1-3 - Electricity at Rest.doc

7/30/2019 1-3 - Electricity at Rest.doc

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ELECTRICITY AT REST

Electrification. According to modern concepts, all matter is made up of atoms, there being as many kindsof atoms as there are isotopes of all the elements.

Atoms are about 10-8 cm in diameter. Each atom is made up of a heavy central part, the nucleus, whichhas almost the entire mass of the atom.

Nucleus has a diameter of about 10 -12 cm and consists of two basic types of particles, the proton and the

neutron, except in the most common isotope of hydrogen which has a single proton for its nucleus.

The protons and neutrons are tightly packed in the nucleus. The nucleus is surrounded by electrons, the

word being derived from the Greek word for amber, orelektron. The electron has a unit negative(-) charge which is often called electronic charge. It is the discrete unit of negative electricity.

Electrons are all alike. One may imagine the atom as a miniature solar system with the nucleus

representing the sun and the electrons representing theplanets. Each proton has a unit positive (+) chargewhile the neutron has no charge. The number of neutrons generally increases as the number of protons is

increased in the nucleus.

The total number of protons in the nucleus is also the total number of positive charges in the atom

and is numerically equal to the atomic number of the element. This number determines the position of the

element in the periodic table. The atom is ordinarily neutral, which means that the number of electrons isequal to the number of protons in the atom. This implies that there are only two kinds of charges: positive

and negative. The total positive charge of the atom resides in the nucleus, while the total negative charge

is carried by the surrounding electrons. Compared to the proton, which is about as massive as the neutron,the electron is very light. The proton is about 1,840 times more massive than the electron. However,

there is some evidence that the size of the electron is of about the same order of magnitude as the size ofthe nucleus.

Even if an atom is electrically neutral, it has a certain amount of affinity for additional electrons.

Example: If a rod of hard rubber is rubbed with fur, there is a net transfer of electrons from the fur to the

rubber rod. Before rubbing, both rubber and fur were electrically neutral. It seems that rubberhas a greater affinity for additional electrons than fur, hence the transfer of electrons from fur to

rubber. A similar case is the process of vigorously passing a plastic comb through the hair.

Electrons are transferred from the hair to the comb. Both the rubber rod and the comb are leftwith a net negative charge, while the fur and hair are left with a net positive charge. Charges

are not created or destroyedin the rubbing process; they are only separated. If a piece of glass

is rubbed with silk, electrons are transferred from the glass to the silk. The affinity foradditional electrons is greater in silk than in glass, and the glass rod is left with a net positivecharge after some electrons are removed .in the process of rubbing with silk.

When a body possesses an excess positive charge or an excess negative charge, it is said to becharged or electrified. The process of charging is electrification. In the examples above given,

electrification is due to rubbing or contact. The hard rubber rodis negatively chargedwhile theglass rod

ispositively charged. The American scientist Benjamin Franklin is credited with having suggested thewordspositive (+) and negative (-) charges for the two types of charges.


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Attraction and Repulsion.

In figure below

(a) a hard rubber rod which has been negatively charged is suspended by means of a string. Asimilar rod is charged in the same manner and is brought near the first rod

(b) the first rod swings in the direction of the arrow, which indicates a repulsive force existingbetween the two similarly charged rods

(c) a glass rod which has been charged by rubbing it with a piece of silk cloth is brought near a

suspended, negatively charged rubber rod. The rubber rod swings in the opposite direction asin. It is attracted by the glass rod

(d) a positively charged glass rod is brought near another positively charged glass rod. The

suspended rod swings in the direction of the arrow, showing that it is repelled by the otherglass rod.

We may state a basic law:Like charges repel each other and unlike charges attract each other.

Insulators and Conductors.

Although the behavior of electrified objects has been known since antiquity, it was not until 1600,

during the times of William Gilbert, that substances were classified in terms of their electrified nature.Gilbert, a court physician to Queen Elizabeth I, published a book in which he systematically classified

substances into electrics and nonelectrics.

Electrics (insulators) - being those which could be electrified by rubbing, andNonelectrics (conductors) - being those which could not.

Insulators, which are nonconductors, are also called dielectrics. Electric charges move more readily about

in conductors than in insulators. In the best insulators, the charges are not free to move. Some vigorousrubbing with some material is needed to electrify these insulators.

Metals are generally good conductors while non-metals are generally poor conductors.

Different metals may also be arranged in the order of their excellence as conductors. In metallic

conductors, the positively charged nuclei are fixed while some electrons are free to move about in themetal.


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In liquid conductors, both the positively charged particles and the negatively charged particles, also called

ions, are free to move.

Silver is one of the best conductors, while other common metals like copper, nickel, iron, and aluminum

are fairly good conductors.

Among the more familiar nonconductors are mica, glass, rubber, porcelain, amber, paraffin, silk, and

sulfur. A thin sheet of any of these nonconductors is translucent, if not transparent, to light. It appears thatthe electrons are tightly bound to the atoms in a nonconductor.

