Chapter 01 Electrostatics Part 1 With Watermark

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Electrostatics:

Part 1

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1. ELECTRIC CHARGE

The electrical nature of matter is inherent in atomic structure. An atom consists of a

small, relatively massive nucleus that contains particles called protons and neutrons.

A proton has a mass of 1.673 × 10

A proton has a mass of 1.673 × 10−−2727 kg, and a neutron has a slightly greater mass of kg, and a neutron has a slightly greater mass of

1.675 × 10

1.675 × 10−−2727_{kg. Surrounding th}_{kg. Surrounding th}_{e nucleus is a diffuse cloud of orbiting particles called electrons,}_{e nucleus is a diffuse cloud of orbiting particles called electrons,}

as the figure suggests. An electron has a mass of 9.11 × 10

as the figure suggests. An electron has a mass of 9.11 × 10−−3131 kg. Like mass, ‘electric charge’ is kg. Like mass, ‘electric charge’ is

an intrinsic property of protons and

an intrinsic property of protons and electrons, and only two types of charge have been discovered,electrons, and only two types of charge have been discovered,

positive and negative. A proton has a positive charge, and an electron has a negative charge. A

neutron has no net electric charge.

Experiment reveals that the magnitude of the charge on the proton exactly equals the

magnitude of the charge on the electron; the proton carries a charge +

magnitude of the charge on the electron; the proton carries a charge +ee, and the electron carries, and the electron carries

a charge

a charge−−e.e. The SI unit for measuring the magnitude of an electric charge is the coulomb The SI unit for measuring the magnitude of an electric charge is the coulomb11 (C), (C),

and

andee has been determined experimentally to have the value, has been determined experimentally to have the value,ee = 1.60 × 10 = 1.60 × 10–19–19 C. C.

(i)

(i) Number of protons = number of electrons.Number of protons = number of electrons.

(ii)

(ii) Protons have the basic +Protons have the basic +ee charge and electrons have the basic – charge and electrons have the basic –ee charge. charge.

(iii)

(iii) Hence a normal atom is electrically neutral.Hence a normal atom is electrically neutral.

Electrons can travel from one atom to another and from one body to another.

If a body loses one electron, it becomes positively charged with +

If a body loses one electron, it becomes positively charged with +ee charge and vice-versa. charge and vice-versa.

2.

2. CHARGING OF CHARGING OF A BODYA BODY

Ordinarily, matter contains equal number of protons and electrons. A body can be charged by the transfer of electrons or redistribution

of electrons. Basically charging can be done by three methods:

(i)

(i) FrictionFriction (ii)(ii) ConductionConduction (iii)(iii) InductionInduction

By friction:

By friction:In friction when two bodies are rubIn friction when two bodies are rubbed together, electrons are transferredbed together, electrons are transferred

from one body to the other. As a result of this one body becomes positively charged

while the other negatively charged, e.g., when a glass rod is rubbed with silk, the rod

becomes positively charged while the silk negatively. However, ebonite on rubbing

with wool becomes negatively charged making the wool positively charged. Clouds

also become charged by friction. In charging by friction in accordance with

Electrostatics: Part 1

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2 _{Electrostatics: Part 1}

conservation of charge, both positive and negative charges in equal amounts appear simultaneously due to transfer of electrons from one body to the other.

Charging by conduction: When a negatively charged ebonite rod is rubbed on a metal object, such as the sphere in Figure (a), some of the excess electrons from the rod are transferred to the object. Once the electrons are on the metal sphere (where they can move readily) and the rod is removed, they repel one another and spread out over the sphere’s surface. The insulated stand prevents them from flowing to the earth, where they could spread out even more. As shown in Figure (b) of the picture, the sphere is left with a negative charge distributed over its surface. In a similar manner, the sphere would be left with a positive charge after being rubbed with a positively charged rod. In this case, electrons from the sphere would be transferred to the rod. The process of giving one object a net electric charge by placing it in contact with another object that is already charged is known as ‘charging by contact’.

By electrostatic induction: If a charged body is brought near an uncharged body, the charged body will attract opposite charge and repel similar charge present in the uncharged body. As a result of this one side of neutral body (closer to charged body) becomes oppositely charged while the other is similarly charged. This process is called electrostatic induction.

It is also possible to charge a conductor in a way that does not involve contact. In the figure negatively charged rod is brought close to, but does not touch, a metal sphere. In the sphere, the free electrons closest to the rod move to the other side, as part (a) of the drawing indicates. As a result, the part of the sphere nearest the rod becomes positively charged and the part farthest away becomes negatively charged. These positively and negatively charged regions have been “induced” or “persuaded” to form because of the repulsive force between the negative rod and the free electrons in the sphere. If the rod were removed, the free electrons would return to their original places, and the charged regions would disappear.

Under most conditions the earth is a good electrical conductor. So when a metal wire is attached between the sphere and the ground, as in the figure, some of the free electrons leave the sphere and distribute themselves over the much larger earth. If the grounding wire is then removed, followed by the ebonite rod, the sphere is left with a positive net charge, as part (c) of the picture shows. The process of giving one object a net electric charge without touching the object to a second charged object is called ‘charging by induction’. The process could also be used to give the sphere a negative net charge, if a positively charged rod were used. Then, electrons would be drawn up from the ground through the grounding wire and onto the sphere.

2.1 Properties of Electric Charge

(i) Charge is transferable: If a charged body is put in contact with an uncharged body, uncharged body becomes charged due

to transfer of electrons from one body to the other.

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Electrostatics: Part 1 3

(iii) Charge is conserved: Charge can neither be created nor be destroyed. For example, in radioactive decay the uranium nucleus

(charge = +92e) is converted into a thorium nucleus (charge = +90e) and emits ana -particle (charge = +2e)

Æ +

238 234 4

92U 90 Th 2 He

Thus the total charge is +92e both before and after the decay.

(iv) Invariance of charge: The numerical value of an elementary charge is independent of velocity. It is proved by the fact that an

atom is neutral. The difference in masses on an electron and a proton suggests that electrons move much faster in an atom than protons. If the charges were dependent on velocity, the neutrality of atoms would be violated.

(v) Charge produces electric field and magnetic field: A charged particle at rest produces only electric field in the space

surrounding it. However, if the charged particle is in unaccelerated motion it produces both electric and magnetic fields. And if the motion of charged particle is accelerated it not only produces electric and magnetic fields but also radiates energy in the space surrounding the charge in the form of electromagnetic waves.

(vi) Charge resides on the surface of conductor: Charge resides on the outer surface of a conductor because like charges repel

and try to get as far away as possible from one another and stay at the farthest distance from each other which is outer surface of the conductor. This is why a solid and hollow conducting sphere of same outer radius will hold maximum equal charge and a soap bubble expands on charging.

(viii) Quantization of charge:When a physical quantity can have only discrete values rather than any value, the quantity is said to

be quantised. The smallest charge that can exist in nature is the charge of an electron. If the charge of an electron ( 1.6 × 10–19C) is taken as elementary unit i.e. quanta of charge the charge on anybody will be some integral multiple of e i.e.,

= ± with =1, 2, 3_

Q ne n

Charge on a body can never be 2 5

, 17.2 10 etc. 3

e e or - e

± ± ±

Note:

• Recently it has been discovered that elementary particles such as proton or neutron are composed of quarks having charge (± 1/3)e

and (± 2/3)e. However, as quarks do not exist in free state, the quanta of charge is stille.

• Quantization of charge implies that there is a maximum permissible magnitude of charge.

Illustration 1 A comb runs through one’s dry hair and attracts small bits of paper. Why? What happens if the hair is wet or it is rainy day?

Solution

When the comb runs through dry hair, it gets charged due to friction. When this comb is brought near small bits of paper, these acquires induced charges whose nature is opposite to that on the comb. Consequently, the bits of paper get attached to the comb. But on a rainy day or when the hair is wet, there is no friction between the hair and the comb and hence the comb remains uncharged.

Illustration 2 Why can one ignore quantization of electric charge when dealing with macroscopic, i.e., large-scale charges? Solution

Large-scale charges contain a very large number of electrons, e.g., 18 19 1 1 6.25 10 1.6 10 C -= = ¥ ¥

electronic charges. On account this of large number, charge flows continuously and the quantization of charge can be ignored.

Illustration 3 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with

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Solution

The positive charge developed on the glass rod is found to have the same magnitude as the negative charge developed on silk cloth. As such, the total charge on the glass rod and the silk cloth is the same before and after rubbing, which is consistent with the law of conservation of charge.

Illustration 4 A glass rod is rubbed with a silk cloth. The glass rod acquires a charge of + 19.2 × 10–19 C.

Find the number of electrons lost by glass rod. Find the negative charge acquired by silk. Is there transfer of mass from glass to silk? Given,m_e = 9 × 1031 kg

Solution

Number of electrons lost by glass rod is

19 19 19.2 19 12 1.6 10 q n e -¥ = = = ¥

(i) Charge on silk = −19.2 × 10–19 C

(ii) Since an electron has a finite mass (m_e = 9 × 10–31 kg), there will be transfer of mass from glass rod to silk cloth.

Mass transferred 31 29

12 (9 10- ) 1.08 10 kg

-= ¥ ¥ = ¥

Note that mass transferred is negligibly small. This is expected because the mass of an electron is extremely small.

Illustration 5 If an object made of substance A rubs an object made of substance B, then A becomes positively charged and B

becomes negatively charged. If, however, an object made of substance A is rubbed against an object made of substanceC , then A becomes negatively charged. What will happen if an o bject made of substance B is rubbed against an object made of substance

C ?

(a) B becomes positively charged andC becomes positively charged. (b) B becomes positively charged andC becomes negatively charged. (c) B becomes negatively charged andC becomes positively charged. (d) B becomes negatively charged andC becomes negatively charged. Solution

When ‘ A’ and ‘ B’ are rubbed. ‘ A’ becomes positively charged and ‘ B’ becomes negatively charged.

• Electrons are loosely bound with ‘ A’ in comparison to ‘ B’ . When ‘ A’ and ‘C ’ are rubbed together. ‘ A’ becomes negatively

charged.

• Electrons are loosely bound with ‘C ’ in comparison to ‘ A’.

• Hence in ‘C ’ electrons are most loosely bound.

• Hence if the object ‘ B’ and ‘C ’ are rubbed together ‘C ’ will lose electrons and ‘ B’ will receive electrons.

• Hence ‘C ’ will become positively charge and ‘ B’ will become negatively charged.

Illustration 6 Objects A, B andC are three identical, insulated, spherical conductors. Originally A and B both have charges

of +3 mC, whileC has a charge of –6 mC. Objects A andC are allowed to touch, then they are moved apart. Then objects B and C are allowed to touch, and they are moved apart. [Neglect effect due to induction].

(A) If objects A and B are now held near each other, they will

(a) attract (b) repel (c) have no effect on each other.

(B) If instead objects A andC are held near each other, they will

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Solution

∑ When the objects A andC are allowed to touch, then move apart

∑ When the objects B andC are allowed to touch, then move apart.

Hence if A and B are now held near each other they will attract each other.

∑ If A andC are now held near each other they will also attract each other.

Illustration 7 Consider figure, a positively charged rod is brought near two uncharged metal spheres A and B attached with

insulated stands and placed in contact with each other.

What would happen if the rod were removed before the spheres are separated?

Would the induced charges be equal in magnitude even if the spheres had different sizes and different conductors?

What will happen when the rod is removed after the spheres are separated?

Solution

(a) When a positively charged rod is brought near A, the free electrons in the sphere A are attracted to the rod, and same move in the

left side of A. This movement leaves unbalanced positive charge on B. If the rod is removed before the spheres are separated,

the excess electrons on sphere A would flow back to B. Both the spheres will become uncharged.

(b) Yes, net charge is conserved. Before the rod is brought near A, both A and B were neutral. They will remain so even if they

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(c) As charge is conserved if rod is removed and spheres are separated, the sphere A will have net negative charge and sphere B

will have net positive charge of same magnitude. 3. ELECTROSCOPE

It is a simple apparatus with which the presence of electric charge on a body is detected (see figure). When metal knob is touched with a charged body, some charge is transferred to the gold leaves, which then diverges due to repulsion. The separation gives a rough idea of the amount of charge on the body. If a charged body brought near a charged electroscope, the leaves will further diverge. If the charge on body is similar to that on electroscope and will usually converge if opposite. If the induction effect is strong enough leaves after converging may again diverge.

(i) Uncharged electroscope

(ii) Charged electroscope

Concept Application Exercise 1

1. One quantum of charge should be at least be equal to the charge in coulomb: (a) 1.6 × 10–17 c (b) 1.6 × 10–19 c

(c) 1.6 × 10–10 c (d) 4.8 × 10–10 c

2. Which one of the following statements regarding electrostatics is wrong? (a) Charge is quantized

(b) Charge is conserved

(c) There is an electric field near an isolated charge at rest

(d) A stationary charge produces both electric and magnetic fields 3. Select the correct alternative/ alternatives:

(a) The charge gained by the uncharged body from a charged body due to conduction is equal to half of the total charge

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(b) The magnitude of charge increases with the increase in velocity of charge (c) Charge cannot exist without matter although matter can exist without charge

(d) Between two non-magnetic substances repulsion is the true test of electrification (electrification means body has net

charge)

4. In the return stroke of a typical lightning bolt, a current of 2.5 × 104 A exists for 20ms. How much charge is transferred in

this event?

5. How many megacoulombs of positive charge are in 1.00 mol of neutral molecular-hydrogen gas (H₂)? (N_A= 6.0 × 1023) 6. Calculate the number of coulombs of positive charge in 250 cm3 of (neutral) water. (Hint: A hydrogen atom contains one

proton; an oxygen atom contains eight protons.)

7. Electrons and positrons are produced by the nuclear transformations of protons and neutrons known as beta decay. (a) If

a proton transforms into a neutron, is an electron or a positron produced? (b) If a neutron transforms into a proton, is an electron or a positron produced?

