Physics Class 12 Chapter 1 Important Questions

With Class 12 Physics Chapter 1 Important Questions, students will get detailed and authentic solutions to questions regarding electric charges and fields. These Class 12 Physics Chapter 1 Important Questions will prepare students for their exams. After studying these important questions, students will be able to solve many CBSE sample papers.

In addition, these important question notes also contain important formulas and CBSE extra questions that can help students test their understanding.

CBSE Class 12 Physics Chapter 1 Important Questions

Study Important Questions for Class 12 Physics Chapter 1 – Electric Charges and Fields

Electric charge refers to a matter’s basic physical property that results in it experiencing a force when it is kept in a magnetic or an electric field. The interaction between the charges causes an electromagnetic force. Here are some important questions that students can that study to prepare them better for their examinations.

Short Questions and Answers (1 Mark)

1.  Name the physical quantity that has these S.I. Units (i) Cm (ii) NC

Ans. (i) Cm is the S.I. unit of electric dipole moment.

(i) NC is the S.I. unit of electric field intensity.

2. Which physical quantity has the S.I. unit of JC-1? Also, state whether it is a vector or a scalar quantity.

Ans. JC-1 is the S.I. unit of electric potential, and it is a scalar quantity.

3. What do you mean by one Coulomb?

Ans. 1 Coulomb refers to the charge on a body if it receives a force of either attraction or repulsion of 9109N  from another equal charge when they have a distance of 1m between them.

4. