Chapter 01 Electrostatics Part 1 With Watermark

Electrostatics: Part 1

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© 2017 Cengage Learning India Pvt. Ltd. ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, without the prior written  permission of the publisher. publisher.

Electrostatics: Part 1

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 × 10−27   kg, and a neutron has a slightly greater mass of 1.675 × 10−27 kg. Surrounding the th e nucleus is a diffuse cloud of orbiting particles called electrons, as the figure suggests. An electron has a mass of 9.11 × 10−31 kg. Like mass, ‘electric charge’ is an intrinsic property of protons and 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 +e, and the electron carries a charge −e. The SI unit for measuring the magnitude of an electric charge is the coulomb1 (C), and e has been determined experimentally to have the value, e = 1.60 × 10–19 C. (i) Number of protons = number of electrons.   (ii) Protons have the basic +e charge and electrons have the basic – e charge.  (iii) 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 + e charge and vice-versa.

2. CHARGING OF A 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)

Friction

(ii)

Conduction

(iii)

Induction

rub bed together, electrons are transferred By friction: In friction when two bodies are rubbed 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

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. (ii) Charge is always associated with mass: i.e., charge cannot exist without mass though mass can exist without charge.

 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 = +90 e) and emits an a -particle (charge = +2e) 92

U

238

Æ

90

Th

234

+

2

He

4

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., Q = ± ne with n

Charge on a body can never be

= 1,

±

2, 3 

2

, ± 17.2 e

e

3

or

± 10

-5

e

etc.

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 still e. • 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.,

1C  =

1 1.6

¥ 10

-19

=

6.25

¥ 10

18

 electronic charges. On account

this of large number, charge flows continuously and the quantization of charge can be ignored. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge. [NCERT]

 Illustration 3

4  Electrostatics: Part 1

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. 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, me = 9 × 1031 kg

 Illustration 4

Solution

Number of electrons lost by glass rod is

n=

q e

19 .2 =

1.6

-19

¥ 19

¥ 10

-19

= 12

Charge on silk = −19.2 × 10–19 C   (ii) Since an electron has a finite mass ( me = 9 × 10–31 kg), there will be transfer of mass from glass rod to silk cloth.   (i)

Mass transferred

=

12 ¥ (9 ¥ 10

-

31

) = 1.08 ¥ 10

-

29

kg

Note that mass transferred is negligibly small. This is expected because the mass of an electron is extremely small. 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 substance C , 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 and C  becomes positively charged.   (b)  B becomes positively charged and C  becomes negatively charged.   (c)  B becomes negatively charged and C  becomes positively charged.   (d)  B becomes negatively charged and C  becomes negatively charged. Illustration 5

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. Objects  A, B and C  are three identical, insulated, spherical conductors. Originally  A and B both have charges of +3 mC, while C  has a charge of –6 mC. Objects  A and C  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 and C  are held near each other, they will Illustration 6

(a)

attract

(b)

repel

(c)

have no effect on each other.

 Electrostatics: Part 1

5

Solution

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

∑ When the objects B and C  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 and C  are now held near each other they will also attract each other. 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.

  Illustration 7

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 have different size or material.

6  Electrostatics: Part 1

  (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:

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 initially present (a)

 Electrostatics: Part 1

 

7

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) In the return stroke of a typical lightning bolt, a current of 2.5 × 104 A exists for 20 ms. How much charge is transferred in this event? How many megacoulombs of positive charge are in 1.00 mol of neutral molecular-hydrogen gas (H 2)? (NA = 6.0 × 1023) 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.) 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? Figure shows four identical conducting spheres that are actually well separated from one another. Sphere W  (with an initial charge of zero) is touched to sphere  A and then they are separated. Next, sphere W   is touched to sphere B (with an initial charge of −32e) and then they are separated. Finally, sphere W  is touched to sphere C   (with an initial charge of +48e), and then they are separated. The final charge on sphere W  is +18e. What was the initial charge on sphere  A? A charged non-conducting rod, with a length of 2.00 m and a cross-sectional area of 4.00 cm 2, lies along the positive side of an x  axis with one end at the origin. The volume charge density r  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 by r = bx 2, where b = −2.00 µC/m5? What is the total charge in coulombs of 72.0 kg of electrons? (The mass of an electron = 9.0 × 10−31 kg) 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) (b)

    4. 5. 6. 7.

8.

9.

  10.

  11.

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.  

 

q1

+F 

–F 

+

–

q2

 –F 

q1

q2

+

+F  +

r 

r 

(a)

(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 charge q1 and particle 2 has charge q2. (These symbols can represent either positive or negative charge.) Let’s also focus on particle 1 and write the force acting on it in terms of a unit vector r   that points along a radial axis extending through the 

8  Electrostatics: Part 1

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 



F

=

k

q1q2  r  (Coulomb's law),   r 2

...(i)

where r is the separation between the particles and k is a positive constant called the electrostatic constant or the Coulomb constant. (We’ll discuss k below.) If between the two charges there is free space then, k  =

1 4pe 0

=

9

¥ 10

9

2

Nm C

-2

 (in SI Units)

where e 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

k 

1 =

4pe 0e r 

 where e r is 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). If q1 and q2 have the same sign, then the product q1q2 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, if q1 and q2 have opposite signs, the product q1q2 gives 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 Let q1 and q2 be two like charges placed at points  A and B in vacuum, separated by a distance r . Due to like nature of charges, they will repel each other. Let F 12  be the force on charge q1 due to q2; 



F 21

ˆ r 

12

 be the force on charge q2 due to q1;

 be the unit vector from q1 to q2;

 be the unit vector from q2 to q1; From figure, it is clear that F21 and r ˆ12  and are in the same direction ˆ r 

21



\



F21

=

1

q1q2

4pe 0

r 



Also,

ˆ F12 and r  21

\

F12



=

2

ˆ r  12

 are in same direction

1

q1q2

4pe 0

r 

2

ˆ21 r 

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.,

F12



=

F 21

=

1

q1q2

4pe 0

r 

2

 Electrostatics: Part 1

Since

ˆ r

12

ˆ and r  21

Put this value in equation, 

From equations,

ˆ r

 are unit vectors, opposite to each other i.e., 1



F12

=



F12

4pe 0

q1q2 r

=

–

=

ˆ – r  12

1

q1q2

4pe 0

r  



(– rˆ12 ); F12

2

21

9

2

rˆ12

 

– 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

F21



1



=

q1q2

4pe 0



r 12

3



ˆ r 

∵

r 12

r  =

12

12

r 

12

Similarly, force on q1 due to q2 is 1



F12

=

q1q2

4pe 0



r21

2

1



rˆ21

or

F12

=

4pe 0

q1q2   r 21 

3



r21

 

The above equations can also be written in the following manner: 1



F21

=

q1q2

4pe 0 r – r 2 1 



1

 

3



(r2 – r1) and F12

=

q1q 2

4pe 0 r –  r 1 2 





3



(r1 – r2 )  

 

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 charge Q Net force on Q will be 



Fnet = F

 Illustration 8



1

+





F2 +  + Fn -1 + Fn

 

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 F 

  Illustration 9

Ê Nm2 ˆ  (2 ¥ 10-7 C) (3 ¥ 10- 7 C) = 9 ¥ 109 Á 2 ˜  = 6 ¥ 10 -3 N  (repulsive) -2 2 (30 ¥ 10 m) Ë C ¯  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? (b) What is the force on the second sphere due to the first?

[NCERT]

10  Electro  Electrostatics: statics: Part Part 1

Solution (a)

As F

=

kq1q2 2

r 

, r 

=

kq1q2 F  1/2

r 

  (b)

Ê 9 ¥ 109 (Nm 2 / C2 ) (0.4 ¥ 10-6 C) (0.8 ¥ 10- 6 C) ˆ  =Á ˜  = 0.2 N Ë ¯ 

144

¥ 10 -4 m 2 = 12 ¥ 10 - 2 m = 12 cm

Since 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.

 Illustration 10

Two insulated charged copper spheres A and B 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 –7 C? The radii of  A and B are negligible compared to the distance of separation.   (b) What is the force of repulsion if:   (i) Each sphere is charged double the above amount, and the distance between them is halved.   (ii) The two spheres are placed in water? (Dielectric constant of water = 80) [NCERT] (a)

Solution (a)

We are given that, q1 = q2 = 6.5 × 10–7 C, r  =  = 50 cm = 0.5 m If F  is  is the force of repulsion, F k e =

or   (b)

F  =

(9 ¥ 10 9 ) ¥

(6.5 ¥ 10

-

7

q1 q2 2

r 

) (6.5 ¥ 10

-

7

)

2

(0.5)

=

1.5

¥

10

-

2

N.