When some electrons are removed from a certain region of the insulator by rubbing, the remaining

electrons do not redistribute themselves.In a conductor, when some electrons are removed by any means, the remaining electrons readily move

about and respond to the influence of attractive or repulsive electrical forces,

In figure below,

(a) a light ball is coated with some metallic paintsay, an aluminum paintto make it conductingand is suspended with a very light string. An uncharged metal rod mounted on an insulating

stand is placed in contact with the ball

(b) when a glass rod is charged and placed in good contact with the other end of the metal rod, the

ball is repelled(c) if the metal rod is replaced with a glass rod as in figure above, the ball is not repelled. When

the charged glass rod is placed in contact with the metal rod, some of the free electrons from

the ball and from the metal rod easily move to the glass rod to neutralize the positive chargeson the latter. Both the metal rod and the ball are left with a positive charge and the ball is

repelled. When the metal rod is replaced with another glass rod, the mere contact of the two

glass rods does not transfer the electrons from one rod to the other.


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Charging by Induction. After a body has charged by rubbing, it may be used to charge another body by

a method which requires no contact between the two bodies. In figure below, an uncharged metalis placed on an insulating stand. When the charged rod is brought near it

(a) the electrons nearer the positively charged rod. There is a greater concentration of negativecharges at the left end and a greater concentration of positive charges at the right end.

(b) when a wire connected to the ground is touched to the right end of the body, there is an almostinstantaneous motion of electrons from the ground to the body. The body is left with a netnegative charge which is still concentrated at the left end while it remains under the influence

of the charged rod.

(c) after the wire used forgrounding the body is removed, there will be no change in thedistribution of the charge on the body until after the charged rod is removed. There will be a

redistribution of the charge on the body after the rod is withdrawn. Thus, the body is

negatively charged without putting it in contact with the charged rod.

The method above of charging or electrifying a body is called charging by induction.

If, instead of bringing a positively charged rod near the conducting body, a negatively charged rod isbrought near it and the same procedure is followed, the body would be left with a positive charge.

Another way to charge bodies by induction is shown in figure below.

(a) Two conducting bodies on insulating stands are put into contact. When a charged body is

brought near the pair, the electrons which are relatively free to move pass through the point ofcontact between the two bodies, and the concentration of charges in the conducting pair

(b) After the bodies are slightly separated, the charged rod is removed. One body retains a positive

charge and the other body retains a negative charge. The net charge on the bodies will beredistributed.


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Electroscope.

- An instrument which is very sensitive in detecting the presence of a charge on a body.

1. In essence, it consists of a thin conducting leaf fastened to the side of another conducting stem

such that the leaf is free to swing about an axis (fig. above).

2. The stem is mounted in a metal container with a transparent window.

3. The stem is passed through the metal case through an insulator.4. At the outer end of the rod is a knob.

5. When the electroscope is uncharged, the thin leaf is vertical.

6. Imagine that a charged body is brought near the knob of the electroscope.7. Since the knob, stem, and leaf of the device are all conducting, the charged body attracts

electrons from the leaf and stem and piles them up on the knob.

(a) The leaf and stem have similar positive charges and the leaf swings as it is repelled by the

stem.

(b) If the rod is now made to touch the knob, the charges on the knob and on the rod neutralizeeach other, leaving the stem and leaf still positively charged. After the rod is withdrawn, the

leaf is still left with some positive charges and is repelled by the similarly charged stem. The

electroscope is said to be charged. Due to leakage of the charge into the air, the angle made bythe leaf with the stem gradually decreases.

(c) The electroscope may be charged negatively by bringing a negatively charged rod toward the

knob.

The electroscope may also be used to determine the kind of charge possessed by a charged body.

Any body with an unknown charge is brought near the knob of the positively charged electroscope and

the magnitude of the angle between the leaf and the stem is observed.- If the angle is increased when the other body is brought near the knob, the charge on the body is


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positive.

- If the angle decreases with the approach of the other body, the unknown charge is negative,opposite to the electroscope charge.

Coulombs Law. Late in the eighteenth century, a French scientist, Charles Agustine de Coulomb,

established the law of repulsion and attraction between electric charges in quantitative terms.

Coulomb utilized a modified torsion balance to study the magnitude of the force between point charges,or charged bodies whose sizes are small compared to the distance between them. He proved by

experiment that: The force between two small charged the line joining them and is directly proportionalto the product of the two charges and inversely proportional to the product of the two charges and

inversely proportional to the square of the distance between them. The relation is known as Coulombs

law for electrostatics. Written as an equation, we have,

2

21

r

qqkF = (similar to Newtons universal law of gravitation)

where: F = force

q1 and q2 = chargesr = distance between the two charges

k = proportionality constant

Note: The force, however, is always attractive in Newtons law while Coulombs law the force is repulsive if the charges arealike in sign, and it is attractive if the charges are unlike. If the charges have the same sign, F is positive; and if they

have different signs, F is negative.

cgs system:k = 1 if the charges are in a vacuum.

