CHAPTER 1: ELECTRIC CHARGES AND FIELDS

  

 

 

 

 

PHYSICS

CHAPTER 1: ELECTRIC CHARGES AND FIELDS

ELECTRIC CHARGES AND FIELDS

Introduction:

 

Study of static charges is called electrostatics and this complete electrostatic will be discussed in two chapters. In this chapter we begin with a discussion of electric charge, some properties of charged bodies, and fundamental electric force between two charged bodies.

What is Electric Charge?

Electric Charge is a fundamental property of a matter which is responsible for electric forces between the bodies. Two electrons placed at small separation are found to repeal each other, this repulsive force (Electric force) is only because of electric charge on electrons.

When a glass rod is rubbed with silk, the rod acquires one kind of charge, and the silk acquires the second kind of charge. This is true for any pair of objects that are rubbed to be electrified. Now if the electrified glass rod is brought in contact with silk, with which it was rubbed, they no longer attract each other.

Types of Electric Charge:

There are two types of charge exist in our nature.

·        Positive Charge

·        Negative Charge

If any object loses their electrons then they get positive charge. It is denoted by (+q) sign. If any object gain electrons from another object, then they get negative charge. It is denoted by (-q) sign. The charges were named as positive and negative by the American scientist Benjamin Franklin. If an object possesses an electric charge, it is said to be electrified or charged. When it has no charge it is said to be neutral.

Basic Properties of Electric Charge:

 

The important properties and characteristic of electric charge are given below.

Attraction and Repulsion: Like charges repel each other while unlike charges attract each other.

Electric Induction: When a charged object brings to contact with another uncharged, it gets opposite charge of charged object. It is called charging by induction.

Charge is Quantized: An object that is electrically charged has an excess or deficiency of some whole number of electrons. Since, electrons cannot be divided into fraction of electrons, it means that the charge of an object is a whole-number multiple of the charge of an electron. For example, it cannot have a charge equal to the charge of 0.5 or 1000.5 electrons.

Mathematically q = ± ne, here n = 1, 2, 3 and e = 1.6 × 10^–19 coulomb.

Electric Charge is Conserved: According to this property, "An electric charge neither can be created nor can be destroyed" i.e., total net charge of an isolated system is always conserved. Thus, when a glass rod rubbed with silk cloth, both glass rod and silk cloth acquire opposite charge in same quantity. Thus, total amount of charge remains same before rubbing as well as after rubbing.

Conductors and Insulators:

Some substances easily allow passage of electricity through them while others do not. Substances which allow electricity to pass through them easily are called ‘conductors. They have electrons that are free to move inside the material. Metals, human and animal bodies, earth etc. are example of conductors. Non-metals e.g., glass, plastic, wood are ‘insulators’ because they do not easily allow passage of electricity through them.

Most substances are either conductors or insulators. There is a third category called ‘semiconductors’ which are intermediate between conductors and insulators because they partially allow movement of charges through them.

Charging by Induction:

Now as we know that two oppositely charged bodies attract each other. But it also has been our observation that a charged body attracts a neutral body as well. This is explained on the basis of charging by induction. In induction process two bodies (at least one body must be charged) are brought very close, but they never touch each other.

Let us examine how a charged body attracts an uncharged body. Imagine a conducting or partially conducting body (sphere here) is kept on an insulating stand and a charged rod (positive, for example) is brought very close to it. It will attract electrons to its side and the farther end of the sphere will become positively charged as it is deficient of electrons.

Coulomb’s Law:

·        In 1785 Charles Coulomb (1736-1806) experimentally established the fundamental law of electric force between two stationary charged particles. He observed that An electric force between two charge particles has the following properties:

·        It is directed along a line joining the two particles and is inversely proportional to the square of the separation distance r, between them.

·        t is proportional to the product of the magnitudes of the charges, |q₁| and |q₂|, of the two particles.

·        It is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.

From these observations, Coulomb proposed the following mathematical form for the electric force between two charges. The magnitude of the electric force F between charges q1 and q2 separated by a distance r is given by

where k is a constant called the Coulomb constant. The proportionality constant k in Coulomb’s law is similar to G in Newton’s law of gravitation. Instead of being a very small number like G (6.67 × 10^–11), the electrical proportionality constant k is a very large number. It is approximately.

k = 8.9875 × 10⁹ N-m²C^–2

The constant k is often written in terms of another constant, ε0, called the permittivity of free space. It is related to k by

Electric Field:

A charge produces something called an electric field in the space around it and this electric field exerts a force on any charge (except the source charge itself) placed in it. The electric field has its own existence and is present even if there is no additional charge to experience the force.