(i) If the charge on each sphere is doubled and the distance between them is halved, the force of repulsion, F ´ would become r2  ). 16 times (as F µ q1q2 and F µ 1/ r  Clearly, F ´ = 16F  =  = 16 (1.5 × 10–2 N) = 0.24 N (ii) If the sphere are are placed in water (e r  = 80),

F w

1.5

F  =

= Œ

r 

¥ 10

-2

= 1.9 ¥ 10

80

-4

N

Suppose the spheres A and B 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 A and B?

 Illustration 11

Solution

We are given that, Charge on sphere A = charge on sphere B = q = 6.5 × 10–7 C Force of repulsion between A and B, i.e., F

=

ke

q¥q r

2

=

k e

q

2

2

r 

-

=

9

(9 ¥ 10 )

7 2

(6.5 ¥ 10 ) (0.5)

2

=

1.5 ¥ 10

-

2

N

Let the third sphere (say C ) of the same size but uncharged be brought in contact with  A. Due to the flow of electrons, the two spheres share the charge equally. Therefore, Charge on A = Charge on C  = q + 0 2

=

q

2

When the sphere C  (having  (having charge q /2) touches the sphere B (having charge q),

 Electrostatics:  Electr ostatics: Part Part 1

11

Total charge on B and C  (  (q + q /2 = 3q /2) is equally distributed between them. Thus, charge on B = charge on C  =  = 3q /4 If F ´ is the new force of repulsion between  A and B, F ¢ = ke

(q / 2) (3q / 4)

r

2

= ke

3 q

2

2

8 r 

=

3 8

F=

3 8

( 1.5 ¥ 10

-2

N) = 5.7 ¥ 10

-3

N

Two particles A and B having charges 8 × 10–6 C and –2 × 10–6C 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?

 Illustration 12 Solution

As the net electric force on C  should   should be equal to zero, the force due to A and  B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C  cannot  cannot be between A and B Also A has larger magnitude of charge than B. Hence, C  should  should be placed closer to B than A. The situation is shown in figure. Suppose  and the charge on C  is  is Q  BC = x  and

and

=

=

1

(2.0 ¥ 10

-

2

4p  Œ0

But



6

-

)Q ˆ i

 x 

FC = FCA + FCB = 

6

(8.0 ¥ 10 ) Q ˆ i 2 (0.2 +  x )

4p  Œ0



FCB

-

1



FCA



-6 -6 È (8.0 ¥ 10 10 ) Q ( 2.0 ¥ 10 10 ) Q ˘ Í ˙i 2 4p  Œ0 ÍÎ (0.2 +  x) 2 x  ˙˚

1



| F C  |

Hence

=

0

-6 -6 È (8.0 ¥ 10 10 ) Q (2.0 ¥ 10 10 ) Q ˘  = 0.2 m Í ˙ = 0  which gives x  = 2 4p  Œ0 ÍÎ ( 0.2 +  x )2 x  ˙˚

 Illustration 13

1

Point charges Q and q are placed at the vertices of a square of side a as shown. What should be sign of charge

q and magnitude of ratio of

q Q

 so that

net force on each Q is zero   (b) net force on each q is zero Is it possible that the entire system could be in electrostatic equilibrium?   (a)

Solution Case I: Let the charge q and Q are of same sign.   (a)

Consider the forces acting on charge Q placed at A (shown in figures (A) and (B))

12  Electro  Electrostatics: statics: Part Part 1

Here

1

F 1

=

4pe 0 1

F 2

=

F 3

=

qQ a

qQ

4pe 0 1

 {force of q at D on Q at A}

2

a

 {force of q at B on Q at A}

2

QQ

4pe 0 2 a

2

 {force of Q at C on Q at A}

In Figure (A), resultant of forces be of opposite signs.





F1 and F 2

 will lie along



F 3

 so that net force on Q cannot be zero. Hence, q and Q have to

Case II: Let the charge q and Q are of opposite sign.

In this case, as shown in Figure (B), resultant of of zero net force. Let us write \









F R and F 3



F  R

 will be opposite to



F 3

 so that it becomes possible to obtain a condition



F12

+

1

F 22

=

qQ

4pe 0

a

2

2



  will be along  AC  (F  R ,   being resultant of forces of equal magnitude, bisects the angle between the two)

are in opposite directions. Net force on Q can be zero if their magnitudes are also equal, i.e., 1

qQ

4pe 0

⇒



F R = F1 + F 2

Fr  =

Direction of



F1 and F 2

q=

a

2

Q 2

2

2

fi

=

1

QQ

4pe 0  2a

q Q

=

1 2 2

2

or

Q 4pe 0 a

2

Ê ÁË

2q

Q ˆ  - ˜  =0 2 ¯ 

{Q π 0}

The sign of q should be negative.   (b)

Consider now the forces acting on charge q placed at B. In a similar manner, as discussed in (a), for net force on q to be zero, q and Q have to be of opposite signs. This is also shown in the given figures.

 Electrostatics: Part 1

Now,

F 1

=

F 2

=

F 3

=

1

Qq

4pe 0

a2

1

Qq

 {force of Q at A on q at B}

2

4pe 0

a

1

q

13

 {force of Q at C  on q at B}

2

4pe 0 2 a

 {force of q at D on q at B}

2

Referring to Figure (D) Let us write F R = F1 + F 2 

\

F R

F12

=



Resultant of



+

1 =

Qq

4pe 0



Qq

4pe 0

F 22



and F2 , i.e., F 3  is

F1

1

\



a

q 4pe 0 a

2

2

2

Ê ÁË

1 =

2Q

a

2

2

opposite to



F 3 .

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

q2

4pe 0  2 a

2

q - ˆ  ˜  = 0 fi 2 ¯ 

Q=

q 2 2

fi

q Q

=2

2

{q π 0}

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 q Q

  (c)

=

2

2

The entire system cannot be in equilibrium since both conditions, i.e., together.

q

Q = -

2 2

and

Q

q = -

  cannot be satisfied

2 2

Three particles, each of mass ‘ m’ and carrying a charge q, 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 charge q on each particle.

 Illustration 14

Solution

In accordance with (shown in Figure (B)), for equilibrium of a particle along a vertical line,

14  Electrostatics: Part 1

T  cos q  = mg 

…(i)

While for equilibrium in the plane of equilateral triangle, T  sin q  = 2F cos 30° So from equations (i) and (ii), we have 3F 

tan q 

Here,

So,

F 

=

 

…(iii)

mg

1 =

q

4pe 0

a

2

and tan q 

2

(a

tan q 

...(ii)

OA =

3)

OP

a

=

=

2

 L

-

2

( 3) L

( a 3)

OA

 and as from Figure (C)

=

2

L

-

OA

OA =

2

2 3

AD

=

2 3

a sin60∞

=

a

3

 {as L >> a}

On substituting the above values of F  and tan q  in equation (iii) we get: a

=

( 3) L

3

q

2

mg 4pe 0 a 2

12

È 4pe 0 a3 mg ˘ i.e., q = Í ˙ 3L Î ˚

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   R2 is given by…………….   13. Three identical positive charges Q  are arranged at the vertices of an equilateral triangle. The side of the triangle is a. Find the intensity of the field at the vertex of a regular tetrahedron of which the triangle is the base. –19 –19   14. Two point charges of +5 × 10  C and +20 × 10  C are separated by a distance of 2 m. The electric field intensity will be zero at a distance d  = ........... from the charge of 5 × 10–19 C.   15. An electron (mass me) falls through a distance ‘d ’ in a uniform electric field of magnitude E .  

 

What are the signs of each of the three charges? Explain your reasoning. (b) At what point(s) is the magnitude of the electric field the smallest? Explain your reasoning. Explain how the fields produced by each individual point charge combine to give a small net field at this point or points. (c) Two point charges q and –q are placed at a distance d  apart. What are the points at which resultant electric field is parallel to the line joining the two charges? 9. Two point charges Q and 4Q are fixed at a distance of 12 cm from each other. Sketch lines of force and locate the neutral point, if any. 10. Is an electric field of the type shown by the electric lines in figure physically possible?

(c)

(a)

Which of the following statement is correct? If  E  = 0, at all points of a closed surface: (a) The electric flux through the surface is zero. (b) The total charge enclosed by the surface is zero.   12. A hollow dielectric sphere as shown in the figure has inner and outer radii of R1 and R2 respectively. The total charge carried by the sphere is + Q, this charge is uniformly distributed between R1 and R2. Then

The direction of the field is reversed keeping its magnitude unchanged and a proton (mass m p) falls through the same distance. If the time taken by electron and proton to fall the distance d  is ‘t electron’ and ‘t proton’ respectively, then the ratio   16.