8. Figure shows four identical conducting spheres that are actually well separated

from one another. SphereW (with an initial charge of zero) is touched to sphere A and then they are separated. Next, sphereW is touched to sphere B (with an

initial charge of −32e) and then they are separated. Finally, sphereW is touched

to sphere C (with an initial charge of +48e), and then they are separated. The

final charge on sphereW is +18e. What was the initial charge on sphere A?

9. A charged non-conducting rod, with a length of 2.00 m and a cross-sectional area of 4.00 cm2, lies along the positive side

of an x axis with one end at the origin. The volume charge densityr is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if r is (a) uniform, with a value of −4.00 µC/m3, and (b) non-uniform, with a

value given byr = bx 2, whereb =−2.00 µC/m5?

10. What is the total charge in coulombs of 72.0 kg of electrons?

(The mass of an electron= 9.0 × 10−31 kg)

11. A 100 W lamp has a steady current of 0.80 A in its filament. How long is required for 1 mol of electrons to pass through

the lamp? ( N _A= 6.0 × 1023)

4. COULOMB’S LAW

The electrostatic force that stationary charged objects exert on each other depends on the amount of charge on the objects and the distance between them. To set the stage for explaining these features in more detail, figure shows two charged bodies. These objects are so small, compared to the distance r between them, that they can be regarded as mathematical points. The “point charges” have

magnitudes|q₁| and|q₂|. If the charges have unlike signs, as in part (a) of the picture, each object is attracted to the other by a force

that is directed along the line between them; +F



is the electric force exerted on object 1 by object 2 and -F



is the electric force exerted on object 2 by object 1. If, as in part (b), the charges have the same sign (both positive or both negative), each object is repelled from the other. The repulsive forces, like the attractive forces, act along the line between th e charges. Whether attractive or repulsive, the two forces are equal in magnitude but opposite in direction. These forces always exist as a pair, each one acting on a different object, in accord with Newton’s action–reaction law.

r q1 +F –F q2 + – (a) +F –F q1 q2 r + + (b)

The equation for the electrostatic forces acting on the particles is called Coulomb’s law after Charles-Augustin de Coulomb, whose experiments in 1785 led him to it. Let’s write the equ ation in vector form and in terms of the particles shown in fi gure, where particle 1 has chargeq₁ and particle 2 has chargeq₂. (These symbols can represent either positive or negative charge.) Let’s also

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two particles, radially away from particle 2. (As with other unit vectors, r  has a magnitude of exactly 1 and no unit; its purpose is

to point, like a direction arrow on a street sign.) With these decisions, we write the electrostatic force as

=   1 2 2 (Coulomb's law), q q F k r r ...(i)

whereris the separation between the particles andkis a positive constant called the electrostatic constant or the Coulomb constant.

(We’ll discusskbelow.)

If between the two charges there is free space then,

9 2 2 0 1 9 10 Nm C 4 k pe -= = ¥ (in SI Units)

wheree ₀ = (8.85 × 1022 C2N–1m–2) is the absolute electric permittivity of the free space. If th ere is medium between the two charges then 0 1 4 _r k pe e

= wheree _ris the relative permittivity of medium which is also known as the dielectric constant. Relative permittivity

(e r ) of vacuum is one. For all other media it is greater than one and for conductors it is inﬁnite.

Let’s first check the direction of the force on particle 1 as given by Eq. (i). Ifq₁andq₂have the same sign, then the productq₁q₂

gives us a positive result. So, Eq. (i) tells us that the force on particle 1 is in the direction of r .

 That checks, because particle 1 is being

repelled from particle 2. Next, ifq₁andq₂have opposite signs, the productq₁q2gives us a negative result. So, now Eq. (i) tells us that

the force on particle 1 is in the direction opposite r .

 That checks because particle 1 is being attracted toward particle 2.

4.1 Coulomb’s Law in Vector Form

Letq₁ andq₂ be two like charges placed at points A and B in vacuum, separated by a distancer . Due to like nature of charges, they

will repel each other. Let F ₁₂



be the force on chargeq₁ due toq₂;

21

F



be the force on chargeq₂ due toq₁;

12

ˆ

r be the unit vector from_q₁ to_q₂;

21

ˆ

r be the unit vector fromq₂ toq₁;

From figure, it is clear that F 21 and r ˆ12



and are in the same direction

\ 1 2 21 2 12 0 1 ˆ 4 q q F r r pe =  Also, F 12 and r ˆ21 

are in same direction

\ 1 2 12 ₂ 21 0 1 ˆ 4 q q F r r pe = 

The above equations give the Coulomb’s law in vector form.

It is also clear from above equations, the magnitude of forces are equal i.e., 12 21 1 2₂ 0 1 4 q q F F r pe = =  

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Since r ˆ 12 and r ˆ21 are unit vectors, opposite to each other i.e., r ˆ 21 =–r ˆ12

Put this value in equation, 1 2 1 2

12 2 12 12 2 12 0 0 1 1 ˆ ˆ (– ); – 4 4 qq qq F r F r r r pe pe = =   From equations, F ₁₂=–F ₂₁  

Thus, the forces exerted by two charges on each other are equal in magnitude and opposite in direction. Therefore, Newton’s third law is obeyed.

Also the forces due to two point charges are parallel to the line joining the point charges; such forces are called central forces and so electrostatic forces are conservative forces.

or 1 2 21 3 12 0 ₁₂ 1 4 q q F r r pe =    ∵ 12 12 12 ˆ r r r = 

Similarly, force onq₁ due toq₂ is

1 2 1 2 12 ₂ 21 12 ₃ 21 0 ₂₁ 0 ₂₁ 1 1 ˆ or 4 4 qq qq F r F r r r pe pe = =     

The above equations can also be written in the following manner:

1 2 1 2 21 3 2 1 12 3 1 2 0 ₂ ₁ 0 ₁ ₂ 1 1 ( – ) and ( – ) 4 _– 4 _– qq qq F r r F r r r r r r pe pe = =          

These equations represent Coulomb’s law in terms of position vectors.

Note:

Coulomb’s law is true for point charges only and the force between any two charges is not affected by the presence of other charges.

Principle of superposition: According to the principle of super position, total force acting

on a given charge due to number of charges is the vector sum of the individual forces acting on that charge due to all the charges.

Consider number of charge Q₁,Q₂,Q₃… are applying force on a chargeQ

Net force onQ will be

net 1 2 n 1 n F F F F F -= + + + +      

Illustration 8 What is the force between two small charged spheres having charges of 2 × 10–7 C and 3 × 10–7 C placed 30 cm

apart in air? [NCERT]

Solution -

-Ê

_{ˆ ¥}

_¥

=

¥

_Á

_˜

=

¥

Ë

¯

2 7 7 9 3 2 2 2 Nm (2 10 C) (3 10 C) 9 10 6 10 N C (30 10 m) F (repulsive)

Illustration 9 The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N.

(a) What is the distance between the two spheres?