In air, k may be taken as equal to unity with no appreciable error.

The unit of charge: the electrostatic unit (e.s.u.) of charge (statcoulomb) is the amount of charge which,when placed at a distance of 1 cm from a similar charge in a vacuum, repels it with a force of 1 dyne.

MKS system:

The unit of charge is the coulomb, which is defined in terms of the ampere, the MKS unit of current. Thecoulomb is the quantity of charge which, when flowing past a given point in a conductor in 1 second,

produces a current of 1 ampere. From precise measurements, 1 coulomb = 2.9979 x 109 stat-coulombs.

This shows that the coulomb is a very large unit of charge. Without any appreciable error, we can take

1 coulomb = 3 109 statcoulombs

e = 4.8022 10-10 statcoulomb = 1.6019 10-19 coulomb

2

29

coulomb

meter-Newton109k

In electricity and magnetism, the equations derived from Coulombs law:

o4

1k

=

where: o =permittivity of free space


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2

212-

ometer-newton

coulomb108.85

k4

1 ==

Coulombs law may, therefore, be written as

=

2

21

o r

qq

4

1F

Example no. 1: Calculate the force between two point charges of +5.0 10-10 coulomb and -6.0 x 10-10coulomb which are 10 cm apart in air.

Soln.:

=

2

21

o r

qq

4

1F

( )( )( )

AnswerN1027-m0.10

C106.0-C105.0CNm109F 7-

2

-10-10229 =

=

The negative sign indicates an attractive force.

Example no. 2: A point charge of +2.0 10 -6 coul is 15 cm distant from a second point charge of -1.5 10-6 coul. Calculate the magnitude and direction of the force on each charge.

Soln.: The situation is shown in figure above. The two forces F12 (force on q1 due to q2)

and F21 (force on q2 due to q1) are also indicated. Because q1 and q2 are ofopposite signs, we conclude that the force between them is attractive. To find the

magnitude of this force, we use Coulombs law:

( )( )

( )

AnswerN.21-

m0.15

C101.5-C102.0CNm109F

2

-6-6229 =

=

Hence, F12 has magnitude 1.2 N and is directed toward q2 and F21 = -F12.

Example no. 3: Figure below shows three charges, q1, q2 and q3. What force acts on q1? Assume that q1 =

+1.0 10-6 coul, q2 = + 25 10-6 coul, and q3 = -1.5 10

-6 coul, r12 = 10 cm, r13 = 15 cm,

and = 30.

Soln.: The force on q1 due to q2 is repulsive and is given by

( )( )( )

AnswerN.252m0.10

C102.5C101.0CNm109F

2

-6-6229 =

=

The direction of this force is indicated in figure above. Note that the force on q l due to q2


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is not changed by the presence of another charge q3. To find the net force on a charge due

to an assembly of discrete charges, we calculate the force on the charge due to theindividual charges and add them vectorially. This is called theprinciple of superposition.

The attractive force between q1 and q3 is

F13 = -0.6 NFor the resultant force on q1, we have

Fnet = (-2.25 N 0.6 cos 60 N)i + (0.6 sin 60 N)j

= (-2.55i + 0.52j)N = 2.60 N

Example no. 4: Two similar balls of mass m are hung from silk threads of length land carry similar

charges q as in figure below. Assume that is so small that tan can be replaced by its

approximate, equal, sin . To this approximation, show that

3

1

o

2

2

=

mg

lqr

where ris the separation between the balls.

Soln.: Let us draw a free-body force diagram of the ball. This is shown in figure below.

The forces acting on this body are its weight mg, the tension T in the thread, and

the force on it due to the other ball F. Since there is no motion along x ory, we

apply the first condition for equilibrium.

0== xy FFAlongy we have

mgT cos =and alongx

2

2

4

1sin

r

q

FT

o

==

or 2

2

4

1

cos

sin

r

q

mg

o

=

for small ,


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2

2

4

1sintan

mgr

q

o

=

Butl

r

2sin =

hence,

2

2

o4

1

2 mgr

q

l

r

=

mg

lqr

o

23

2

=

And we finally have

3

1

o

2

2

=

mg

lqr

.

The Electric Field. Any body, by virtue of sis mass, is surrounded by a gravitational field, a regionextending from the body toward infinity. Any other body placed anywhere in the gravitational

field is acted upon by a gravitational force given by the universal law of gravitation. Force is

exerted upon one body by the other through this field, which could fill all empty space.

Similarly, any electric charge is surrounded by an electric field, and any other charge placedanywhere in the electric field of the first charge will be acted upon by an electrical force as given

by Coulombs law.

The second charge is also surrounded by an electric field which exerts a force upon the first charge. We

can also think of the electric field as a property of space brought about by the presence of a charge or of

charges in the neighborhood, such that any other charge placed anywhere in this space will experience anelectrical force.

An electric field exists at a point if a test charge placed at the point experiences a force.

The intensity of the electric field at a point may be defined as the force per unit of positive test

charge placed at the point. Since force is a vector quantity, the intensity of the electric field involves both

magnitude and direction; that is,

oq

FE =

where: E = electric field intensity

qo small test charge placed at the point under consideration.

The arrows over E and F mean that E and F are vector quantities. Rewriting the equation, we have

EqF o=

MKS system:

The electric field intensity is expressed in N/coul. The direction of the electric field at a point is

the same as the direction of the force on a positive test charge which is placed at the point. If the

electric field at a point is due only to a single charge q, then, from Coulombs law, the force whichwould act on a test charge qo placed at the point is


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2

o

o

r4

qqF

=

where: r is the distance of the point from the charge qIf we put this value of the force in equation above, we note that qo cancels out and we have

the magnitude of the intensity of the electric field at a point P given by the following

expression:

2o

p r4

q

E =

The intensity of the electric field at the point P may be represented by an arrow (see fig. below), and the

field around an isolated point charge is represented by the arrows in the figure. When the electric field at a

point is due to several point charges, it is obtained by superposition of the several separate fields, and thisis just finding the resultant of vector quantities.

In figure below, the electric field surrounding two neighboring point charges which are equal in

magnitude are shown for two like charges and for two unlike charges. Note that in the first diagram, thearrows are directed away from both charges. In the other figure, the arrows start from the positive charge

and end on the negative charge. These arrows also represent the imaginary lines of force. A line of force

is the trace of the path taken by a positive charge which is free to move and which is placed at any pointon the line. We can think of as many lines as we can. But no matter how the lines are traced, they never

cross each othernot even if the electric field is very complicated in pattern. The spacing of theseimaginary lines of force is a measure of the magnitude of the electric field in the region.


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A uniform electric field, or electrostatic field as it is often called, is represented by uniformly spaced

parallel lines of force. Figure below shows a uniform electric field near the center between two oppositelycharged plates placed close and parallel to each other. Near the ends of the plates, the field is not uniform

anymore as indicated by the fringing of the lines of force.

Although these lines allow us to visualize the electric interaction between charges, they are simply

fictitious constructions (mental constructs); they have no real physical existence and we shall not employthem quantitatively.

Let us enumerate the properties of electric lines of force:

1. A line of force emanates from a positive charge and terminates at a negative charge. A line of forcewhich seems to stop at a point in empty space, really extends to infinity.

2. Left alone, a positive charge would not go along a curved line of force as may have been implied

above. Because of its inertia, the charged body goes from one line to another and its trajectory will,in general, not be a straight line.

3. The tangent to a line of force gives the direction of the force that would act on a positive charge if

placed at the point.4. The spacing of the lines of force is a measure of the magnitude of the electric field in the region.

Where the lines are close together, E is large; where they are far apart, E is small.

The lines of force for two positive point charges are as shown in (a) and those for an electric dipole are asshown in figure (b).

Example no. 1: Find the force on an electron if it is placed at a point where the electrostatic field is 200

N/coul directed to the right.Soln.: F = qoE

AnswerN103.204-CN200C101.602--17-19 ==

The negative sign means that F is opposite E, hence this force is directed to the left.


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Example no. 2: (Falling charge). Figure below shows a positive charge q projected with a speed v0 at

right angles to a uniform field E. Describe its motion.

Soln.: The motion is like that of a projectile fired horizontally in the earths gravitational

field. For the motion along x and y.

x = v0tand

22

2

1

2

1 t

m

qEaty

==

Taken simultaneously, the two equations yield

2

2

02 x

mv

qEy

=

This is an equation of a parabola and gives the equation of the trajectory. As soon

as the charge leaves the region where the E field exists, it travels along a straightline tangent to the parabola at point A. Answer

Example no. 3: At the three consecutive corners of a square 10 cm on each side are point charges of 50 10-9 coul, 100 10-9 coul, and -100 10-9 coul, respectively. Find the electrostatic field at

point P of the square.

Soln.: The electrostatic field at point P can be obtained by taking the vector sum of thethree component electrostatic fields at the point. Due to the charge at A, the

electrostatic field at point P is

( )( )

rightthetoCN1045m0.10

C1050

C

Nm109E 32

-9

2

29

A ==

Due to the charge at B, the field at P

( )( )

horizontalthebelowCN1045m0.141

C10100

C

Nm109E 3

2

-9

2

29

B =

=

Due to the charge at C, the field at P

( )( )

CpointtowardCN1090m0.10

C10100

C

Nm109E 3

2

-9

2

29

C =

=


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By the method of components, the x-component of the field at P is

( ) CN1076CN1045cos4545E 33x =+=The y-component is

( ) CN1058.2CN1045sin45-90E 33y ==The electrostatic field at point P is

AnswerCN1096CN1058.276EEE 33222

y

2

xp =+=+=

Let be the angle made by the field with the x-axis.

=

===

37.5

0.76676

58.2

Ex

Eytan

Once the electrostatic field at a point is known, the magnitude and direction of

the force on any charge placed at the point is easily obtained.

Seat Work no. 1: A charge of 20 10-8 C is 20 cm from another charge of 180 10 -8 C (a) Find the force

between them. (b) What is the potential at the point exactly midway between the two?

(c) What is the electric field intensity at the same point?

Answers: (a) 8.1 10-2

N (b) 1.8 105

V (c) 14.4 105

N/C to the left

Electric Potential. When two charged bodies, A and B (see fig. below), are connected by a conducting

wire and there is a flow of electrons from A to B, the body B is considered to be at a higherelectric potential than body A. Or B is at a positive potential with respect to A, and A is at anegative potential with respect to B. It does not matter whether both A and B are negatively

charged or both are positively charged before they are brought into electrical contact.

The flow of electrons from A to B is equivalent to the flow of positive charges from B to A. If weassume that positive charges are free to move, they will flow from a body at higher potential to a body at

lower potential when the two are placed in electrical contact. Connecting the two bodies momentarily by a

conductor is equivalent to putting them in electrical contact.


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When a body is grounded (see fig. below) and there is a flow of electrons from the body to the

ground, the body is at a negative potential with respect to the ground.

Similarly, if the flow of electrons is from the ground to the body, the latter is at a positive potentialwith respect to the ground. For convenience, the potential of the earth is assumed to be zero, so the body

shown in figure above is said to be at a negative potential. After it is grounded, it rises to zero potential.

Any body that is grounded is said to be at zero potential.

Note the similarity between the flow of positive charges and bodies falling from higher to lower

gravitational potentials.

We define the electric potential V of a body as numerically equal to the amount of work done byan external agent to bring a unit positive charge from a point of zero potential to the body .

MKS system of units:

q

WV =

where: V = potential of the body; volt

W = work; joule needed to transfer a charge of q coulombs from the earth, or fromany other point at zero potential, to the body.

If the workis in ergand the charge is in statcoulomb, thepotentialis in statvolt.

1 statvolt = 300 volts

Since work is a scalar quantity, potential is also a scalar quantity. However, the workneeded may either be positive or negative depending upon whether the body is at a positive

or at a negative potential.Negative workmeans work that is given up or that it is the electric field which does thework. No external source of energy is needed to perform negative work.

We may also speak of the electric potential at a point and define it as the work done by an

external agent in bringing a unit positive charge from a point of zero potential to the point. Near anisolated point charge q, we consider a point P at a distance r from q (see fig. below).

To estimate the work done in bringing a unit positive charge from infinity, which is also


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considered to be a point of zero potential, to the point P, it will be necessary to divide the total distance

into a very large number of short distances. The total work done is found by adding the small incrementsof work done corresponding to these short distances. The task is long and tedious, but calculus offers a

short and easy method of doing it. Putting down the result obtained by use of calculus, we have for the

potential at point P due to the charge q

r

qkV =

where: k = 1 in the ESU system of units

MKS system,o4

1k

=

The sign of the potential at the point follows the sign of q. Note that while the electric field

intensity at a point due to a point charge varies inversely as the square of the distance of the point fromthe charge, the potential at a point due to a point charge varies inversely as the first power of the distance.

The potential approaches zero as the distance approaches infinity.

If the charged body is small, it may be considered as a point charge. For an isolated charged

spherical conductor, the potential at any point outside the conductor is obtained by assuming the totalcharge to be concentrated at the center of the sphere, and using the distance of the point to the center for r

in equationr

qkV = . The potential at any point inside the sphere is equal to the potential at any point on

the surface and is equal tor

qk where r is the radius of the sphere.

Potential Difference. Consider a point charge q and two points Pl and P2 in its neighborhood at distancesr1 and r2, respectively, from q (see fig. below).

Let the potentials at the two points be V1 and V2, respectively. From equationq

WV = , V1 is

numerically equal to the work done in moving a unit charge from infinity to P1. A charge q1 placed at Plwill have a potential energy equal to q1V1. Any other charge q2 placed at Pl will have a potential energy

equal to q2V1. If q2 is in coulombs and V1 is in volts, the potential energy possessed by the charge at the

point is in joules. The difference in potential between the two points P1 and P2 isV = V2 V1

and it may be defined as the work done in carrying a unit positive charge from the point of lower


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potential to the point of higher potential. If a charge of ql coulombs is carried from P1 to P2,

W = qV

Example no. 1: Two point charges of 2.0 10 -7 coul and -3.0 10-7 coul are placed at two corners A and

B of an equilateral triangle ABC, respectively. The side of the triangle is 20 cm. Howmuch work is needed to transfer a third charge from the third corner to a point exactly

midway between A and B?

Soln.: Let q be the charge to be transferred from C to P (fig. below). Disregarding q fora while, find the difference in potential between points C and P due to the chargesat A and B. The potential at point C is VC, where VC is equal to

( )VorCJ4,500-CNm4,500-

m0.20

C103.0-102.0CNm109V

-7-7229

C ==

=

The potential at point P is

( )V9,000-CNm9,000-

m0.10

C103.0-102.0CNm109V

-7-7229

P ===

The difference between the potentials at P and C isVP - VC = -9000 volts - (-4 500 volts) = -4500 volts

Therefore, the work needed isW = (VP - VC) volts q coul

= (VP - VC) q volt-coulomb or joules

= -4,500q J Answer

Note that to find the potential at a point due to several point charges, the separate

potentials are added algebraically.