Intensity of Electric Field:

Intensity of electric field due to a charge configuration at a point is defined as the force acting on a unit positive charge at this point. Hence if a charge q experiences an electric force F at a point then intensity of electric field at this point is given as

It has S.I. units of newtons per coulomb (N/C).

Electric Field due to a Point Charge:

To determine the direction of an electric field, consider a point charge q as a source charge. This charge creates an electric field at all points in space surrounding it. A test charge q0 is placed at point P, a distance r from the source charge. According to Coulomb’s law, the force exerted by q on the test charge is.

This force is directed away from the source charge q, since the electric field at P, the position of the test charge, is defined by

we find that at P, the electric field created by q is

Electric Field Lines:

Electric field lines are a way of pictorially mapping the electric field around a configuration of charges. An electric field line is, in general, a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point.

The field lines follow some important general properties:

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

·        In free space, they are continuous curves which emerge from positive charge and terminate at negative charge

·        They do not intersect each other. If they do so, then it would mean two directions of electric field at the point of intersection, which is not possible.

·        Electrostatic field lines do not form any closed loops. This follows from the conservative nature of electric field.

Electric Dipole:

A configuration of two charges of same magnitude q, but of opposite sign, separated by a small distance (say 2a) is called an electric dipole.

Dipole moment for an electric dipole is a vector quantity directed from the negative charge to the positive charge and its magnitude is p = q × 2a (charge × separation). The SI unit of dipole moment is C^-m (coulombmeter).

Gauss’s Law:

·        The flux of electric field through any closed surface S is 1/ε_0 times the

·        Total charge enclosed by S.

·        Electric field outside the charged shell is as though the total charge is concentrated at the center. The same result is true for a solid sphere of uniform volume charge density.

·        The electric field is zero at all points inside a charged shell.

Deduction of Coulomb’s law from Gauss' Law:

Consider a charge +q in place at origin in a vacuum. We want to calculate the electric field due to this charge at a distance r from the charge. Imagine that the charge is surrounded by an imaginary sphere of radius r as shown in the figure below. This sphere is called the Gaussian sphere.

Consider a small area element dS on the Gaussian sphere. We can calculate the flux through this area element due to charge as follows:

Using this in Gauss theorem we get

We know that

This is the required Coulomb’s law obtained from Gauss theorem.

Dipole Placed in Uniform External Field:

Since the impact of an external electric field on charges is already known to us; a dipole too will experience some form of force when introduced to an external field. It is interesting to learn that, a dipole placed in an external electric field acquires a rotating effect. This rotating effect is termed as ‘torque’ felt by the dipole. Excitingly, the net torque can be calculated on the opposite charges present in a dipole for estimating the overall rotation.

Torque on dipole:

Consider a dipole located in the same position ‘E’ to calculate the torque received by the dipole when positioned outside. The compulsory charge will be placed below the ‘qE’ magnitude as you go up, while the negative charge will be placed below the ‘qE’ magnitude as you go down.

Since the absolute power is zero, it can be seen that the dipole is in the equation at the moment. But what is the rotation rate? In this case, the dipole may remain stable but rotates at a certain angular velocity. This fact has been demonstrated by experimentation, and it shows that both electrostatic forces (qE) act as clock-related torque.

As a result, when a dipole is inserted into the same external electrical circuit, it rotates. Torque always works with external force applied which will be in pairs. Moreover, its size is a result of its strength and arm. The arm can be thought of as the distance between the point of force applied and the point at which rotation occurs at the dipole.

Torque

Torque (τ) = Force × distance separating forces

Torque is a vector whose direction is determined by the force acting on the axis. The magnitude of the torque vector is determined as follows:

T = F r sinθ

Which means,

F - force acting on the axis

r - temporary arm length

θ - angle between force vector and temporary arm

τ - is the vector of torque

Derivation of Torque

Consider a dipole with the angles of + q and q forming a dipole because they are separated by a distance of d. Positioned in the same electric field of power E, the dipole axis forms an θ angle with an electric field.

Charging power, F = ± q E

Elements of power perpendicular to dipole, F = ± q E sinθ

Since ‘qd’ is the magnitude of the dipole moment (p), and the direction of the dipole moment ranges from positive to negative; torque is the product of a dipole moment cross and an electric field. When the direction of the electric field is positive, the torque is in the clock (therefore negative) in the image above.

So,

τ = - pE sinθ

An incorrect sign indicates that the torque is in the clockwise direction.

Potential energy of dipole:

Consider a dipole with charges q₁ = +q and q₂ = -q placed in a uniform electric field as shown in the figure above. The charges are separated by a distance d and the magnitude of an electric field is E. The force experienced by the charges is given as –qE and +qE, as can be seen in the figure.

As we know that, when a dipole is placed in a uniform electric field, both the charges as a whole do not experience any force, but it experiences a torque equal to τ which can be given as,

τ = p × E

Consider a dipole with charges q₁ = +q and q₂ = -q placed in a uniform electric field as shown in the figure above. The charges are separated by a distance d and the magnitude of an electric field is E. The force experienced by the charges is given as –qE and +qE, as can be seen in the figure.

As we know that, when a dipole is placed in a uniform electric field, both the charges as a whole do not experience any force, but it experiences a torque equal to τ which can be given as,

 = pE(cosθ₀ – cosθ₁)

As we know that the work done in bringing a system of charges from infinity to the given configuration is defined as the potential energy of the system, hence the potential energy U(Ɵ)  can be associated with the inclination Ɵ of the dipole using the above relation.

U(θ) = pE(cosθ₀ – cosθ₁

From the above equation, we can see that the potential energy of dipole placed in an external field is zero when the angle Ɵ is equal to 90° or when the dipole makes an angle of 90°.

Considering the initial angle to be the angle at which the potential energy is zero, the potential energy of the system can be given as,

U(θ) = pE(cos 

Workdone by dipole:

A pair of force which is equal in magnitude, with opposite direction, and displaced by perpendicular distance or moment is known as the couple.

When a couple acts on a dipole

Work done to rotate a dipole is given by

Total work done is given by

When the dipole is rotated from 0 to θ

degrees, work done is given by

Hence, the work done to rotate a dipole in an external electric uniform field is

Work done if dipole rotated from 90 degree

Work done if dipole rotated from 180 degree

When

The dipole is said to be stable when the dipole is aligned in the direction of the electric field.

Stable and unstable equilibrium:

Stable equilibrium

“A body is said to be in stable equilibrium if after a slight tilt it returns to its previous position.”stable equilibrium state

Consider a book lying on the table. Tilt the book slightly about its one edge by lifting it from the opposite side. It returns to its previous position when sets free. Such a state of the body is called a stable equilibrium.

When a body is in stable equilibrium, its center of gravity is at the lowest position. When it is tilted, its center of gravity rises. It returns to its stable equilibrium as long as the center of gravity acts through the base of the body.

Examples of stable equilibrium:

·        Chair lying on the floor

·        The heavy base of the vehicle

·        Table lying on the ground

·        Cone lying on its base by lowering its center of gravity

·        Bottle lying on its base

Unstable equilibrium:

“If a body does not return to its previous position when sets free after the slightest tilt is said to be in unstable equilibrium.”unstable equilibrium state

Take a pencil and try to keep it in the vertical position on its tip. Whenever you leave it, the pencil topples over about its tip and falls down. This is called an unstable equilibrium. In an unstable equilibrium, a body may be made to stay only for a moment. Thus a body is an unstable equilibrium.

The center of gravity of the body is at its highest position in the state of unstable equilibrium. As the body topples over about its base (tip), its center of gravity moves towards its lower position and does not return to its previous position.

Example of unstable equilibrium:

·        When the ice cream cone is made to rest on its apex on a book, the movement of the book will disturb the position of the ice cream cone. This is an example of unstable equilibrium.

·        When the weather changes from freezing to hot to freezing rapidly and without reason, this is an example of a time when it is unstable. When a person has a bad temper that can explode or flare up with no provocation at all, this is an example of a person who would be described as unstable. Not firmly placed; unsteady.

Electric Field on axial and equatorial line

Axial line: Axial line is the line which is passing through the positive and negative charges and the point lies on that line is called the axial point.

Electric field on axial line of dipole is given by:

Equatorial line: Equatorial line is the perpendicular line to the line passing through the positive and negative charges and the point lies on that line is known as the equatorial point.

Electric field on equatorial line of dipole is given by:

Relation between electric field at axial and equatorial line

Electric field due to an electric dipole at points situated at a distancer r along its axial line is given as,

 

Electric field due to an electric dipole at points situated at a distance r along its equatorial plane is given as,

From (i) and (ii)

Therefore, ratio is 2 : 1

Relation between E and V:

The electric field exists if and only if there is a electric potential difference. If the charge is uniform at all points, however high the electric potential is, there will not be any electric field. Thus, the relation between electric field and electric potential can be generally expressed as – “Electric field is the negative space derivative of electric potential.”

Electric Field And Electric Potential

The relation between Electric field and electric potential is mathematically given by-

Where,

E is the Electric field.

V is the electric potential.

dx is the path length.

– Sign is the electric gradient

Direction of Electric Field

If the field is directed from lower potential to higher then the direction is taken to be positive.

If the field is directed from higher potential to lower potential then the direction is taken as negative.

Test charge	Formula	Electric gradient
Positive		Higher as you go closer towards test charge.
Negative		Higher as you go move away from test charge.
Equipotential surface		Electric potential is perpendicular to Electric field lines.

 

 

 

 

Electric Field And Electric Potential Relation Derivation:

Where,

·        F is the force applied

·       dl is the short element of the path while moving it from a to b.

The force can be written as charge times electric field.

Dividing both sides by test charge q₀

Workdone by the test charge is the potential V_a - V_b

For equipotential surface, V_a = V_b thus,

Important Questions

Multiple Choice questions-

1. The surface considered for Gauss’s law is called

(a) Closed surface

(b) Spherical surface

(c) Gaussian surface

(d) Plane surface

2. The total flux through the faces of the cube with side of length a if a charge q is placed at corner A of the cube is

3. Which of the following statements is not true about Gauss’s law?

(a)   Gauss’s law is true for any closed surface.

(b)  The term q on the right side of Gauss’s law includes the sum of all charges enclosed by the surface.

(c)   Gauss’s law is not much useful in calculating electrostatic field when the system has some symmetry.

(d)  Gauss’s law is based on the inverse square dependence on distance contained in the coulomb’s law

4. A charge Q is placed at the center of the line joining two point charges +q and +q as shown in the figure. The ratio of charges Q and q is

(a) 4

(b) 1/4

(c) -4

(d) -1/4

5. The force per unit charge is known as

(a) electric flux

(b) electric field

(c) electric potential

(d) electric current

6. Electric field lines provide information about

(a) field strength

(b) direction

(c) nature of charge

(d) all of these

7. Which of the following figures represent the electric field lines due to a single negative charge?

8. The SI unit of electric flux is

(a) N C^-1 m^-2

(b) N C m^-2

(c) N C^-2 m²

(d) N C^-1 m²

9. The unit of electric dipole moment is

(a) newton

(b) coulomb

(c) farad

(d) Debye

10. Consider a region inside which, there are various types of charges but the total charge is zero. At points outside the region

(a)   the electric field is necessarily zero.

(b)  the electric field is due to the dipole moment of the charge distribution only.

(c)   the dominant electric field is inversely pro-portional to r3, for large r (distance from ori-gin).

(d)  the work done to move a charged particle along a closed path, away from the region will not be zero.

Very Short:

1.         What is the value of the angle between the vectors  and  for which the potential energy of an electric dipole of dipole moment , kept in an external electric field , has maximum value.

2.                 Define electric field intensity at a point.

3.                 Two equal point charges separated by 1 m distance experience force of 8 N. What will be the force experienced by them, if they are held in water, at the same distance? (Given: K_water = 80) (CBSE Al 2011C)

4.                 A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube? (CBSE AI 2012) (CBSE Sample Paper 2019)

5.                 Why do the electric field lines not form closed loops? (CBSE Al 2012C)

6.                 Two equal balls having equal positive charge ‘q’ coulomb are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two? (CBSE AI 2014)

7.                 What is the electric flux through a cube of side l cm which encloses an electric dipole? (CBSE Delhi 2015)

8.                 Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (CBSE Al 2015C)

9.                 What is the amount of work done in moving a point charge Q. around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located? (CBSE Al 2016)

10.            How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? (CBSE Delhi 2016)

Short Questions:

1.

