This article deals with NCERT Solutions for Class 12 Physics Chapter 1. Physics is certainly a complicated branch of science. However, NCERT Solutions makes this subject pretty easy. Most noteworthy, these solutions clear all fundamental concepts. Hence, the details of Physics become crystal clear for those who learn from NCERT Solutions
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Topics and Subtopics in NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges And Fields:
| Section Name | Topic Name |
|---|---|
| 1 | Electric Charges And Fields |
| 1.1 | Introduction |
| 1.2 | Electric Charge |
| 1.3 | Conductors and Insulators |
| 1.4 | Charging by Induction |
| 1.5 | Basic Properties of Electric Charge |
| 1.6 | Coulomb’s Law |
| 1.7 | Forces between Multiple Charges |
| 1.8 | Electric Field |
| 1.9 | Electric Field Lines |
| 1.10 | Electric Flux |
| 1.11 | Electric Dipole |
| 1.12 | Dipole in a Uniform External Field |
| 1.13 | Continuous Charge Distribution |
| 1.14 | Gauss’s Law |
| 1.15 | Applications of Gauss’s Law |
Q1.1. What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7 C placed 30 cm apart in the air?
Solution:-
Q1.2. The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in the 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?
Solution:-
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
Q1.3. Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Solution:-
Question 1.4:
- Explain the meaning of the statement ‘electric charge of a body is quantised’.
- Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
ANSWER:-
(a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.
(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.
Q1.5. 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.
ANSWER:- Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.
Q1.6. Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
ANSWER:- The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = cm
AO = OC = DO = OB = cm
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.
Question 1.7:
- An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
- Explain why two field lines never cross each other at any point?
ANSWER:-
(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.
Q1.8: Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?
Q1.9: A system has two charges qA = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?
ANSWER:- Both the charges can be located in a coordinate frame of reference as shown in the given figure.
At A, amount of charge, qA = 2.5 × 10−7C
At B, amount of charge, qB = −2.5 × 10−7 C
Total charge of the system,
q = qA + qB
= 2.5 × 10−7 C − 2.5 × 10−7 C
= 0
Distance between two charges at points A and B,
d = 15 + 15 = 30 cm = 0.3 m
Electric dipole moment of the system is given by,
p = qA × d = qB × d
= 2.5 × 10−7 × 0.3
= 7.5 × 10−8 C m along positive z-axis
Therefore, the electric dipole moment of the system is 7.5 × 10−8 C m along positive z−axis.
Q1.10: An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 N C−1. Calculate the magnitude of the torque acting on the dipole.
ANSWER:-
Electric dipole moment, p = 4 × 10−9 C m
Angle made by p with a uniform electric field, θ = 30°
Electric field, E = 5 × 104 N C−1
Torque acting on the dipole is given by the relation,
τ = pE sinθ
Therefore, the magnitude of the torque acting on the dipole is 10−4 N m.
Q1.11: A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
ANSWER:-
(a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.
Amount of charge on the polythene piece, q = −3 × 10−7 C
Amount of charge on an electron, e = −1.6 × 10−19 C
Number of electrons transferred from wool to polythene = n
n can be calculated using the relation,
q = ne
= 1.87 × 1012
Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.
(b) Yes.
There is a transfer of mass taking place. This is because an electron has mass,
me = 9.1 × 10−3 kg
Total mass transferred to polythene from wool,
m = me × n
= 9.1 × 10−31 × 1.85 × 1012
= 1.706 × 10−18 kg
Hence, a negligible amount of mass is transferred from wool to polythene.
Question 1.12:
(a) Two insulated charged copper spheres A and B have their centers 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 each sphere is charged double the above amount, and the distance between them is halved?
Solution:-
Therefore, the force between the two spheres is 0.243 N.
Q1.13: Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Solution:-
Q1.14: Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
Solution:-
Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.
Q1.15: Consider a uniform electric field E = 3 × 103 îN/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Solution:-
Q1.16: What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
ANSWER:- All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.
Q1.17: Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 N m2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?
Solution:-
Q1.18: A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
Q1.19: A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
The net electric flux through the surface is 2.26 ×105 N m2C−1.
Q1.20: A point charge causes an electric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
Solution 20:-
Q1.21: A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
Solution:-
Q1.22: A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Solution:-
Q1.23: An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
Solution:-
Q1.24: Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10−22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Solution:-
Q1.25: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C).
Solution:-
Q1.26: Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?