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10

10 _Electro_Electro_{statics: P}_{statics: P}_{art 1}_{art 1}

Solution Solution (a) (a) AsAs 11 22 11 22 2 2 ,, kkq q q q kkq q qq F F r r F F r r = = == - - --- -

--Ê

Ê

_¥

_¥¥

ˆ

=

_Á

_˜

=

¥

=

¥

==

Ë

¯

1/2 1/2 9 9 22 22 66 66 4 4 2 2 22 9 9 110 0 ((NNm / C m / C ) ) ((00..4 4 110 0 CC) ) ((00..8 8 110 0 CC)) 1 1444 4 110 0 m m 112 2 110 0 m m 112 2 ccmm 0.2 N 0.2 N r r (b)

(b) Since the two charges are of opposite signs, the force between them is attractive in nature. Further, according to Newton’sSince the two charges are of opposite signs, the force between them is attractive in nature. Further, according to Newton’s

third law, the force on the second sphere due to first is also attractive and has a magnitude 2 N.

third law, the force on the second sphere due to first is also attractive and has a magnitude 2 N. Illustration 10

Illustration 10

(a)

(a) Two insulated charged copper spheresTwo insulated charged copper spheres A A and and B B have their centres separated by a distance of 50 cm. What is the mutual force have their centres separated by a distance of 50 cm. What is the mutual force

of electrostatic repulsion if the charge on each is 6.5 × 10

of electrostatic repulsion if the charge on each is 6.5 × 10–7–7 C? The radii of C? The radii of A A and and B B are negligible compared to the distance are negligible compared to the distance

of separation.

of separation. (b)

(b) What is the force of repulsion if:What is the force of repulsion if:

(i)

(i) Each sphere is charged double the above amount, and the distance between them is halved.Each sphere is charged double the above amount, and the distance between them is halved.

(ii)

(ii) The two spheres are placed in water? (Dielectric constant of water = 80)The two spheres are placed in water? (Dielectric constant of water = 80) [NCERT][NCERT]

Solution

(a)

(a) We are given that,We are given that,qq₁₁ = =qq₂₂ = 6.5 × 10 = 6.5 × 10–7–7 C, C,r r = 50 cm = 0.5 m = 50 cm = 0.5 m

If

IfF F is the force of repulsion, is the force of repulsion, F F k k _ee q q q1 21 ₂q₂2 r r = = or or 9 9 7 7 77 22 2 2 ((66..5 5 110 0 ) ) ((66..5 5 110 0 )) ((9 9 110 0 ) ) 11..5 5 110 0 NN.. (0.5) (0.5) F F - - - -¥ ¥ ¥¥ = = ¥¥ ¥¥ == ¥ ¥ (b)

(b) (i) (i) If the charge on each sphere is doubled and the distance between them If the charge on each sphere is doubled and the distance between them is halved, the force of repulsion,is halved, the force of repulsion,F F ´ would become´ would become

16 times (as

16 times (asF F µµqq₁₁qq₂₂ and andF F µµ 1/ 1/ r r 22).).

Clearly,

Clearly,F F ´ = 16´ = 16F F = 16 (1.5 × 10 = 16 (1.5 × 10–2–2 N) = 0.24 N N) = 0.24 N

(ii)

(ii) If the sphere aIf the sphere are placed in water re placed in water ((e e _r_r = 80), = 80),

2 2 4 4 1 1..5 5 1100 1 1..9 9 110 0 NN 80 80 w w r r F F F F -¥ ¥ = = == == ¥¥ Œ Œ Illustration 11

Illustration 11 Suppose the spheresSuppose the spheres A A and and B B in previous illustration have identical sizes. A third sphere of the same size but in previous illustration have identical sizes. A third sphere of the same size but

uncharged is brought in contact with the first, then in contact with the second, and finally removed from both. What is the new

force of repulsion between

force of repulsion between A A and and B B??

Solution

We are given that, Charge on sphere

We are given that, Charge on sphere A A = charge on sphere = charge on sphere B B = =qq = 6.5 × 10 = 6.5 × 10–7–7 C C

Force of repulsion between

Force of repulsion between A A and and B B, i.e.,, i.e., 2 2 7 7 22 9 9 22 2 2 2 2 22 ((66..5 5 110 0 )) ((9 9 110 0 ) ) 11..5 5 110 0 NN (0.5) (0.5) e e ee q q q q qq F F k k k k r r r r -¥ ¥ ¥¥ = = == == ¥¥ == ¥¥

Let the third sphere (say

Let the third sphere (sayC C ) of the same size but uncharged be brought in contact with) of the same size but uncharged be brought in contact with A A. Due to the flow of electrons, the two. Due to the flow of electrons, the two

spheres share the charge equally. Therefore,

Charge on

Charge on A A = Charge on = Charge onC C 00

2 2 22 q q ++ qq = = ==

When the sphere

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Electr

Electrostatics: Postatics: Part 1art 1 1111

Total charge on

Total charge on B B and andC C ( (qq + +qq /2 = 3 /2 = 3qq /2) is equally distributed between them. /2) is equally distributed between them.

Thus, charge on

Thus, charge on B B = charge on = charge onC C = 3 = 3qq /4 /4

If

IfF F ´ is the new force of repulsion between´ is the new force of repulsion between A A and and B B,, 2 2 2 2 33 2 2 22 ( ( // 22) ) ((3 3 / / 44) ) 3 3 3 3 33 ((11..5 5 110 0 NN) ) 55..7 7 110 0 NN 8 8 8 8 88 e e ee q q q q qq F F kk kk FF r r r r - - - -= = == == == ¥¥ == ¥¥ ¢¢ Illustration 12

Illustration 12 Two particlesTwo particles A A and and B B having charges 8 having charges 8×× 10 10–6–6 C and –2 C and –2×× 10 10–6–6C respectively are held fixed with a separationC respectively are held fixed with a separation

of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force?

of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force? Solution

Solution

As the net electric force on

As the net electric force on C C should be equal to zero, the force due to should be equal to zero, the force due to A A and and B B must be opposite in direction. Hence, the particle must be opposite in direction. Hence, the particle

should be placed on the line

should be placed on the line AB AB. As. As A A and and B B have charges of opposite signs, have charges of opposite signs,C C cannot be between cannot be between A A and and B B

Also

Also A A has larger magnitude of charge than has larger magnitude of charge than B B. Hence,. Hence,C C should be placed closer to should be placed closer to B B than than A A. The situation is shown in figure. Suppose. The situation is shown in figure. Suppose

BC

BC== x x and the charge on and the charge onC C is isQQ

6 6 2 2 0 0 1 1 ((88..0 0 110 0 )) _ˆˆ 4 4 ((00..2 2 )) CA CA Q Q F F ii x x p p -¥ ¥ = = Œ Œ ++   and and p p -- - ¥¥ = = Œ Œ   66 2 2 0 0 1 1 ((22..0 0 110 0 )) _ˆˆ 4 4 CB CB Q Q F F ii x x 6 6 66 2 2 22 0 0 1 1 ((88..0 10 10 ) 0 ) ((22..0 10 10 )0 ) 4 4 ((00..2 2 )) C C CCA A CCBB Q Q QQ F F FF FF ii x x x x p p - - - -È È _¥_¥ _¥_¥ ˘˘ = = + + = = Í Í -- ˙˙ Œ Œ Í Í _Î_Î ++ ˙˙_˚˚      But But _{| |} _{| |} ₀₀ C C F F ==   Hence Hence 6 6 ₂₂ ₂₂ 66 0 0 1 1 ((88..0 10 10 ) 0 ) ((22..0 10 10 )0 ) 0 0 4 4 ((00..2 2 )) Q Q QQ x x x x p p - - - -È È _¥_¥ _¥_¥ ˘˘ - - == Í Í ˙˙ Œ

Œ Í Í _Î_Î ++ ˙˙_˚˚ which gives which gives x x = 0.2 m = 0.2 m

Illustration 13

Illustration 13 Point chargesPoint chargesQQ and andqq are placed at the vertices of a square of side are placed at the vertices of a square of side aa as shown. What should be sign of charge as shown. What should be sign of charge

q

q and magnitude of ratio of and magnitude of ratio of qq

Q

Q so that so that (a)

(a) net force on eachnet force on eachQQ is zero is zero

(b)

(b) net force on eachnet force on eachqq is zero is zero

Is it possible that the entire system could be in electrostatic equilibrium?

Solution

Case I:

Case I: Let the charge Let the chargeqq and andQQ are of same sign. are of same sign.

(a)

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12

12 _Electro_Electro_{statics: P}_{statics: P}_{art 1}_{art 1}

Here Here 11 22 0 0 1 1 4 4 qQ qQ F F a a pe pe =

= {force of {force ofqq at at D D on onQQatat A A}}

2 2 22 0 0 1 1 4 4 qQ qQ F F a a pe pe =

= {force of {force ofqq atat B B on onQQatat A A}}

3 3 ₂₂ 0 0 1 1 4 4 22 QQ QQ F F a a pe pe =

= {force of {force ofQQatatCC ononQQatat A A}}

In Figure (A), resultant of forces

In Figure (A), resultant of forces _F_F₁₁ andand _F_F₂₂



 

will lie along

will lie along _F_F₃₃

so that net force on

so that net force onQQ cannot be zero. Hence, cannot be zero. Hence,qq and andQQ have to have to

be of opposite signs.

be of opposite signs. Case II:

Case II:Let the chargeLet the chargeqq and andQQ are of opposite sign. are of opposite sign.

In this case, as shown in Figure (B), resultant of

In this case, as shown in Figure (B), resultant of F F 1 1 andandF F 22

   will be opposite to will be opposite toF F ₃₃  

so that it becomes possible to obtain

so that it becomes possible to obtain a conditiona condition

of zero net force.

Let us write Let us write 1 1 22 R R F F = = F F ++F F      \ \ 2 2 22 1 1 22 22 0 0 1 1 2 2 4 4 r r qQ qQ F F F F F F a a pe pe = = + + == Direction of Direction of F F _R_R   will be along

will be along AC AC ( ( F F _R_R,,

 

being resultant of forces of equal magnitude, bisects the angle between the two)

3 3 and and R R F F F F    are in

are in opposite directions. Net force opposite directions. Net force ononQQ can be zero if their magnitudes are also equal, i.e., can be zero if their magnitudes are also equal, i.e.,

2 2 2 2 22 0 0 00 00 1 1 11 2 2 oorr 22 00 4 4 4 4 2 2 44 22 q qQQ QQQQ QQ QQ q q a a a a aa p pee ppee pe pe

Ê

ˆ

=

_Á

_Ë

-

_˜

_¯

==

⇒ ⇒ 11 { { 00}} 2 2 2 2 22 22 Q Q qq q q QQ Q Q = = ﬁﬁ == ππ The

The sign sign ofofqq should be negative. should be negative.

(b)

(b) Consider now the forces acting on chargeConsider now the forces acting on chargeqq placed at placed at B B..

In a similar manner, as discussed in (a), for net force on

In a similar manner, as discussed in (a), for net force onqq to be zero, to be zero,qq and andQQ have to be of opposite signs. This is also shown have to be of opposite signs. This is also shown

in the given figures.

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Electrostatics: Part 1 13 Now, 1 ₂ 0 1 4 Qq F a pe

= {force ofQ at A onq at B}

2 ₂ 0 1 4 Qq F a pe

= {force ofQ atC onq at B}

2 3 ₂ 0 1 4 2 q F a pe

= {force ofq at D onq at B}

Referring to Figure (D) Let us write 1 2 R F = F +F    \ 2 2 1 2 2 0 1 2 4 R Qq F F F a pe = + =

Resultant of F 1 and F 2 , i.e.,F 3

  

is opposite to _F₃.



Net force can become zero if their magnitudes are also equal i.e.,

2 2 2 0 0 1 1 2 4 4 2 Qq q a a pe pe = \ ₂

{ }

0 2 0 2 2 0 2 4 2 2 q q q q Q Q q Q a pe

Ê

_{- = ﬁ =}

ˆ

_{ﬁ =}

_π

Á

˜

Ë

¯

The sign of ‘q’ should be negative.

In this case we need not to repeat the calculation as the present situation is same as previous one; we can directly write

2 2

q Q =

(c) The entire system cannot be in equilibrium since both conditions, i.e., and

2 2 2 2

Q q

q =- Q=- cannot be satisfied

together.

Illustration 14 Three particles, each of mass ‘m’ and carrying a chargeq, are suspended from a common point by insulating

massless strings, each ‘ L’ long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side

‘a’, calculate the chargeq on each particle. Solution

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T cosq =mg …(i)

While for equilibrium in the plane of equilateral triangle,

T sinq = 2F cos 30° ...(ii)

So from equations (i) and (ii), we have

3 tan F mg q = …(iii) Here, 2 2 ₂ ₂ 0 1 and tan 4 q OA OA F OP a _L _OA q pe = = =

- and as from Figure (C) 2 2 sin60

3 3 3 a OA = AD = a ∞= So, 2 2 ( 3) tan ( 3) ( 3) a a L L a q = = - {as L >>a}

On substituting the above values ofF and tanq in equation (iii) we get:

1 2 3 2 0 2 0 4 3 i.e., 3 4 ( 3) a mg a q q mg a L L pe pe È ˘ = = Í ˙ Î ˚

Illustration 15 A thin fixed ring of radius ‘a’ has a positive charge ‘q’ uniformly distributed over it. A particle of mass ‘m’

and having a negative charge ‘Q’ is placed on the axis at a distance of x ( x <<a) from the centre of the ring. Show that the motion

of the negatively charged particle is approximately simple harmonic. Calculate the time period of oscillation.

Solution

The force on the point chargeQ due to the elementdq of the ring

2 0 1 4 dqQ dF r pe = along AB

As for every element of the ring there is symmetrically situated diametrically opposite element, the components of forces along the axis will add up while those perpendicular to it will cancel each other. Hence, net force on the charge –Q is

2 0 1 cos cos ; 4 x Qdq F dF dF F r r q q pe È ˘ = = = _Í- _˙ Î ˚

Ú

So, 3 2 2 32 0 0 1 1 4 4 ( ) Qx Qqx F dq r a x pe pe = - = +

Ú

…(1) 2 2 1 2 {as r a x = ( + ) and

Ú

dq q= }

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As the restoring force is not linear, the motion will be oscillatory. However, if x <<a so that x 2 <<a2,

3 3 0 0 1 with 4 pe 4pe = =- = Qq Qq F x kx k a a

i.e., the restoring force will become linear and so the motion is simple harmonic with time period

3 0 4 2 2 m 2 ma T k qQ pe p p p w = = =

Concept Application Exercise 2

1. Two similar charge of +Q , as shown in figure are placed at A and B. –q charge is placed at pointC midway between A and B.