17.

  18.

  19.

20.

the electric field for r d . Show your results in a graph of the radial component Æ

of  as a function of r . What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell ?   30. Which of the following statements is/are correct?   (a) Electric field calculated by Gauss law is the field due to only those charges, which are enclosed inside the Gaussian surface.   (b) Gauss law is applicable only when there is a symmetrical distribution of charge.   (c) Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only.  E 

 

(c)

 Electrostatics: Part 1

71

STATE LEVEL EXERCISES attraction, while pairs (2, 3), (4, 5) show repulsion. Therefore ball 1 must be (a) neutral (b) made of metal (c) positively charged (d) negatively charged 8. There are two charges +1 micro coulomb and 5 micro coulomb. The ratio of the forces acting on them will be (a) 1 : 1 (b) 1 : 25 (c) 1 : 5 (d) 5 : 1 9. An electric charge q exerts a force F  on a similar electric charge q separated by a distance r . A third charge q /4 is placed mid-way between the two charges. Now, th e force F  will….

 YPE SINGLE CORRECT  ANSWER T 

Two identical metal balls with charges +2Q and −Q are separated by same distance, and exert a force F   on each other. They are joined by a conducting wire, which is then removed. The force between them will now be  /2 (a) F  (b) F   /4  /8   (c) F  (d) F  2. A cube of side ‘b’ has charge ‘q’ at each of its vertices. The electric field at the centre of the cube will be 1.

(a)

q b

 

(c) 3.

32q

b

(d)

2

q 2b

2

(c)

3Q

(b)

r  Q 2

2r 

(d)

 

(c)

become

  10.

A charged particle moves with a speed v in a circular path of radius R around a long uniformly charged conductor

2

r 

Two similar conducting spheres A and B are brought in contact and insulated from earth. A negatively charged ebonite rod is brought near A. Now (a) A will have +ve charge and B will have –ve charge.   (b) A will have –ve charge and B will have +ve charge. (c) Both will acquire negative charge. (d) Both will remain uncharged. 5. Two identical metallic spheres X and Y have exactly equal masses. X is given a positive charge q coulomb and Y is given an equal negative charge. Then after charging. (a) Masses of X and Y are equal   (b) Mass of Y is greater than X (c) Mass is not involved (d) Mass of X is greater than Y 6. A charge q1 exerts some force on a second charge q2. If a third charge q3 is brought near, then the force of q1 exerted on q2   (a) will increase in magnitude (b) will decrease in magnitude (c) will remain unchanged (d) will increase if q3 is of the same sign as q1 and will decrease if q3 is of opposite sign 7. Five balls numbered (1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4) and (4, 1) show electrostatic 4.

F 

become

3Q

zero

F 

(a)

Zero

Three small spheres, each carrying a positive charge Q, are placed on the circumference of a circle of radius ‘ r ’ to form an equilateral triangle. The electric field intensity at the centre of the circle will be (a)

 

(b)

2

 

(a)

v

(c)

v

µ

µ

3 F 

27

(b)

become

(d)

remain F 

(b)

R 1

v

µ

9

1  R

(d) v is independent of R

 R

A positive point charge Q is brought near an isolated metal cube (a) The cube becomes negatively charged (b) The cube becomes positively charged (c) The interior becomes positively charged and the surface becomes negatively charged (d) The interior remains charge free and the surface gets non-uniform charge distribution 12. A simple pendulum of time period T   is suspended above a large horizontal metal sheet with uniformly distributed positive charge. If the bob is given some negative charge, its time period of oscillation will be (a) > T    (b) < T  11.

   

(c) T  (d) proportional

to its amplitude 13. Charge Q is divided into two parts which are then kept some distanced apart. The force between them will be maximum if the two parts are (a) Q /2 each (b) Q /4 and 3Q /4 (c) Q /3 and 2Q /3   (d) e and (Q-e), where e = electronic charge 14. Two charges q1 and q2 separated by a dielectric of dielectric constant 4 repel each other with a force of 10 N. Another

72  Electrostatics: Part 1

charge q3 is placed between q1 and q2 such that the distance of q3 from q1 is 1/4 times the distance of q3 from q2. Now, the force of repulsion between q1 and q2 is (a) 10 N (b) 10 q1 (c)

4 q2

q1

¥

10  N 

(d)

10 q1 q2 Æ

An electron moves with velocity v in  x -direction. An electric field acts on it in  y-direction. The force on the electron acts in   (a) +ve direction of Y -axis (b) –ve direction of Y -axis (c) +ve direction of Z-axis (d) –ve direction of Z-axis 16. An uncharged metal object  M   is insulated from its surroundings. A positively charged metal sphere S  is then brought near to  M . Which diagram best illustrates the resultant distributions of charge on S  and M  ?   15.

17.

(a)

(b)

(c)

(d)

A thin metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another charge q1 is placed outside it as shown in the figure. All the three charges are positive. The force on the charge at the centre is

the system can never be in equilibrium the system will be in equilibrium if the charges rotate about the centre of the triangle (c) the system will be in equilibrium if the charges have different magnitudes and different signs   (d) the system will be in equilibrium if the charges have the same magnitude but different signs 20. A point charge A of charge +4 mC and another point charge  B of charge –1 mC are placed in air at a distance 1 metre apart. Then the distance of the point on the line joining the charges and from the charge B, where the resultant electric field is zero, is (in metre) (a) 1.5 (b) 0.5   (c) 1 (d) 2   21. Two equal negative charges (– q  each) are fixed at the points [0, a] and [0, –a] on the  y-axis. A positive charge Q is released from rest at the point [2a, 0] on the  x -axis. The charge Q will (a) execute simple harmonic motion about the origin (b) move to the origin and remain at rest (c) move to infinity (d) execute oscillatory but not simple harmonic motion   22. Three charges 4 q, Q and q are placed in a straight line of length l at point 0, l/2 and l respectively. What should be Q in order in make the net force on q to be zero? (a) (b)

 

(a)

−q

towards left (b) towards right upward (d) zero 18. Three small spheres, each carrying a charge q are placed on the circumference of a circle of radius  R, forming an equilateral triangle. If we place another charge Q at the centre of the circle, then the force on Q will be (a)

    19.

(c)

zero

(b)

1 4pe 0

¥

2qQ

 R

2

(d)

1 4pe 0

¥

1 4pe 0

¥

2

3qQ

 R

2

Three point charges are placed at the corners of an equilateral triangle. Assuming only electrostatic forces are acting:

2

q

(c)

q

(b) (n - 1) k 

2

r 

n

k 

q

n -1

k 

r 2 q

(c)

  24.

A charge Q is placed at each of two opposite corners of a square. A charge q is laced at each of the two opposite corners of the square. If the resultant electric field on Q is zero, then

2

n - 1 r 

(a) Q

q = -

2

2

(d)

q

 

qQ  R

 1

−2q (d) 4q   23. In a regular polygon of n sides, each corner is at a distance r   from the centre. Identical charges of magnitude q are placed at (n – 1) corners. The field at the centre is  

(a) k 

(a) (c)

(b) -

n

(b) Q

= -

(d) Q

=

2

r 

2

2

q

 

(c) Q = –2q

  25.

The charge per unit length of the four quadrant of the ring is 2l , –2l , l  and –l  respectively. The electric field at the centre is

 2

2

q

 Electrostatics: Part 1

 

(a) -

 

(c)

  26.

l  2pe 0 R

2 l  ˆ i 4pe 0 R

iˆ

(b)

(d)

l  2pe 0 R

 

(a) T = 2 p  r 

 

(c) T 

m 2 k l q

 jˆ

None

Two identical positive charges are fixed on the y-axis. At equal distance from the origin O. A negatively charged particle starts on the  x -axis at a large distance from O, moves along the x -axis, passes through O and moves for away from O. Its acceleration a is taken as positive along its direction of motion. The particle’s acceleration a is plotted against its  x -coordinate. Which of the following best represents the plot?

where

2p r

k  =

(d) T 

m

=

4p 

2

m

2 k l q 1

m

2p r

2 k l  q

1 4p 

Π0

ARCHIVES 1.

 

A total charge Q is broken in two parts Q1 and Q1 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when [EAMCET 2011] Q

(a) Q2

=

(b) Q2

=

 

(c) Q2

=

 

(d) Q1

(a)

(b)

2 k l  q

1 =

(b) T 2 =

73

 R Q 4

Q 4

Q =

2

Q

, Q1

=

Q

-

, Q1

=

Q

-

, Q1

=

, Q2

=

R 2Q 3

3Q 4

Q 2

An isolated solid metallic sphere is given +Q charge. The charge will be distributed on the sphere [MH CET 2011] (a) Uniformly but only on surface (b) Only on surface but non-uniformly (c) Uniformly inside the volume (d) Non-uniformly inside the volume 3. Two small spheres each having the charge + Q   are suspended by insulating threads of length  L  from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between the two suspensions and the tension in each will be 2.