(a) Electric field inside a conductor is zero. Explain.

(b) The electric field due to a point charge at any point near it is given as

what is the physical significance of this limit?

2. Define the electric line of force and give its two important properties.

3. Draw electric field lines due to (i) two similar charges, (ii) two opposite charges, separated by a small distance.

4. An electric dipole is free to move in a uniform electric field. Explain what is the force and torque acting on it when it is placed

(i) parallel to the field

(ii) perpendicular to the field

5. A small metal sphere carrying charge +Q. is located at the centre of a spherical cavity in a large uncharged metallic spherical shell. Write the charges on the inner and outer surfaces of the shell. Write the expression for the electric field at the point P₁ (CBSE Delhi 2014C)

6. Two-point charges q and -2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero. (CBSE Al 2014C)

7. Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to the electric field () directed along +X direction, what will be the magnitude and direction of the torque acting on this? (CBSE Delhi 2011)

8. A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the centre of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface? (NCERT Exemplar)

Long Questions:

1.

(a) State Gauss theorem in electrostatics. Using it, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

(b) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged? (C8SE Delhi 2012)

2. Use Gauss’s law to derive the expression for the electric field () due to a straight uniformly charged infinite line of charge λ Cm^-1. (CBSE Delhi 2018)

Assertion and Reason Questions-

1. For two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.

a)     Both A and R are true, and R is the correct explanation of A.

b)     Both A and R are true, but R is not the correct explanation of A.

c)     A is true but R is false.

d)     A is false and R is also false.

Assertion (A): The electric flux emanating out and entering a closed surface are 8 × 10³ and 2 × 10³Vm respectively. The charge enclosed by the surface is 0.053μC.

Reason (R): Gauss's theorem in electrostatics may be applied to verify.

2. For two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.

a)     Both A and R are true, and R is the correct explanation of A.

b)     Both A and R are true, but R is not the correct explanation of A.

c)     A is true but R is false.

d)     A is false and R is also false.

Assertion (A): Charge is quantized.

Reason (R): Charge which is less than I C is not possible.

Case Study Questions-

1. Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in the given figure, the electric field at P is stronger that at Q.

(i)    Electric lines of force about a positive point charge are:

a)     Radially outwards.

b)     Circular clockwise.

c)     Radially inwards.

d)     Parallel straight lines.

(ii)  Which of the following is false for electric lines of force?

a)     They always start from positive charges and terminate on negative charges.

b)     They are always perpendicular to the surface of a charged conductor.

c)     They always form closed loops.

d)     They are parallel and equally spaced in a region of uniform electric field.

(iii)    Which one of the following pattern of electric line of force in not possible in filed due to stationary charges?

a)
b)
c)
d)

(iv)    Electric lines of force are curved:

a)     In the field of a single positive or negative charge.

b)     In the field of two equal and opposite charges.

c)     In the field of two like charges.

d)     Both (b) and (c).

(v)  The figure below shows the electric field lines due to two positive charges. The magnitudes E_A, E_B and E_C of the electric fields at points A, Band C respectively are related as:

a)     E_A > E_B > E_C

b)     E_B > E_A > E_C

c)     E_A = E_B > E_C

d)     E_A > E_B = E_C

2. Net electric flux through a cube is the sum of fluxes through its six faces. Consider a cube as shown in figure, having sides of length L = 10.0cm. The electric field is uniform, has a magnitude E = 4.00 × 10³N C^-1 and is parallel to the xy plane at an angle of 37º measured from the + x - axis towards the + y - axis.

(i)    Electric flux passing through surface S₆ is:

a)     -24N m² C^-1

b)     24N m² C^-1

c)     32N m² C^-1

d)     -32N m² C^-1

(ii)  Electric flux passing through surface S₁ is:

a)     -24N m² C^-1

b)     24N m² C^-1

c)     32N m² C^-1

d)     -32N m² C^-1

(iii)    The surfaces that have zero flux are:

a)     S₁ and S₃

b)     S₅ and S₆

c)     S₂ and S₄

d)     S₁ and S₂

(iv)    The total net electric flux through all faces of the cube is:

a)     8N m² C^-1

b)     -8N m² C^-1

c)     24N m² C^-1

d)     Zero.