Consider the problem of bringing a unit positive test charge from point A to point B in figure


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below.

Let us imagine that the two points A and B are immersed in a uniform electric field as suggested

by the evenly spaced, parallel lines of force. In order to accomplish this feat, an external agent must exerta force that opposes the electric force. Simplifying matters a bit, we shall consider the case where the

force exerted by our external agent is just enough to balance the electric force. The unit (positive) test

charge is consequently brought from A to B at constant velocity.

Accordingly, the work done by the external agent in bringing the positive charge from A to B isW = q(VB - VA) = Fd = -qEd = -qEd cos 180 = qEd, so that the electric potential difference between A

and B is ( ) Ed.q

WVVA-VB ===

From equationr

qkV = all points equidistant from a given point charge are at the same potential.

If we draw any circle with center at the point charge, it is an equipotential line. Similarly, any sphere withcenter at the point charge represents an equipotential surface.

Figure below shows the lines of force and some equipotential lines around two equal positivecharges. At any point of intersection of an equipotential line with a line of force, there is no component ofthe electric field along the equipotential line. Hence, the two lines intersect each other at right angles.

From this geometrical relation, one can easily draw the lines of force from the equipotential line pattern. It

is often easier to trace the equipotential lines or surfaces ahead of the lines of force.

Figure below shows the lines of force (solid lines) and the equipotential lines (broken lines)


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surrounding two equal but unlike charges.

No work is done in bringing a charge between two points on an equipotential surface. This can

readily be verified with reference to equation W = qV. Since V = 0, then W = 0.

Like the gravitational field, the electric field (strictly speaking, the electrostatic field) is

conservative. It possesses similar properties as a conservative force field. In figure below, we have a

family of equipotential surfaces at potentials V1, V2 and V3 with V1 < V2 < V3. The work done in bringinga unit positive charge from point A to point B along path I is zero; along path II it is (V B - VA), along path

III it is (VB - VA), etc.

Note again that from the definition of an equipotential surface, the electric line of force can

intersect the surface only at right angles. If this is not so, then the electric field will have a component

along the surface and work must be done to move a charge about the surface. This is illustrated in figurebelow.

Example no. 2: Potential due to an electric dipole. Figure below shows a configuration known as an


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electric dipole; two point charges equal in magnitude but opposite in sign separated a

small distance s.

Soln.: The electric potential at point P is just the algebraic sum of the potentials due to

each charge.

( )

21

12

o21o2o1o

Prr

r-r

4

q

r

1-

r

1

4

q

r

q-

4

1

r

q

4

1V

=

=+=

But if point P is very far away, r1 and r2 are approximately parallel to r and r1r2 r2. From figure below,

r2 - r1 = s cos Hence,

2

o

Pr

coss

4

qV

=

The product qs in the above equation is known as the electric dipole moment,

and we shall designate it by the letter p.

2

o

Pr

cosp

4

1V

= Answer Potential due to a dipole for r >> s.

Seat Work no. 1: Three identical charges of 0.030 C are placed at three corners of a square 10 cm onthe side. Calculate (a) the electrostatic potential at the fourth corner, (b) the

electrostatic field at the same corner, and (c) the work needed to transfer a charge of q

coulomb from this fourth corner to the center of the square.Answer: (a) 7.3 103 V (c) 9(4.27 103) J

(b) 5.13 104 N/C along E3

Electric Potential Energy. We know that a body at a certain distance from the surface of the earth


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possesses some form of energy (strictly speaking, the earth-body system) called potential energy.

If the body is released, it accelerates downward, thereby converting the (stored) potential energyof the system to kinetic energy of the body.

We can have a similar situation in electrostatics. Consider a very large charge Q and near it asmall charge q shown in figure below. Let Q be positive and q be negative. Left alone, the two charges

will accelerate toward each other; but since Q >> q, we can neglect the acceleration of the larger charge.

At a distance of separation r, we say that the configuration possesses a kind of potential energyelectrical potential energy in this case. To increase the separation between the charges, an external agentmust do work which is positive in this case since Q and q have opposite signs.

Electric Potential Energy of a system of point charges is defined as the work mat has to be done by an

external agent to assemble this system of charges by bringing them to their present positions frominfinity.

To avoid complications, we shall assume that at infinity, the charges are at rest, meaning that they

have zero kinetic energy.

Now consider a system of two point charges ql and q2 separated a distance r shown in figure

below.

Imagine that q2 is removed and brought to infinity. The electric potential at the original site of q 2,

from equation

r

qkV = , is coming from equation

q

WV = , if q2 is moved from infinity to its original

location, the work required is W= Vq2. Therefore, this work which, from our definition, is also the electric

potential energy, is .r

qq

4

1WU

12

21

o==

It is important to emphasize that the distance involved is that between the point charges q l and q2.

