along the equatorial line, = 90 so V = 0

Potential due to System of Charge: Since potential is a scalar quantity, the potential due to a

system of charge is the algebraic sum of potential due to each charge at a point,

The potential inside the conducting shell is constant

Equipotential Surface: The surface with the same potential at all points

Any point on a sphere of radius r enclosing a point charge q is at equipotential

No work is done in moving a body from one point to other in a equipotential surface

The electric field line is normal to equipotential surface

r

Q

TI4V !

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**2V

r

Cosqa

TIU

!

24

*2V

r

qa

TI!

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