(c)

       

(d)

[KCET 2011]

27.

A particle of charge – q and mass m moves in a circle of radius r  around an infinitely long line of charge of linear charge density +l . Then time period will be:

 

(a) 180 ,

 

(c) 180 ,

∞

∞

4.

1

Q

2

4pe 0 (2 L )2 1

Q

(b) 90 , ∞

1

(d) 180 , ∞

2

4pe 0  L2

2

4pe 0 2 L2

Q

1

Q

2

4pe 0  L2

Three equal charges are placed on the three corners of a square. If the force between q1  and q2 is F 12 and that

74  Electrostatics: Part 1

between q1 and q3 is F 13, the ratio of magnitudes

F 12 F 13

 is

[MP PET 2011]  

(a)

1/2

 

(c)

1/

(b) (d)

2

2 2

Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4) and (4, 1) show electrostatic attraction, while pair (2, 3) and (4, 5) show repulsion. Therefore ball 1 must be [UP SEAT 2012]   (a) Positively charged   (b) Negatively charged (c) Neutral   (d) Made of metal 6. Equal charges q are placed at the four corners A,B,C,D of a square of length a. The magnitude of the force on the charge at B will be [EAMCET 2012] 5.

 

(a)

3q

2

4pe 0 a

 

(c)

Ê1 + 2 Á 2 Ë

(b)

2

2

ˆ  q ˜  4pe  a ¯  2

0

2

(d)

4q

2 2

ˆ  q ˜  2 ¯  4pe  a

1

2

2

0

Two identical conductors of copper and aluminum are placed in an identical electric fields. The magnitude of induced charge in the aluminum will be [MP CET 2012] (a) Zero (b) Greater than in copper (c) Equal to that in copper (d) Less than in copper 8. When a body is earth connected, electrons from the earth flow into the body. This means the body is….. 7.

       

[KCET 2012]

Unchanged (b) Charged positively Charged negatively (d) An insulator 9. The charges on two sphere are +7m C and – 5m C respectively. They experience a force F . If each of them is given and additional charge of – 2m C,  the new force of attraction will be [MH CET 2013]   (a) F  (b) F  / 2   (c) F  / 3 (d) 2F     

(a) (c)

10.

 

[UPSEAT 2014]    

(a) F   /2 (c) F 

 /4 (b) 3F   /4 (d) F 

Two charges of equal magnitudes and at a distance r  exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is [MH CET 2014]   (a) F / 8 (b) F  / 4   (c) 4 F  (d) F  / 16   14. An infinite number of charges, each of charge 1 mC, are placed on the x -axis with co-ordinates x  = 1, 2, 4, 8, ....• . If a charge of 1 C is kept at the origin, then what is the net force acting on 1 C charge? [EAMCET 2014]   (a) 9000 N (b) 12000 N (c) 24000 N (d) 36000 N 15. The number of electrons in 1.6 C charge will be   13.

4pe 0 a

Ê ÁË 2 +

Two point charges 3 × 10 –6 C and 8 × 10 –6 C repel each other by a force of 6 × 10–3 N. If each of them is given an additional charge –6 × 106 C, the force between them will be [KCET 2013] –3   (a) 2.4 × 10  N (attractive)   (b) 2.4 × 10–9 N (attractive) (c) 1.5 × 10–3 N (repulsive)   (d) 1.5 × 10–3 N (attractive)   12. Two equally charged, identical metal spheres  A  and  B repel each other with a force ‘ F ’. The spheres are kept fixed with a distance ‘r ’ between them. A third identical, but uncharged sphere C   is brought in contact with A and then placed at the mid-point of the line joining A and  B. The magnitude of the net electric force on C  is 11.

The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance 5 × 10–11m, will be (Charge on electron = 1.6 × 10–19 C, mass of electron = 9.1 × 10–31 kg, mass of proton = 1.6 × 10–27 kg, G = 6.7 × 10–11 Nm2 /kg2) [EAMCET 2013] 39 (a) 2.36 × 10 (b) 2.36 × 1040 41 (c) 2.34 × 10 (d) 2.34 × 1042

[MPPET 2014] 19

20

10 (b) 10 19 2 1.1 × 10 (d) 1.1 × 10   16. Four metal conductors having different shapes    

(a) (c)

[Kerala 2015]

A sphere 2. Cylindrical Pear 3. Lightning conductor are mounted on insulating stands and charged. The one which is best suited to retain the charges for a longer time is [KCET 2014]   (a) 1 (b) 2   (c) 3 (d) 4   17. Identify the wrong statement in the following. Coulomb’s law correctly describes the electric force that 1. 3.

[WB JEE 2015]    

(a) (b)

Binds the electrons of an atom to its nucleus Binds the protons and neutrons in the nucleus of an atom

 Electrostatics: Part 1

Binds atoms together to form molecules Binds atoms and molecules together to form solids   18. A cylinder of radius R and length L is placed in a uniform electric field E  parallel to the cylinder axis. The total flux for the surface of the cylinder is given by EAMCET 2015] 2 2  R  E   R  /   E    (a) 2p  (b) p  2  R  – p   R)/   E    (c) (p  (d) Zero   19. Total electric flux coming out of a unit positive charge put in air is [KCET 2015]    

   

(c) (d)

  24.

Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is [MP PET 2016]

 

(a)

(b) e 0–1 (d) 4pe 0 20. Eight dipoles of charges of magnitude e are placed inside (a) e 0 (c) (4pe 0)–1

a cube. The total electric flux coming out of the cube will be [MP PET 2015]  

(a)

8e

(b)

e 

e

(c)

16e

(d)

Zero

0

A point charge +q is placed at the centre of a cube of side  L. The electric flux emerging from the cube is

 

(a)

q

(b)

e 

(c)

0

10Q

100Q

(d)

(pe 0 )

(pe 0 )

  25. q1, q2, q3  and q4  are

point charges located at points as shown in the figure and S  is a spherical Gaussian surface of radius  R. Which of the following is true according to the Gauss’s law? [KCET 2016]

[UPSEAT 2016]  

100Q

(b)

0

e 

  21.

Q e 

e 0

0

Zero

e 

0 2

 

(c)

6qL

(d)

e 0

  22.

 

 

q 2

6 L

e 0

A charge q  is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is [EAMCET 2016] Zero 

(a)

q

(c)

2e 0

 

(b)

(d)

e 0

0

 

(a) r 

(b)

1 2

r 

1 3 r 

(d)

1 r 







 

(b)  ( E1 + E2 + E3 ). dA = s

 

e 



(a)  ( E1 + E2 + E3 ). dA = s

Ú 

2q

According to Gauss’ theorem, electric field of an infinitely long straight wire is proportional to [MH CET 2016]

(c)

Ú 

 

q

  23.

 

75

 

Ú 

















(c)  ( E1 + E2 + E3 ). dA = s (d)

None of the above

q1 + q2 + q3 2e 0 (q1

+ q2 + q3 ) e 

0

(q1

+ q2 + q3 + q4 ) e 0

76  Electrostatics: Part 1

JEE EXERCISES  YPE SINGLE CORRECT  ANSWER T 

5.

Level 1 1.  A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. A and B are far away from the dipole and at equal distances from it. The fields at A and B are Æ

 

(a)

 E A

(b)

 E A

(c)

 

(d)

 E A

=

2.

.

not equal to the total flux through S  due to charges q3 and q4 (b) equal to the total flux through S  due to charges q3 and (a)

 

q4

2 E B

   

Æ

= - 2 E 

B

Æ

 E B

 E  B

Æ

Æ

 

 and

E B

Æ

 

Æ

Æ

 E  A

Æ

=

=

1 2

Æ

E A

Æ

, and

 E  B

Æ

 is perpendicular to  E  A

In a gravity-free space, uniform electric field  E  has been applied in the vertical direction as shown. A particle of mass m and charge q is projected horizontally with velocity V . The time after which the velocity vector makes an angle 45° with the initial velocity vector, is equal to

zero if q1 + q2 = q3 + q4 twice the total flux through S   due to charges q3 and q4 if q1 + q2 = q3 + q4 6. Three charges of q1 = 1 × 10–6 C, q2 = 2 × 10–6 C and q3 = –3 × 10–6 C have been placed as shown. Then the net electric flux will be maximum for the surface (c) (d)

   

(a) S 1 (c) S 3 7.

 / qE  (a) mV   /2qE  (c) 3mV 

    3.

The figure shows four charges q1, q2, q3, q4 fixed in space. Then the total flux of electric field through a closed surface S , due to all charges q1, q2, q3 and q4 is

 / qE  (b) 2mV   /4qE  (d) 3mV 

A cylinder of length L and radius b had its axis coincident

(b) S 2 (d) same for all three

Two charges q  and –4q  are held at a separation r   on a frictionless surface. Another charge is kept in such a way that they do not move if released. The value of the third charge and its position from –4q is

Æ

with the x-axis. The electric field in this region is  E = 200 iˆ . Find the flux through the left end of cylinder.   (a) 0 (b) 200p b2   (c) 100p b2 (d) –200p b2 4. Charge on an originally uncharged conductor is separated by holding a positively charged rod very closely nearby, as in figure. Assume that the induced negative charge on the conductor is equal to the positive charge q on the rod. Then, flux through surface S 1 is

   

(a) (c)

zero –q / e 0

(b) q / e0  (d) None of these

(a) –2q, 2r  (c) q, r 

    8.

(b) –4q, 2r  (d) 4q, 2r 

An electron of mass me, initially at rest, moves through a certain distance in a uniform electric field in time t 1. A proton of mass mp, also, initially at rest, takes time t 2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t 2 / t1  is nearly equal to   (a) 1 (b) (mp / me)1/2   (c) (me / mp)1/2 (d) 1836 9. Four point charges are placed at the corners of a square with diagonal 2a as shown. What is the total electric field at the centre of the square?

 Electrostatics: Part 1

centre of triangle. The ratio in equilibrium is:

q Q

77

 so as to make the system

2

(a) kq / a  at an angle 45° above the + x -axis (b) kq / a2 at an angle 45° below the – x -axis 2 (c) 3kq / a  at an angle 45° above the – x -axis 2 (d) 3kq / a  at an angle 45° below the + x -axis

        10.

Four electrical charges are arranged on the corners of a 10 cm square as shown. What would be the direction of the resulting electric field at the centre point P?

→ (b) ↑ ←   (d) ↓   11. Two pith balls each with mass m  are suspended from insulating threads. (a) (c)

 

(a)

 

(c)

1: 3 3 :1

(b)

1:

3

(d)

2:

3

The maximum electric field at a point on the axis of a uniformly charged ring is E 0. At how many points on the axis will the magnitude of electric field be  E 0 /2   (a) 1 (b) 2 (c) 3 (d) 4   14. An electric charge q exerts a force F  on a similar electric charge q separated by a distance r . A third charge q /4 is placed mid-way between the two charges. Now, th e force F  will   13.

F 

F 

 

(a)

become

 

(c)

become

  15.

The electric field intensity at the centre of a uniformly charged hemispherical shell is E 0. Now two portions of the hemisphere are cut from either side and remaining portion is shown in figure.

When the pith balls are given equal positive charge Q, they hang in equilibrium as shown. We now increase the charge on the left pith ball from Q to 2 Q while leaving its mass essentially unchanged. Which of the following diagrams best represents the new equilibrium configuration?

If a

=

b 

p  =

3

3 F 

27

(b)

become

(d)

remain F 

9

, then electric field intensity at centre due to

remaining portion is

(a)

(c)

12.

 

(a)

 

(b)

 

(c)

(b)

(d)

Three charges each Q are placed at the three corners of an equilateral triangle. A fourth charge q is placed at the

 E 0

3  E 0

6  E 0

2

Information insufficient   16. A block of mass m  is suspended in vertical orientation with a spring of spring constant k . The block is made to oscillate in gravitation field. Its time period is found to be  

(d)

78  Electrostatics: Part 1

T . Now the space between the plates is made gravity free and an electric field  E   is produced in vertical downward direction. Now the block is given charge q. The new time

 

period of oscillation is

(a) (c) -

PkQ r 

cosq  iˆ

2 PkQ 3

r 

(b)

cos q  iˆ

(d)

PkQ r 

cos q  iˆ

2 PkQ 3

r 

cos q  iˆ

Level 2   20. Two charges q1 and q2 are kept on x -axis and electric field at different points an  x -axis is plotted against  x . Choose correct statement about nature and magnitude of q1 and q2. (b) T  + 2p 

(a) T   

(c) 17.

2p 

qE  md 

(d)

qE  md 

none of the above

The figure below shows the forces that three charged particles exert on each other. Which of the four situations shown can be correct?

(I)

(II)

(III) (a) (c)

(IV) (b) (d)

all of the above none of the above II, III II, III and IV   18. An electroscope is given a positive charge, causing its foil leaves to separate. When an object is brought near the top plate of the electroscope, the foils separate even further. We conclude

(a) (b) (c) (d)

  21.

Three identical point charges, each of mass m and charge q, hang from three strings as shown in figure. The value of q in terms of m, L and q.

 

(a) q

=

(16/5) pe 0 mg L sin q tan q

 

(b) q

=

2 2 (16/15) pe 0 mg L sin q tan q

 

 

(c) q

=

(15/16) pe 0 mg L2 sin 2 q tan q

 

that the object is positively charged that the object is electrically neutral that the object is negatively charged none of these   19. A point negative charge – Q is placed at a distance r  from a dipole with dipole moment P in  x − y plane as shown in figure. The x  component of force acting on the charge – Q is (a) (b) (c) (d)

2

2

 

none of these   22. A metallic shell has a point charge ‘q’ kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces?  

 

q1 +ve, q2 –ve ; |q1| > |q2| q1 +ve, q2 –ve ; |q1| < |q2| q1 –ve, q2 +ve ; |q1| > |q2| q1 –ve, q2 +ve ; |q1| < |q2|

       

(d)

(a)

(b)

(c)

(d)

 Electrostatics: Part 1



  23.

The direction (q ) of  E   at point P due to uniformly charged finite rod will be

(a)

 

(c)

0

(b)

s l l e0

2

sin q

(d)

s l l 2e0

79

2

sin q

s l l 2e0

A sphere of radius  R carries charge such that its volume charge density is proportional to the square of the distance from the centre. What is the ratio of the magnitude of the electric field at a distance 2 R from the centre to the magnitude of the electric field at a distance of  R /2 from the centre? (a) 1 (b) 2   (c) 4 (d) 8   27. An uncharged conducing large plate is placed as shown. Now an electric field  E  towards right is applied. Find the induced charge density on right surface of the plate.   26.

at angle 30° from x -axis 45° from x -axis 60° from x -axis none of these     24. Find the force experienced by the semicircular rod charged with a charge q, placed as shown in figure. Radius of the wire is R and the line of charge with linear charge density λ is passing through its centre and perpendicular to the plane of wire.    

(a) (b) (c) (d)

l  q

(a)

 

(c)

  25.

A large sheet carries uniform surface charge density s . A rod of length 2l has a linear charge density l  on one half and –l  on the second half. The rod is hinged at mid-point O and makes an angle q  with the normal to the sheet. The torque experienced by the rod is

2p

2

e 0  R

l q 4p

2

e 0  R

(b)

l  q

 

(d)

2

p e 0  R l  q 4 p e 0  R

– e 0 E  (b) e 0 E  – 2e 0 E  (d) 2e 0 E    28. An uncharged aluminium block has a cavity with in it. The block is placed in a region where a uniform electric field which is directed upwards. Which of the following is a correct statement describing conditions in the interior of the block’s cavity? (a) The electric field in the cavity is directed upwards   (b) The electric field in the cavity is directed downwards   (c) There is no electric field in the cavity   (d) The electric field in the cavity is of varying magnitude and is zero at the exact centre   29. The diagram shows a uniformly charged hemisphere of radius R. It has volume charge density r . If the electric field at a point 2 R distance above its centre is E  then what is the electric field at the point which is 2 R below its centre?    

(a) (c)

 

 R /6e 0 + E  (a) r   R /6e 0 + E  (c) –r 

  30.

Consider an infinite line charge having uniform linear charge density and passing through the axis of a cylinder.

(b) ρ R /12e 0 – E   R /12e 0 + E  (d) r 

80  Electrostatics: Part 1

What will be the effect on the flux passing through curved surface if the portions of the line charge outside the cylinder is removed?

Decreases (b) Increases Remains same (d) Can’t say   31. One fourth of a sphere of radius  R is removed as shown. An electric field  E   exists parallel to  x − y plane. Find the flux through remaining curved part.    

 

(a) (c)

(a)

p  R

2

(b)

E 

2

2p  R E 

None of these   32. A non-conducting sphere of radius R is filled with uniform volume charge density –r . The centre of this sphere is displaced from the origin by d  . The electric field  E   at any point P having position vector, inside the sphere is  

(c)

p  R

2

E  /

(a)

 

(c)

  33.

r 

(b)

3 Œ0

r 

3 Œ0





d 



(d



-

r )

ˆ 

1

0

–2 2 3.6 p  × 10–2 m2 (b) 0.9 p × 10  m 1.8 p  × 10–2 m2 (d) none   34. A positively charged sphere of radius r 0 carries a volume charge density r  (figure). A spherical cavity of radius r 0 /2 is then scooped out and left empty, as shown. C 1 is the centre of sphere and C 2 that of cavity. What is the direction and magnitude of the electric field at point  B?

   

(a) (c)

 

(a)

17r r 0 54

Π0

 left

(b)

r r 0 6

 left

Π0

(d)

2



 

Ê

Use Á = 9 × 109 Nm2 /C2 ˜  4 pe  Ë ¯ 

(d)

r 

3 Œ0 r 

3 Œ0

 

(r

-

d )



(r  )

A charged large metal sheet is placed into un iform electric field, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is E 1 = 5 × 105 V/m and on the right it is  E 2 = 3 × 105 V/m. The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume external field to remain constant after introducing the large sheet.

    35.

(c)

17r r 0 54

Π0

 right

(d)

r r 0 6 Œ0

 right

Four very large metal plates are given the charges as shown in figure. The middle two are then connected through a wire. Find the charge that will flow through the wire.

5Q from A to B (b) 5Q /2 from A to B 5Q from B to A (d) no charge will flow   36. A conic surface is placed in a uniform electric field  E  as shown such that field is perpendicular to the surface on the side AB. The base of the cone is of radius  R and height of the cone is h.    

(a) (c)

 Electrostatics: Part 1

The angle of cone is q   as shown. Find the magnitude of that flux which enters the cone curved surface from left side. Don’t count the outgoing flux. (q  < 45°).   (a)  ER[h cos q  + p ( R /2) sinq ]   (b)  ER[h sinq  + p ( R /2) cosq ]  R sinq ]   (c)  ER[h cos q  + p    (d) none of these   37. Flux passing through shaded surface of sphere when a point charge q is placed at the centre is (radius of the sphere is  R)

   

(a) q / e0  (c) q /4e 0

  38.

A uniformly charged and infinitely long line having a linear charge density ‘l ’is placed at a normal distance y from a point O. Consider a sphere of radius  R with O as centre and R > y. Electric flux through the surface of the sphere is

 

 

(a)

 

(b)

s

 

 E1 = E 2 =

 

2 2 + s 2 1

2e 0 s

 

 E 2 = E 4 =

 

π E 4

2 2 + s  1 2

2e 0 2

 

   

(c)

 

81

 

 

 E1 = E 2 = E3 = E 4 =

2

s 1 + s 2

2e 0

None of the above 40. Three large identical conducting parallel plates carrying charge +Q, -Q and +2Q respectively are placed as shown in the figure. If E A, E B and E C refer the magnitude of electric field at points A, B and C respectively then: (d)

(b) q /2e 0 (d) zero  

(a)  E A > E B > E C (c)  E A = 0 and E B > E C

(b)  E A = E B > E C (d)  E A = 0 and E B = E C

  41.

The number of electric field lines crossing an area DS    is n1 when D S  E  while number of field lines crossing same area is n2 when D E   makes an angle of 30º with  E  then: (a) n1 = n2 (b) n1 > n2  < (c) n1 n2 (d) cannot say anything ,

 

  

,

   

MULTIPLE CORRECT  ANSWERS T  YPE zero

 

(a)

 

(c)

  39.

Two infinite sheets having charge densities s 1and s 2 are placed in two perpendicular planes whose two-dimensional view is shown in the figure. The charges are distributed uniformly on sheets in electrostatic equilibrium condition. Four points are marked I, II, III and IV. The electric field intensities at these points are  E1 , E 2, E 3 and  E 4 respectively. The correct expression for electric field intensities is

2l 

(b) 2

R -y e 0

2

(d)

e 0 l   R

2

+

y

2

e 0

 

When an electron moves in a circular path around a stationary nucleus charge at the centre:   (a) the acceleration of the electron changes   (b) the velocity of the electron changes   (c) electric field due to the nucleus at the electron changes   (d) none of these 2. Two point charges (Q  each) are placed at (0,  y) and (0, – y). A point charge q of the same polarity can move along x -axis. Then: (a) the force on q is maximum at x  = + y /  2   (b) the charge q is in equilibrium at the origin   (c) the charge q performs an oscillatory motion about the origin   (d) for any position of q  other than origin the force is directed away from origin 3. A conducting ball is positively charged and another positive point charge is brought closer to the ball. (a) the ball may attract the point charge   (b) the ball may repel the point charge 1.

2l R

 

 

 

82  Electrostatics: Part 1

   

there may be no force between them the ball will only repel the point charge and in no condition it can attract the point charge 4. We have two electric dipoles. Each dipole consists of two equal and opposite point charges at the ends of an insulating rod of length d . The dipoles sit along the x -axis a distance r  apart, oriented as shown below. Their separation r  >> d . The dipole on the left: (c) (d)

will feel a force to the left will feel a force to the right will feel a torque trying to make it rotate counterclockwise   (d) will feel no torque 5. Imagine a short dipole is at the centre of a spherical surface. If magnitude of electric field at a certain point on the surface of sphere is 10 N/C, then which of the following cannot be the magnitude of electric field anywhere on the surface of sphere   (a) 4 N/C (b) 8 N/C (c 16 N/C (d) 32 N/C 6. Consider a Gaussian spherical surface, covering a dipole of charge q and –q, then      

(a) (b) (c)

(a) q in   =

0 (Net charge enclosed by the spherical surface)   (b) f net = 0 (Net flux coming out the spherical surface)   (c)  E  = 0 at all points on the spherical surface   (d)  E · d s  = 0   (Surface integral of over the spherical Ú  surface) 7. Two large thin conducting plates with small gap in between are placed in a uniform electric field ‘ E ’ (perpendicular to the plates). Area of each plate is A and charges +Q and –Q are given to these plates as shown in the figure. If points  R, S  and T  as shown in the figure are three points in space, then the

(c)

Ê

ˆ 

field at point T  is Á E  + Q ˜  Œ  A ¯  Ë 0

 

(d)

Ê

ˆ 

Field at point S  is Á E  + Q ˜   A Œ ¯  Ë 0

Charges Q1 and Q2 lie inside and outside respectively of a closed surface S . Let E  be the field at any point on S  and f  be the flux of E  over S .   (a) If Q1 changes, both E  and f  will change   (b) If Q2 changes, E  will change but f  will not change   (c) If Q1 = 0 and Q2 ≠ 0 then E ≠ 0 but f  = 0 (d) If Q1 ≠ 0 and Q2 = 0 then E  = 0 but f ≠ 0 –6 9. The figure shows a point charge of 0.5 × 10  C at the centre of a spherical cavity of radius 3 cm of piece of metal. Th e electric field at 8.

A (2 cm from the charge) is 0 A (2 cm from the charge) is 1.125 × 10 7 N/C B (5 cm from the charge) is 0   B (5 cm from the charge) is 1.8 × 10 6 N/C     10. A right circular imaginary cone is shown in the adjacent figure.  A,  B and C   are the points in the plane containing the base of the cone, while  D is the point at the vertex of the cone. If f  A, f  B, f C  and f  D represent the flux through curved surface of the cone when a point charge Q is at point A, B, C  and D, respectively, then (a) (b) (c) (d)





 

(a) f A

 

(c) f  B

=

f C 

π 0

Q =

2e 0

(b) f  D (d) f A

π 0

=

fC

=

f D

=

0

LINKED COMPREHENSION T  YPE

(a) (b)

field at point R is E  field at point S  is E 

For Problems (1–2) 1. A simple pendulum of mass m  charged negatively to q coulomb oscillates with a time period T in a downward electric field  E   such that mg > qE . If the electric field is withdrawn, the new time period

 Electrostatics: Part 1

83

For Problems (6–7) Positive and negative charges of equal magnitude lie along the symmetry axis of a cylinder. The distance from the positive charge to the left end-cap of the cylinder is the same as the distance from the negative charge to the right end-cap.        

= T  > T  < T  any of the above three is possible 2. At equilibrium of the bob the change in tension in string will be (assuming rest condition) (a) (b) (c) (d)

   

(a) mg (c) 2qE 

(b) qE   /2 (d) qE 

For Problems (3–5) There is an insulator rod of length L and of negligible mass with two small balls of mass m and electric charge Q attached to its ends. The rod can rotate in the horizontal plane around a vertical axis crossing it at an L /4 distance from one of its ends. 3. At first the rod is in unstable equilibrium in a horizontal uniform electric field of field strength E . Then we gently displace it from this position. Determine the maximum velocity attained by the ball which is closer to the axis in the subsequent motion. (a)

 

(c) 4.