For a system of more than two point charges, the procedure is to compute the electric potentialenergy for every pair of charges and add the result algebraically.

Example no. 1: Calculate the electric potential energy of the system of charges shown in fig. below (rl2 =


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r24 = r13 = r34 = 10 cm), q1 = q, q2 = q, q3 = 2q, q4 = -4q, q = 10-6 coul.

Soln.:( )( )

J0.09-J10

10-109

r

qq

4

1U

1-

-129

12

21

o

12 =

==

( )( )J0.18J

10

102109U

1-

-129

13 =

=

( )( )J0.25-J

102

104-109U

1-

-129

14 =

=

( )( )J0.127-J

102

102-109U

1-

-129

23 =

=

( )( )J0.36J

10

104109U

1-

-129

24 =

=

( )( )J0.72-J

10

108-109U

1-

-129

34 =

=

Therefore,

U = U12 + U13 + U14 + U23 + U24 + U34

U = (-0.09 + 0.18 - 0.25 - 0.127 + 0.36 - 0.72) J = -0.647 J Answer

Seat Work no. 1: What is the work done in interchanging the positions of charges q 1 and q4 of the configuration

in figure below (left)? The final configuration is as shown in figure below (right).

Answer: 0

Seat Work no. 2: Two point charges, +0.04 C and +6.16 C, are 20 cm apart in air. (a) What is the


22/27

potential energy of the system? (b) Find the work required to bring them to a distance

of 10 cm from each other.Answer: (a) 2.810-3 J (b) 2.910-4 J

Potential Gradient. Let us take two points P1 and P2 in an electrostatic field and assume that the distance

s between the two points is very small (see fig. below).

Let the difference in potential between the two points be V, and let the intensity of the electricfield E in this small region be uniform. By definition, the difference in potential between the two pointsV is the work done in carrying a unit positive charge from the point of lower potential to the other

point. Let P2 be the point of lower potential. To transfer a unit positive charge from P 2 to P1 a force Fequal to qE but opposite in direction to E must be applied to the unit charge as it is moved along the

displacement s. The work is

W = Fs cos = -qEs cos or

sE-cossE-q

WV s==

=

s-VEs =

or the component of the electric field along the direction Pl to P2 is the same as the negative of the

space rate of change of the potential along that direction. In the small region where E is uniform, we can

take s to be along any direction. The value of V along the direction of s will vary from zero, when s

is along an equipotential line, to maximum, when s is perpendicular to the equipotential line or along thedirection of E. For this latter case,

ely.approximats

VE

max

s

=

In the limit, as s becomes very small,

potential.theofgradientds

dV-

s

Vlimit-E

maxmax0s

s ==

=

In words, the electric field intensity at a point is the negative of the maximum space rate ofchange of the potential at the point. This maximum space rate of change of the potential has direction

which is opposite the direction of the electric field at the point. This important relation finds usefulness inmore advanced work. The concept of potential gradient is similar to that oftemperature gradientin heat

flow and that ofpressure gradientin liquid flow.


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In the MKS system, the potential gradientis in volts per meter. Since the electric field intensity E is innewtons per coulomb,

.mV1CN1 =

The Van de Graaff Generator. High potentials may be mechanically generated by separation of

charges. A common device for this purpose is the Van de Graaff generator, developed at PrincetonUniversity in 1931 by the American physicist Robert Van de Graaff. The essential parts are shown in

figure below.

A large, hollow sphere is mounted on insulated supports. A silk conveyor belt passes over two

pulleys. Electrons are sprayed on the belt through the action of the generator and needle points A. Theseare collected at the needle points B and passed over on to the surface of the sphere, thus building up its

potential to as much as 10,000,000 V.

The limit of the potential on the metal sphere is governed by the insulation. Small-scale Van deGraaff generators are found in any modest physics laboratory.

Gausss Law. The electric field intensity due to a point charge is given by equation 2o

pr4

qE

= . In

vector form, we may write E as (see fig. below)

rr

q

4

1E

2

o=


24/27

where: r is a unit vector in the direction of increasing r.

To obtain the resultant E at a point due to a number of point charges as in figure below, we make

use of the principle of super position; that is, the resultant E is the vector sum of the individual Es.

Eresultant = E1 + E2 + + En

For a continuous distribution of chargesay, charge spread uniformly over the spherical surface oruniformly over a spherical surface or over a long wire or cylinderthe electric intensity may still be

calculated but the process is somewhat involved, although straight forward, and we often have to use

methods more advanced than elementary calculus. There is, however, an elegant manner of calculating E,especially so when there is symmetry in the problem.

By area vector.

It may seem odd, at first glance, how an area can be though of as a vector since it does not havethe properties of what we normally think of as vector. This really presents no major problem if we all

agree to represent a surface by a directed line segment erected perpendicular to it having magnitude

numerically equal to its area. This requirement will not be of any particular usefulness unless we specifywhich direction; there are two directions perpendicular to a surface element [see fig. (a) below]. Which of

these two directions do we take? A surface element is usually part of a larger surface that encloses a large

volume [fig. (b) below].

For the direction of our area vector, we can take that out of the volume enclosed by the surface.

Let us now consider a point charge located at the center of an imaginary spherical surface of

radius r. A small portion of the spherical surface is not flat; but if we choose one whose length andwidth are small enough, then it approximates a flat surface. Since all points on the spherical surface are

equidistant from the point charge, E will have a constant magnitude over the surface and it will be

perpendicular to every small area element at the point. We then form the product of the magnitudes of E

and S for each surface element and call this .SE=


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Let us next take the sum of all these s over the surface to get :

== SE over surfaceor S

r

q

4

1SE

2A oA

=

where: the index of summation A indicates that the sum is over the total area bounding the

spherical volume

Since the factor 2o r

q

4

1

over which we are taking the sum is constant, it can be placed before the

summation symbol. Now we shrink the length and width of each S to zero. This gives us a better

estimate of the value of over the spherical surface:

==

A2

A o0S

A0S

Sr

q

4

1limitSElimit

or

== dSrq

4

1EdS

2A o

The circle over reminds us to take the sum (integral) over the closed spherical surface. Now

dS on the right-hand side of equation == dSrq

4

1EdS

2A o

is just the sum of all the Ss over the

sphere, and this is just its area, 2r4 .

Hence, ( )o

2

2

o

qr4

r

q

4

1EdS

=== or

qEdS =

Equation qEdS = is in the form known as Gausss law (named after a Germanmathematician and astronomer, Johann Gauss) which states: The (surface) integral dStimes o equals the charge enclosed by the surface. In equation qEdS = , q is the netcharge. For example, ifthe surface encloses two equal but opposite charges, the right-hand side of equation qEdS = is zeroalthough this does not necessarily mean that E is zero in the region.

Example no. 1: Coulombs Law from Gausss Law.Consider two point charges q1, and q2 separated a distance r. Let us enclose ql with a

spherical surface of radius r so that the charge q2 is just on the surface of this sphere as

shown in figure 21.38. Applying Gausss law,

.qEds 1o =


26/27

Soln.: E is constant over the surface, = 1o qEdsand once more we recognize that dS is just the total area of the surface,

2

1

o r

q

4

1E

=

This is just the electric field intensity experienced by the charge q2. Therefore, the

force acting on it is

,r

qq

4

1EqF

2

21

o

2

== or Coulombs Law Answer

As it appears in equation qEdS = , Gausss law is not in its most general form. This form ofGausss law applies only to cases where the electric intensity is everywhere parallel (or antiparallel, as inthe case for a negative charge) to the area vector. In cases where is not parallel to S, Gausss law takes

the following form:

= qEdso

Distribution of Charge on a Conductor. The excess charge that is given a solid spherical conductor

resides mainly on its surface and is uniformly distributed over it. The first part of this statement

can be proved by Gausss law. Consider a cross section of an insulated conductor that carries anexcess charge.

Let us now imagine a closed Gaussian surface just beneath the conductor surface as shown infigure below.

We can take this imaginary surface to be as close to the conductor surface as possible; the

important thing here is that the Gaussian surface is inside the conductor. Now E inside such a conductor

must be zero, for otherwise the free electrons of the conductor would move about, resulting in collisionswith the atoms that make it up. Hence, we should expect that after some time the conductor would

become hot. As this never happens, E inside a conductor is zero, and from Gausss law, if E = 0, then q =0. This means that the net charge enclosed by the Gaussian surface is zero. We have put an excess charge

on it which we do not find inside; hence, it must be on the conductor surface.

If the solid is irregular in shape, the total charge will still be on the surface but the distribution of

the charge will not be uniform. It is shown in more advanced work that the surface density of the charge is

greater where the curvature is greater (see fig. below).


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When the solid is tapered to a point, the surface density of the charge at the point becomes very large, theelectric field near the point becomes intense, and the air particles near the point become ionized and

conducting. There is an electrical discharge from the point, and the charge leaks from the point at a rapid

rate. This is the principle of the lightning rod. During thunderstorms, convection currents are set up in theair. Some clouds may acquire a larger charge, and an opposite charge may be induced in a neighboring

cloud or on the ground just below the charged cloud. If the difference in potential between the cloud and

the ground is large, a discharge might suddenly occur and this causes the lightning we see. To prevent the

accumulation of a large amount of charge on the ground or on buildings, the charge is allowed to leakrapidly from sharp points of lightning arresters placed at the highest points of the building. The leakage

must be rapid enough; the rods might attract a discharge if the accumulated charge is not allowed to leak

rapidly.

Whether a conductor is solid or hollow, the charge is on the outer surface. There is no electric

field inside, and the interior is an equipotential region. When a hollow conductor is placed in anelectrostatic field, the region inside is field-free. This is the principle of electric shielding. The conductor

acting as a shield may have a

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