(v) The dimensional formula of surface integral ∮ d of an electric field is:

a)     [M L² T^-2 A^-1]

b)     [M L³ T^-3 A^-1]

c)     [M L^-1 T³ A^-3]

d)     [M L^-3 T^-3 A^-1]

ü Answer Key:

Multiple Choice Answers-

1.    Answer: c

2.     Answer: a

3.     Answer: c

4.     Answer: d

5.     Answer: b

6.     Answer: d

7.     Answer: b

8.     Answer: d

9.     Answer: d

10.Answer: c

Very Short Answers:

1.     Answer:

P.E. = -pEcos θ

P.E. is maximum when cos θ = - 1, i.e.

θ = 180°

2.     Answer: Electric field intensity at a point is defined as the force experienced by a unit test charge placed at that point. Mathematically

we have

3.     Answer: The force in water is given by

4.     Answer: Φ = q/6ε₀

5.     Answer: It is due to the conservative nature of the electric field.

6.    Answer: It decreases because force ∝ =  and k > 1.

7.     Answer: Zero

8.     Answer: So that no net force acts on the charge at the equipotential surface, and it remains stationary.

9.     Answer: Zero.

10.Answer: No change, as flux does not depend upon the size of the Gaussian surface.

Short Questions Answers:

1.     Answer:

(b) It indicates that the test charge should be infinitesimally small so that it may not disturb the electric field of the source charge.

2.     Answer:

It is a line straight or curved, a tangent to which at any point gives the direction of the electric field at that point.

(a) No two field lines can cross, because at the point of intersection two tangents can be drawn giving two directions of the electric field which is not possible.

(b) The field lines are always perpendicular to the surface of a charged conductor.

3.     Answer:

(a) The diagram is as shown.

(b) The diagram is as shown.

4.     Answer:

(i) When an electric dipole is placed parallel to a uniform electric field, net force, as well as net torque acting on the dipole, is zero and, thus, the dipole remains in equilibrium.

(ii) When the dipole is placed perpendicular to the field, two forces acting on the dipole form a couple, and hence a torque acts on it which aligns its dipole along the direction of the electric field.

5.     Answer:

6.     Answer:

Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is

Solving for x we have

x = d

7.     Answer:

The resultant dipole moment of the combination is

since cos 120° = -1/2

This will make an angle of 30° with the X-axis, therefore torque acting on it is

τ=PE sin 30° = pE/2 (Along Z-direction)

Answer: The induction of charges is as shown.

Therefore, surface charge density on the inner and the outer shell is on the outer surface is

Long Questions Answers:

1.     Answer:

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore, by the definition of electric flux, the flux Linked with the Gaussian surface is given by

(b)

(i) directed outwards

(ii) directed inwards.

2.     Answer:

Consider an infinitely Long, thin wire charged positively and having uniform Linear charge density λ. The symmetry of the charge distribution shows that must be perpendicular to the tine charge and directed outwards. As a result of this symmetry, we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary Length L. with its ends perpendicular to the wire as shown in the figure. Applying Gauss’s theorem to curved surface ΔA₁ and circular surface ΔA₂.

This is the expression for the electric field due to an infinitely long thin wire.

The graph is as shown.

Assertion and Reason Answers-

1. (a) Both A and R are true, and R is the correct explanation of A.

Explanation:

According to Gauss's theorem in electrostatics,

2. (c) A is true but R is false.

Explanation:

The charge q on a body is given as q = ne where n is any integer positive or negative. The charge on the electron is q = 1.6 × 10^-19C which is less than 1C.

Case Study Answers-

1. Answer :

(i)    (a) Radially outwards.

(ii)  (c) They always form closed loops.

Explanation:

Electric lines of force do not form any closed loops.

(iii)    (c)

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​​​​​​​​​​​​​​^{​​​​​​​}Explanation:

Electric field tines can't be closed.

(iv)    (d) Both (b) and (c).

(v)  (a) E_A > E_B > E_C

2. Answer :