2QEL

m QEL 5m

(b)

(d)

What is the flux of the electric field through the closed cylinder? (a) 0 (b) + Q / e 0 (c) + 2Q / e 0 (d) – Q / e 0 7. What is the sign of the flux through the right end-cap of the cylinder?   (a) Positive   (b) Negative   (c) There is no flux through the right end-cap. 6.

For Problems (8–9) There are two non-conducting spheres having uniform volume charge densities r   and –r . Both spheres have equal radius  R. The spheres are now laid down such that they overlap as shown Æ

in the figure. Take

4QEL 5m

 

The electric field  E   in the overlap region is (a) non-uniform (b) zero

 

(c)

8.

9.



 

(d)

(c)

What is the time period of the SHM as mentioned in above question? (a)

2p 

(c)

2p 

mL QE  5mL 8QE 

(b)

2p 

(d)

2p 

2 mL 3QE 

5mL 4QE 

r 

Æ

d 

3 Œ0

(d)

Æ

- r 

d 

3 Œ0

The potential difference DV  between the centres of the two spheres for d  = R is (a)

5.

O1O2

5m

(b)

(c)

Æ

=

2QEL

In what position is the rod to be set so that if displaced a little from that position it begins a harmonic oscillation about the axis A?

(a)

d

r  3 Œ0

zero

2

d 

(b)

r 

2

d 

Π0

(d)

2 r 

2

d 

Π0

For Problems (10–13) Gauss’s law and coulomb’s law expressed in different forms although are equivalent ways of describing th e relation between charge and electric field in static conditions. Gauss’s law is e 0f  = qencl in which qencl is the net charge inside an imaginary closed    surface called Gaussian surface f  = Ú  E.d   a gives electric flux

84  Electrostatics: Part 1

of through Gaussian surface. The two equations hold only when the net charge is in vacuum or air. 10. A Gaussian surface encloses two of the 4 positively charged particles. The particles which contribute to the electric field at point p on the surface are

(a) q1 and q2 (c) q4 and q3

(b) (d)

Charge q3 applies equal force on both the charges Charge q3 applies no force on any of the charges   15. If q1 is displaced from its centre slightly (being always inside the cavity) then the correct representation of field lines inside the same cavity is:    

(c) (d)

(a) (c)

q2 and q3 q1, q2, q3 and q4

The net flux of the electric field through the surface is   (a) due to q1 and q2 only   (b) due to q3 and q4 only   (c) equal due to all the four charges (d) cannot say   12. The net flux of the electric field through the surface due to q3 and q4 is (a) zero (b) positive (c) negative (d) can’t say 13. If the charge q3 and q4 are displaced (always remaining outside the Gaussian surface), then consider the following two statements A: Electric field at each point on Gaussian surface will remain same.

(b)

Then will be no field lines inside cavity

  11.

  

     

B: The value of Ú  E. d  A  for the Gaussian surface will remain same. (a) Both A and B are true (b) Both A and B are false (c) A is true but B is false (d) B is true but A is false

For Problems (14–16) A spherical conductor A contains two spherical cavities as shown in figure. The total charge on conductor itself is zero. However, there is a point charge q1 at centre of one cavity and q2 at the centre of other cavity. Another charge q3  is placed at large distance ‘r ’ from the centre of the spherical conductor.

 

(d) 16.

The force acting on conductor A will be

 

(a)

 

(c)

  17.

   

3 ( q1 + q2 ) 2 4pe 0 r 

-q

(d)

q3 ( q1 + q2 ) 2

4pe 0 r 

q3 q1 + q2 q3 + q1 q2 2

4pe 0 r 

Find the tensions in the string AB. (a)

zero

(b) q

4pe 0

d 

(c)

  18.

Find the tensions in the string BC .

(c)

2

3

q

4pe 0

d 2

2

9

 

 

Which of the following statements are true? (a) Charge q3 applies larger force on charge q2 than on charge q1 (b) Charge q3 applies smaller force on charge q2 than on charge q1

(b)

For Problems (17–20) Three particles  A,  B, C   having charges +q, +2 q  and +4 q are connected by strings as shown in the figure. Assuming equilibrium of the particle.

(a)

  14.

zero

2

zero

(d)

(b)

none of these

2

3

q

4pe 0

d 2

2

9

q

4pe 0

d 

2

(d)

none of these

  19.

If the particle C  is discharged then tension in string BC 

 

(a)

 

(c)

zero

(b)

3

q2

4pe 0

d 2

2

9

q

4pe 0

d 2

(d)

none of these

 Electrostatics: Part 1

  20.

If the particle  B is discharged then the tension in string

(s)

 AB

(a)  

(b)

 

(c)

 

(d)

becomes zero 2

5

q

4pe 0

d 

2

3.

2

7

q

4 p e0

d 

2

If position of Q2 is changed outside keeping Q1 fixed inside at any point

(d)

85

Electric field outside changes

Axis of a hollow cone shown in figure is vertical. Its base radius is R. It is kept in a uniform electric field  E  parallel to its axis.

is equal to the tension in BC .

 YPE MATCHING COLUMN T  1.

Four large parallel identical conducting plates of area A are arranged as shown. The charges on each plate are given in diagram and separation between plates is d  ( d   is very small). Surface of plates are numbered (1), (2), ……… (8). Column I (p) (q)

(r)

Column I (p) (q) (r) (s)

2.

Column II

Surfaces having charges of same magnitude and sign Surfaces having positive charges Uncharged surfaces Charged surfaces

(a)

1 and 8

(b)

3 and 5

(c)

2 and 3 6 and 7

(d)

(s)

4.

Magnitude of flux through base of cone Magnitude of flux through curved part of cone Magnitude of flux through curved part  MN QP of cone Net flux through entire cone

Column II (a)

p  R

(b)

2

p  R

E 

2

E 

2

(c)

zero

(d)

non-zero

Consider three point charges –2Q, +Q, –Q and four closed surfaces S 1, S 2, S 3 and S 4.

Inside a neutral metallic spherical shell a charge Q1 is placed and outside the shell a charge Q2 is placed

Column I (p) (q)

(r)

If Q1 is at the centre of shell If Q1 is not at the centre of shell If position of Q1 is changed within the shell keeping Q2 fixed

Column I

Column II (a) (b)

(c)

Electric field inside the shell remains zero Electric field inside the shell remains non-zero Electric field inside changes

(p)

Net flux through S 1

Column II (a)

Q

(q)

Net flux through S 2

(b)

2Q

(r) (s)

Net flux through S 3 Net flux through S 4

(c) (d)

e  0

e  0

zero non-zero

86  Electrostatics: Part 1

5.

Following figure shows for Gaussian surface S 1, S 2, S 3 and S 4.

(ii)

(iii)

(iv)

Electric field strength is non-zero Magnitude of charge is 3Q

(q)

Charge is + 8Q

(s)

(r)

On inner surface of the larger spherical shell On outer surface of the smaller spherical shell For a < r  < b

Now match the given columns and select the correct option from the codes given below. Codes

Now, match the following columns. Column I (i)

(a)

Column II (p)

f S 

1

+

(b)

Q

(c)

e  0

(ii)

f S 

(q)

f S 

(r)

2

(iii)

3

(iv)

(s)

3

1.

e  0

2Q

-

e  0

Codes

(a) (b) (c) (d)

6.

ii.

iii.

iv.

p s q, r p, q, r

q p, q, r p s

s q, r p, q, r r

p, q, r p s p

INTEGER ANSWER T   YPE

Q

-

f S 

(d)

0

i.

i.

ii.

iii.

iv.

s q q r

r s r s

q r s p

p p p q

In the figure shown a conducting spherical shell of inner radius  x   and outer radius  y  is concentric with a larger conducting spherical shell of inner radius a and outer radius b. The inner shell has a total charge +3Q and the outer shell has a total charge +5Q. Let r  be the distance of any point from the common centre O. Match column I and column II.

Four charge particles each having charge Q = 1 C are fixed at corners of base (at  A, B, C  and D) of a square pyramid with slant length ‘a’ ( AP = BP = BP = PC  = a = 2 m ), a charge -Q is fixed at point P. A dipole with dipole moment  p = 1 C-m is placed at centre of base and perpendi cular to its plane as shown in figure. If the force on dipole due to charge particles is Ω . Find the value of W.  N 

4pe 0 P 

D

C 

2

a O

 A

2.

a

B

A ring of radius  R, has charge –Q distributed uniformly over it. Calculate the charge (q) that should be placed at the centre of the ring such that the electric field becomes zero at a point on the axis of the ring distant ‘ R’ from the centre of the ring. If the value of q of W.