(a)
(b)
(c)
(d)
(e)
ANSWER:-
(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.
(b) The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.
(c) The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.
(d) The field lines showed in (d) do not represent electrostatic field lines because the field lines should not intersect each other.
(e) The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.
Q1.27: In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10−7 Cm in the negative z-direction?
Solution:-
Question 1.28:
(a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)].
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Solution:-
Q1.29: A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is 
n, where is the unit vector in the outward normal direction, and is the surface charge density near the hole.
Solution:-
Q1.30: Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Solution:-
Q1.31: It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Solution:-
Question 1.32:
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
ANSWER:-
(a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.
Q1.33: A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/ (2m).
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
Solution:-
Q1.34: Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (| e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)
Solution:-
Velocity of the particle, vx = 2.0 × 106 m/s
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between the two plates, E = 9.1 × 102 N/C
Charge on an electron, q = 1.6 × 10−19 C
Mass of an electron, me = 9.1 × 10−31 kg
Let the electron strike the upper plate at the end of plate L, when deflection is s.
Therefore,
Therefore, the electron will strike the upper plate after travelling 1.6 cm.
Important Topics covered in NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and fields
Electric Charge
The chapter talks about the accumulation of NCERT Solutions for Class 12 Physics Chapter 1 which occurs due to the rubbing of insulating bodies. This charge is known as an electric charge and was first discovered in the 600 BC. When a body acquires charge due to rubbing, it is said to be electrified. The chapter talks about the accumulation of charges which occurs due to the rubbing of insulating bodies. This charge is known as an electric charge and was first discovered in the 600 BC. When a body acquires charge due to rubbing, it is said to be electrified. In the case of two electrified objects, the like charges repel each other whereas the unlike charges attract each other. Polarity is the property by which we are able to differentiate between the two types of charges.
Conductors, Insulators and Semiconductors
The substances which allow electricity to pass through them are known as insulators and those which do not allow electricity to pass through them are known as insulators. This chapter will help you understand the flow of charges which is responsible for the substance to function as an insulator or a conductor. There is another category of substances, with respect to the flow of charges and passage of electricity; these are known as semiconductors. They lie between the conductors and the semiconductors, in terms of the flow of charge and electricity.
Quantization of Charge
This chapter will introduce you to the concept of quantization of charges. The charge q on a body is denoted as q = ne. here n is supposed to be any integer and e denotes the charge it carries. Here you will be introduced to the fact that the electric charge is always an integral multiple of e and is known to be the quantization of charge.
Electric Dipole
You will learn about electric dipoles in the chapter. A pair of equal and opposite point charges q and –q, which are separated by a distance of 2a is known as an electric dipole. The direction between –q and q is known to be the direction of the dipole and the central point between these two is known as the centre of the dipole.
Exercise Wise Discussion of NCERT Solutions for Class 12 Chapter 1 Electric Charges and fields
- The entire chapter contains one exercise and thirty-three questions in total. All of these questions aim at testing your observational and analytical skills. They revolve around the major concepts and the fundamentals that have been tackled in the chapter.
- The exercises, in the beginning, tackle simple questions which are numerical questions. The first question 1.1 demands the use of the electrostatic relation and its mathematical formula. Similar questions are discussed in further exercises 1.2, 1.3 etc. They all discuss electrostatic force, charge ratios and their signs. These questions will test your application skills. All the formulas used have been introduced in the chapter along with their theories and proper explanations.
- There are exercises which require a theoretical and conceptual approach. For example, question 1.4 demands a theoretical explanation of electric charges and quantization. There are situational questions which require you to apply reasoning and logic in order to derive the solutions. For example question 1.16 where you need to find the net flux of the uniform electric field based on the given conditions and scenarios.
- The chapter also includes diagram based questions, like question 1.14 where the figure represents the direction and density of charges, based on which you will have to determine the mass ratio and intensity of the charges. Question 1.24 is also a diagrammatic question where two plates are drawn parallel to each other and their charge densities have been given, based on the description given in the question, you are required to derive the electric field between the plates, in the outer region of the first place and in the outer region of the second plate.
Benefits of NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and fields
- The expert team at Instasolv has formulated all the solutions in an absolutely simple and student-friendly language. We ensure that you develop a firm understanding of the chapter and all its core concepts.
- Each and every question has been tackled by our team with step by step answers and a well-furnished explanation.
- All the answers are in accordance to the latest CBSE syllabus
Conclusion
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