–q charge will oscillate if

(a) It is moved towards A. (b) It is moved towards B.

(c) It is moved upwards AB. (d) Distance between A and B is reduced.

2. When the distance between two charged particle is halved, the force between them becomes

(a) One fourth (b) One half

(c) Double (d) Four times

3. Two point charges in air at a distance of 20 cm. from each other interact with a certain force. At wh at distance from each other

should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction

(a) 8.94 × 10–2 m (b) 0.894 × 10–2 m (c) 89.4 × 10–2 m (d) 8.94 × 102 m

4. Two small balls having equal positive chargeQ (Coulomb) on each are suspended by two insulating strings of equal length L metre, from a hook fixed to a stand. The whole set up is taken in a satellite in to space where there is no gravity (state of

weightlessness). Then the angle (q) between the two strings is

(a) 0° (b) 90°

(c) 180° (d) 0° <q < 180°

5. Three equal charges (q) are placed at corners of an equilateral triangle. The force on any charge is

(a) Zero (b) 2 2 3 Kq a (c) 2 2 3 Kq a (d) 2 2 3 3 Kq a

6. Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with

their diameters (Fig. (a)). The electrostatic force acting on sphere 2 due to sphere 1 is F .



. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1 (Fig. (b)), then to sphere 2 (Fig. (c)) and finally removed (Fig. (d)). The electrostatic force that now acts on sphere 2 has magnitudeF’. What is the ratioF’ / F ?

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(c) (d)

7. Two equally charged particles are held 5.0 mm apart and then released from rest. The initial acceleration of the first particle

is observed to be 6.0 m/s2 and that of the second to be 10.0 m/s2. If the mass of the first particle is 6.0 × 10−7 kg, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?

8. In the figure, three charged particles lie on an x -axis. Particles 1 and 2 are fixed in place. Particle 3 is free to move, but the

net electrostatic force on it from particles 1 and 2 happens to be zero. If L₂₃ = L₁₂, what is the ratioq₁ /q₂?

9. In figure, particles 1 and 2 are fixed in place, but particle 3 is free to move. If the net electrostatic force on particle 3 due

to particles 1 and 2 is zero and L₂₃ = 2.00 L₁₂, what is the ratioq₁ /q₂?

10. Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 N when their

centre-to-centre separation is 50.0 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other?

11. In the figure four particles form a square. The charges areq₁ =q₂ =Q andq₂= q₁= q. (a). What isQ / q if the net electrostatic

force on particles 1 and 4 is zero? (b) Is there any value ofq that makes the net electrostatic force on each of the four particles

zero? Explain.

12. In the figure, particle 1 of charge +1.0mC and particle 2 of charge −3.0mC are held at separation L = 10.0 cm on an x -axis.

If particle 3 of unknown chargeq₃ is to be located such that the net electrostatic force on it from particles 1 and 2 is zero,

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13. Three particles are fixed on an x axis. Particle 1 of chargeq₁ is at x =−a, and particle 2 of chargeq₂ is at x = +a. If their net

electrostatic force on particle 3 of charge +Q is to be zero, what must be the ratioq₁ /q₂ when particle 3 is at (a) x = +0.500a and (b) x = +1.50a?

14. In Fig. (a), particle 1 (of chargeq₁) and particle 2 (of chargeq₂) are fixed in place on an x -axis, 8.00 cm apart. Particle 3

(of chargeq₃ = +8.00 × 10−19 C) is to be placed on the line between particles 1 and 2 so that they produce a net electrostatic

force F 3,net



on it. Figure (b) gives the x component of that force versus the coordinate x at which particle 3 is placed. The

scale of the x -axis is set by x _s= 8.0 cm. What are (a) the sign of chargeq₁ and (b) the ratioq₂ /q₁?

(a) (b)

5. ELECTRIC FIELD

‘The electric field intensity’ (or strength of electric field or electric intensity) at a point in an electric field is the force experienced by a unit positive charge placed at that point, provided the presence of this charge does not disturb the field.

‘Electric field’ due to a point charge is the space surrounding it, within which electric force can be experienced by another charge.

Superposition of electric field’ (electric field at a point due to various charges): The resultant electric field at any point is equal to the vector sum of electric fields at that point due to various charges.

1 2 3

E = + + +E E E

   

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The magnitude of the resultant of two electric fields is given by

2 2

1 2 2 1 2 cos

E ₌ E ₊E ₊ E E q

and the direction is given by 2 1 2 sin tan cos E E E q a q = +

‘Electric field strength or Electric intensity’ ( ) E



at a point is the electric force experienced by a unit positive charge at that point. Mathematically, 0 , F E q =  

whereq₀ is positive test charge. In vector form, electric field at B due to chargeq at A,

2 1 3 2 1 ( ) q E k r r r r = -    

So for a point-chargeq, electric field at position r 

in free space

3 2 0 0 1 1 or 4 4 q q E r E r r pe pe = =  

If instead of single charge, field is produced by a charge distribution, by ‘principle of super-position’ for discrete distribution: 1 2 3 0 1 with 4 i i i i q E E E E E r r pe = + + =

Â

=      _ 

Electric field intensity is vector quantity. SI unit of electric intensity is NC–1 (newton/coulomb) In CGS system, unit of E



is dyne/stat-coulomb. Dimensional formula of E



: Force 2 2 3 1

[ ]

Charge ampere × time .

MLT MLT MLT A A T - -- -= = =

Illustration 16 Four point charges all have the same magnitude, but they do not have the same sign. These charges are fixed to the corners of a rectangle in two different ways, as the figure shows.

Consider the net electric ﬁeld at the centreC of the rectangle in each case. In which of the statements, if either, is the net electric

ﬁeld greater?

Statement (i): It is greater in Figure (a). Statement (ii): It is greater in Figure (b).

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Solution

The net electric field atC is the vector sum of the individual fields created there by

the charges at each corner. Each of the individual fields has the same magnitude, since the charges all have the same magnitude and are equidistant from C . The

directions of the individual fields are different, however. The field created by a positive charge points away from the charge, and the field created by a negative charge points toward the charge.

Statements (i) and (iii) are incorrect. To see why these answers are incorrect, note that the charges on corners 2 and 4 are identical in both parts of figure. Moreover, in Fig. (a) the charges at corners 1 and 3 are both +q, so they contribute

individual fields of the same magnitude at C that have opposite directions and,

therefore, cancel. However, in Fig. (b) the charges at corners 1 and 3 are –q and

+q, respectively. They contribute individual fields of the same magnitude atC that

have the same directions and do not cancel, but combine to produce the field E 13



shown in Fig. (b). The fact that this contribution to the net field atC is present

in Fig. (b) but not in Fig. (a) means that the net fields in the two cases are different and that the net field in Fig. (a) is less than (not greater than) the net field in Fig. (b).