3.

Column I (i)

Electric field strength is zero

Column II (p)

On outer surface of the larger spherical shell

a

Q =

4

Ω . Find the value

Two identical small equally charged conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is a (the length of thread L >> a). One of the balls is then discharged. Again for the certain value of distance b (b  E 1 > E 2 (d)  E 3 > E 2 > E 1 (JEE Advanced 2014)

The figure below depict two situations in which two infinitely long static line charges of constant positive line charge density l   are kept parallel to each other. In their resulting electric field, point charges q and –q are kept in equilibrium between them. The point charges are confined to move in the  x -direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is (are):

→

16.

Consider an electric field  E = E  ˆx , where E 0 is a constant. The flux through the shaded area (as shown in the figure) due to this field is 0



(,a, a, a)

(,a, 0, a)

Both charges execute simple harmonic motion. Both charges will continue moving in the direction of their displacement.   (c) Charge +q  executes simple harmonic motion while charge –q  continues moving in the direction of its displacement.   (d) Charge –q  executes simple harmonic motion while charge +q  continues moving in the direction of its displacement. (JEE Advanced 2015)  20. Consider a uniform spherical charge distribution of radius  R1 centred at the origin O. In this distribution, a spherical cavity of radius  R2, centred at P with distance OP = a =  R1 – R2 (see figure) is made. If the electric field inside the cavity at position r  is  E(r ), then the correct statement(s) is(are)    



2

(0, 0, 0)

(b)

2

 

(a)

2 E0 a

 

(c)

 E0 a

 17.

A wooden block performs SHM on a frictionless surface with frequency n 0. The block carries a charge + Q on its surface. If a uniform electric field  E   is switched on as shown, then the SHM of the block will be

2

 

(,0, a, 0)

 

(d)

 E  +Q

2 E0 a  E0 a

2

2

(IIT-JEE 2011)

(a) (b)







 Electrostatics: Part 1

93

 

at a finite distance to the right of Q2, the electric field is zero (IIT-JEE 2010) 4. A cubical region of side a has its center at the origin. It encloses three fixed point charges, –q  at (0, –a /4, 0), + 3q at (0, 0, 0), and – q at (0, +a /4, 0). Choose the correct option(s).

 

The net electric flux crossing the plane  x  = +a /2 is equal to the net electric flux crossing the plane  x  = –a /2. (b) The net electric flux crossing the plane  y = +a /2 is more than the net electric flux crossing the plane  y = –a /2. (c) The net electric flux crossing the entire origin is q / e 0. (d) The net electric flux crossing the plane  z = +a /2 is equal to the net electric flux crossing the plane  x  = +a /2. (IIT-JEE 2012) 5. Two nonconducting solid spheres of radii R and 2 R, having uniform volume charge densities r 1 and r 2, respectively, touch each other. The net electric field at a distance 2 R from the center of the smaller sphere, along the line joining the center of the spheres is zero. The ratio r1 / r 2  can be (a) –4  (b) –32/25 (c) 32/25  (d) 4

(d)



 is uniform, its magnitude is independent of R2 but its direction depends on r  (b)  E    is uniform, its magnitude depends of  R2  and its direction depends on r  (c)  E   is uniform, its magnitude is independent of a but its direction depends on a (d)  E    is uniform, and both its magnitude and direction depends on a (JEE Advanced 2015)

 

(a)

 E 





 





 





 



Multiple Correct Answers Type 1. A positively charged thin metal ring of radius R is fixed in the  xy plane with its center at the origin O. A negatively charged particle P  is released from rest at the point (0, 0,  z 0 ), where  z 0

> 0. Then the motion of P is

 

(a)

periodic, for all values of

 

(b)

simple harmonic, for all values of 0

 z 

0

 satisfying

0

<

 z 

0

z 

0

b, electric field is zero, because no charge inside. (ii) for c > r  > b, electric field will be similar to that as if a point charge 2q is placed at centre. (iii) for d  > r  > c, electric field will be zero, as no charge inside. (iv) for r  > d , electric field will be similar to that as if a point charge 6q is placed at centre. The resulting plot is as shown in figure. (b) (i) 0 (ii) 0 (iii) (v)

q 2

2 p e 0 r  3q 2

, radially outward

(iv) 0

(d) (a) (b) (d) (a) (a)

3. 8. 13. 18. 23.

(d) (a) (a) (a) (a)

4. 9. 14. 19. 24.

(a) (d) (a) (a) (b)

5. 10. 15. 20. 25.

(b) (d) (b) (c) (a)

2. 7. 12. 17. 22.

(a) (c) (a) (b) (c)

3. 8. 13. 18. 23.

(a) (b) (d) (d) (d)

4. 9. 14. 19. 24.

(b) (a) (b) (b) (b)

5. 10. 15. 20. 25.

(c) (a) (a) (d) (b)

ARCHIVES (d) (c) (d) (a) (a)

2 p e 0 r 

(iii) –2q

Level 1 1. (c) 6. (a) 11. (d)   16. (a)

2. 7. 12. 17.

(a) (b) (b) (c)

3. 8. 13. 18.

(d) (b) (d) (a)

4. 9. 14. 19.

(b) (b) (d) (c)

5. 10. 15.

(b) (b) (c)

Level 2   20. (c)   25. (b)   30. (a)   35. (a)   40. (d)

21. 26. 31. 36. 41.

(a) (b) (c) (a) (b)

22. 27. 32. 37.

(c) (b) (c) (c)

23. 28. 33. 38.

(a) (c) (a) (c)

24. 29. 34. 39.

(b) (b) (a) (c)

5. 10. 15. 20.

(d) (d) (b) (d)

MULTIPLE CORRECT  ANSWERS T  YPE 1. 5. 9.

(a, b, c) 2. (a, b, d) 3. (a, b, c) 4 (a, d) (a, d) 6. (a, b, d) 7. (a, d) 8. (b, c) (b, c) 10. (c, d)

LINKED COMPREHENSION T  YPE

, radially outward

(c) (i) 0 (ii) +2q   30. (a) False (b) False (c) True

2. 7. 12. 17. 22. 27.

 YPE SINGLE CORRECT  ANSWER T 

  19.   21.   24.

=

(d) (c) (d) (d) (d) (b)

 JEE EXERCISES

m p

(b) 107 m/s

4f 

       

1. 6. 11. 16. 21. 26.

1. 6.   11.   16.   21.

2

3Q

  14.

  17.

1

SINGLE CORRECT  ANSWER T   YPE

(iv) +6q

1. 6.   11. 16.

(c) (a) (a) (b)

2. 7. 12. 17.

(b) (a) (a) (b)

3. 8. 13. 18.

(c) (c) (d) (c)

4. 9. 14. 19.

(a) (a) (d) (a)

 Electrostatics: Part 1

 YPE MATCHING COLUMN T  1. 2. 3. 4. 5.

 JEE (ADVANCED)

(p – a) (q – a, b) (r – d) (s – a, b, c) (p – b) (q – b) (r – b, c) (s – b, d) (p – a, d) (q – a, d) (r – b, d) (s – c) (p – a, d) (q – c) (r – b, d) (s – c) (a) 6. (b)

Single Correct Answer Type 1. (d) 2. (b) 3. (d) (c) (c) 6. 7. 8. (d)   11. (b) 12. (a) 13. (b)   16. (c) 17. (a) 18. (c)

INTEGER ANSWER T  YPE 1. 6.   11.

(6) (2) (2)

2. 7.

(2) (3)

3. 8.

(4) (9)

4. 9.

(2) (2)

5. 10.

(9) (5)

Matching Column Type 1. (a)

 JEE (MAIN) Single Correct Answer Type 1. (b) 2. (d) 3. (b) 6. (a) 7. (d) 8. (b)   11. (c) 12. (c) 13. (a)   16. (c) 17. (a) 18. (c)   21. (a) 22. (a)

4. 9. 14. 19.

(c) (a) (c) (c)

5. 10. 15. 20.

(b) (c) (b) (d)

(b) (c) (a) (c)

5. 10. 15. 20.

(c) (d) (d) (d)

Multiple Correct Answers Type 1. (a, c) 2. (a, c) 3. (a, d) 4. (a, c, d) 5. (b, d) 6. (a, b, c) Linked Comprehension Type 1. (a) 2. (b) 3. (c)

 JEE ARCHIVES

4. 9. 14. 19.

Integer Answer Type 1. (2) 2. (6) 3. (6)

97