‘Statement (ii) is correct.’ To assess the net field at C , we need to consider the contribution from the charges at corners 2 and 4,

which are −q and +q, respectively, in both cases. This is just like the arrangement on corners 1 and 3, which was discussed previously.

It leads to a contribution to the net field atC that is shown E 24



as in both parts of the figure. In Fig. (a) the net field atC is just E 24,



but in Fig. (b) it is the vector sum of E 13 and E 24,

 

which is clearly greater than either of these two values alone.

Illustration 17 A charge 10−9 coulomb is located at origin in free space and another chargeQ at (2, 0, 0). If the x -component

of the electric field at (3, 1, 1) is zero, calculate the value ofQ. Is the y-component zero at (3, 1, 1)? Solution

As electric field due to a point chargeq_i at position r i



in vector from is given by

3 0 1 4 i i i q E r r pe Æ _Æ = here: Æ_r₁ ₌_{(3 0)}_- _iˆ₊_{(1 0)}_- ˆ_j₊_{(1 0)}_- _kˆ ₌₃_iˆ₊ _jˆ₊ ˆ_k with 2 2 2 1 (3 1 1 ) 11 r ₌ ₊ ₊ ₌ m 2 (3 2 ) ˆ (1 0) ˆ (1 0) ˆ ˆ ˆ ˆ r i j k i j k Æ = - + - + - = + + with 2 2 2 2 (1 1 1 ) 3 r ₌ ₊ ₊ ₌ m So that, ₁ _3/2 ₂ _3/2 0 0 1 _ˆ _ˆ _ˆ 1 _ˆ _ˆ _ˆ [3 ] and [ ] 4 (11) 4 (3) Q Q E i j k E i j k pe pe Æ Æ = + + = + + And hence, 9 9 9 1 2 0 1 3 10 _ˆ 10 _ˆ 10 _ˆ 4 11 11 3 3 11 11 3 3 11 11 3 3 Q Q Q E E E i j k pe - - -Æ Æ Æ

_È

_Ê

_¥

_ˆ

_Ê

_ˆ

_Ê

_ˆ

_˘

=

+

=

Í

_Á

+

_˜

+

_Á

+

_˜

+

_Á

+

_˜

˙

Ë

¯

Ë

¯

Ë

¯

Í

˙

Î

˚

According to given problem: E _x = 0,

i.e., 9 0 1 3 10 0 4 11 11 3 3 Q pe -È _¥ ˘ + = Í ˙ Í ˙ Î ˚

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20 _{Electrostatics: Part 1} So that, 3/2 9 3 3 10 coulomb 11 Q= - È ˘_{Í ˙} ¥ ¥ -Î ˚

And for this value ofQ

9 3/2 9 9 0 0 1 10 (3 / 11) 3 10 1 2 10 0 4 11 11 3 3 4 11 11 y E pe pe - - -È _{¥ ¥} ˘ _¥ = _Í - _˙=- π Í ˙ Î ˚

i.e., E _y is not zero. Illustration 18

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the

configuration. Show that the equilibrium of the test charge is necessarily unstable.

(b) Verify this result for the simple configuration of two charges of the same magnitude and sig n placed a certain distance apart. [NCERT] Solution

(a) In case, the equilibrium is stable, then the test charge will experience a restoring force towards the

null point when displaced slightly in any direction. In such a situation, all electric field lines would be directed inwards towards the null point. Consequently, there would be a net inward electric flux through a closed surface around the null point. But as per Gauss’s law this is not allowed because this closed surface does not enclose any charge, hence the equilibrium cannot be at all stable, i.e., it is necessarily unstable.

(b) For the simple configuration of two charges (+q, +q), the midpoint (O) of the line AB joining the

charges is the null point. If the test charge (q₀) is displaced along AB in any direction, there acts a

restoring force on it. But if the test charge is displaced alongOP (a line normal to AB atO), the net

forceF will take it away from the null point alongOP. But for the stability of the equilibrium, there should be a restoring force

in all directions. Hence, the equilibrium is unstable.

Illustration 19 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N/C in Millikan’s oil drop experiment. The density of the oil is 1.26 g/cm3. Estimate the radius of the drop.

(g = 9.81 m/s2;e = 1.60 × 10–19 C). [NCERT]

Solution

Here,q (charge on the drop) = 12e = 12(1.6 × 10–19 C) = 19.2 × 10–19 C E = 2.55 × 104 V/m

r (density of oil) = 1.26 g/cm3= 1.26 × 103 kg/m3

g = 9.81 m/s2

Letr be the radius of the drop. As the drop is held stationary under constant electric field E , force on the drop due to electric

field (qE ) = weight of the drop (mg)

or 4 3 3 qE

= Á

Ê

p r g

ˆ

_˜

r

Ë

¯

or 1/3 1/3 ₁₉ ₄ 6 4 3 3 2 3 3(1.92 10 C) (2.55 10 V/m) 0.981 10 m 9.81 10 mm 4 4 3.14 (1.26 10 kg/m ) (9.81 m/s ) qE r g pr -- -È ˘ È ˘ ¥ ¥ =Í ˙ =Í ˙ = ¥ = ¥ ¥ ¥ Í ˙ Î ˚ Î ˚

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6. ELECTRIC FIELD DUE TO CONTINUOUS DISTRIBUTION OF CHARGE

To find the field of a continuous charge distribution, we divide the charge into infinitesimal charge elements. Each infinitesimal charge element is then considered, as a point charge and electric field dE



is determined due to this charge at given point. The net field at the given point is the summation of fields of all the elements, i.e., E =

Ú

dE

 

Continuous charge distribution

Linear charge distribution Surface charge distribution Volume charge distribution

In this the distribution charge is distributed on a line.

For example: Charge on a wire, charge on

a ring etc. Relevant parameter isl which is

called linear charge density i.e.,

charge length l = 2 Q R l p =

Circular charged ring

In this the distribution charge is distributed on the surface.

For example: Charge on a conducting

sphere, charge on a sheet etc. Relevant parameter isswhich is called surface charge

density i.e., charge area s = s p = 2 4 Q R Spherical shell

In this the distribution charge is distributed in the whole volume of the body.

For example: Non-conducting charged

sphere. Relevant parameter isr which is called

volume charge density i.e.,

charge volume r = r p = 3 4 3 Q R

Non conducting sphere Each infinitesimal charge element is then considered as a point charge and its field is given by: 2

0 4 dq dE r pe = .

At a point distantr from the element, the net field is the summation of fields of all the elements, E =

Ú

dE

 

.

Taking ˆ _{ˆ, we have} _and _.

x y x x y y

dE dE i dE j = + E =ÚdE E =ÚdE



7. FIELD OF RING CHARGE

A ring-shaped conductor with radiusa carries a total chargeQ uniformly distributed around it. Let us calculate the electric field at

a pointP that lies on the axis of the ring at a distance x from its centre.

As shown in the figure, we imagine the ring divided into infinitesimal segments of lengthds. Each segment has chargedQ and

acts as a point-charge source of electric field. Let dE



be the electric field from one such segment; the net electric field at P is then the sum of all contributions dE



from all the segments that make up the ring. (This same technique works for any situation in which charge is distributed along a line or a curve.) The calculation of

E



is greatly simplified because the field pointP is one the symmetry axis of the

ring. If we consider two ring segments at the top and bottom of the ring, we see that the contributions dE



to the field at P from these segments have the same x -component but opposite y-components. Hence that total y-component of field due

to this pair of segments is zero. When we add up the contributions from all such pairs of segments, the total field E



will have only a component along the ring’s

symmetry axis (the x -axis), with no component perpendicular to that axis (that is, no y-component or z-component). So the field at P is described completely by its x -component E _x.

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To calculate E x note that the square of the distancerfrom a ring segment to the pointP isr 2 = x 2 +a2. Hence the magnitude of

this segment’s contribution to the electric field atPis ₂ ₂

0 1 4 dQ dE x a pe = + Using cos ₂ _{2 1 2}, ( ) x x r x a a = = +

the componentdE _x of this field along the x -axis is

2 2 ₂ ₂ 2 2 3/2 0 0 1 1 cos 4 4 ( ) x dQ x xdQ dE dE x a _x _a x a a pe pe = = = + ₊ +

To find the total x-component Exof the field atP, we integrate this expression over all segments of the ring:

2 2 3/2 0 1 4 ( ) x xdQ E x a pe = +

Ú

Since xdoes not vary as we move from point to point around the ring, all the factors on the right side except dQ are constant

and can be taken outside the integral. The integral of dQ is just the total chargeQ is just the total chargeQ, and we finally get

2 2 3/2 0 1 ˆ ˆ 4 ( ) x xQ E E i i x a pe Æ = = + …(i)

∑ Electric field is directed away from positively charged ring.

∑ For x = 0, E = 0, this conclusion may be arrived at the symmetry consideration.

∑ At a large distance, from the ring the electric field will be zero, hence it should have certain maximum value between x = 0

and x =• (or x = –•)

∑ If we maximize the equation (i) we can get the value of x _m as well as E _max..

For maximum value of E _x;

2 2 3/2 0 1 0 4 ( ) d x Q dx pe x a Ï ¸ Ô Ô = Ì ˝ + Ô Ô Ó ˛ 2 2 3/2 2 2 1/2 2 2 3 3 ( ) 1 ( ) 2 2 ₀ ( ) x a x x a x x a + ◊ - ◊ + ◊ = + 2 2 2 ( ) 3 0 2 a x + a - x = ﬁ = ±x

and the maximum value of the electric field is _(max) ₂

0 1 2 4 3 3 a Q E R pe

Ê

ˆ

=

_Á

_˜

Ë

¯

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Illustration 20 A uniformly charged wire, linear charge densityl , is laid in the form of a semicircle of radius R. Find the

electric field generated by the semicircle at the centre?

Solution

We consider a differential element dl on the ring, that subtends an angled q at the centre of the ring,dl =r d q.

This element creates a fielddE which makes an angleq at the centre as shown in figure. For each differential element in the upper half of the ring, there corresponds a symmetrically placed charge element in the lower half plane. The y-components of field due to these

symmetric elements cancel out, and x -components remain.

2 2 0 0 ( ) cos cos cos 4 4 x dQ R d dE dE R R l q q q q pe pe = = =

On integrating the expression fordE _x, w.r.t. angleq , in limitsq = –p /2 toq = +p /2, we obtain /2 2 0 0 /2 cos 2 4 R E d R R p p l l q q pe pe + -=

Ú

= In terms of charge, ₂ ₂ 0 2 Q Q E R R l p p e = ﬁ =

If we consider the wire in the form of an arc as shown in the figure, the symmetry consideration is not useful in cancelling out

x and y components of the fields, ifq ₁ andq ₂ are different. We will integratedE _x as well asdE _y in limitsq = –q ₁ toq = +q ₂.

2 1 1 2 2 0 0

cos (sin sin )

4 4 x R E d R R q q l l q q q q pe pe + -=

_Ú

= + 2 1 1 2 2 0 0

sin (cos cos )

4 4 y R E d R R q q l l q q q q pe pe + -=

_Ú

=

- For a symmetrical arc,q 1 =q 2. Thus E y vanishes and

0 sin 2 x E R l q pe =

8. FIELD OF LINE CHARGE

Positive electric chargeQ is distributed uniformly along a line, lying along the y-axis. Let us find the electric field at point D on the x -axis at a distance r ₀

from the origin.

We divided the line charge into infinitesimal segments, each of which acts as a point charge; let the length of a typical segment at height l bedl. If the charge is distributed

uniformly with the linear charge densityl . Hence the chargedQin a segment of length dl isdQ =l dl. At point D the differential electric fielddE created by this element,

2 2 0 0 4 4 dQ dl dE r r l pe pe = = …(i)

In triangle AOD;OA =OD tanq

l =r ₀ tanq ; Differentiating this equation with respect toq dl =r ₀ sec2q d q

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Substituting the value ofdl in equation (i);

0 0 4 d dE r l q pe =

FielddE has componentsdE _x,dE _y given by

0 0 0 0 cos sin and 4 4 l q q l q q pe pe = = x y d d dE dE r r

On integrating expression for dE _x and dE _y in limits to

2 2

p p

q _{= -} q _{= +} we obtain_E_x and_E_y. Note that as the length of wire

increases, the angleq increases; for a very long wire (infinitely long wire), it approaches p /2.

/2 /2 0 0 00 0 0 /2 /2 cos sin and 0 4 2 4 x y d d E E r r r p p p p l q q l l q q pe pe pe + + - -=

Ú

= =

Ú

= 0 0 Thus 2 x E E r l pe = =

Using a symmetry argument, we could have guessed that E y would be zero; if we place a positive test charge at D, the upper

half of the line of charge pushes downward on it, and the lower half pushes up with equal magnitude.

∑ “If the wire has finite length and the angle subtended by ends of wire at a point are q ₁ andq ₂, the limits of integration would change.”

2 1 1 2 0 0 00 cos (sin sin ) 4 4 x d E r r q q l q q l q q pe pe + -=

Ú

= + 2 1 1 2 0 0 0 0 sin (cos cos ) 4 4 y d E r r q q l q q l q q pe pe + -=

_Ú

=

-∑ “If we wish to determine field at the end of a long wire, we may substituteq ₁ = 0 andq ₂

=p /2 in the expressions for E _x and E _y.”

l p l pe pe l p l pe pe

È

Ê ˆ

˘

=

Í

_Î

+

Á ˜

_{Ë ¯}

˙

_˚

=

È

Ê ˆ

˘

=

_Í

+

_{Á ˜}

_{Ë ¯}

_˙

=

Î

˚

0 0 0 0 00 0 0 sin(0) sin 4 2 4

and cos (0) cos

4 2 4 x y E r r E r r

Magnitude of resultant field E

Æ : 2 2 0 0 2 | | 4 x y E E E r l pe Æ = + = E Æ

makes an angleq with the x -axis, where

| | tan 1; 45 | | y x E E q ₌ ₌ q ₌ _∞

Illustration 21 What is the electric field at any point on the axis of a charged rod of length L and linear charge densityl '? The point is separated from the nearer end bya.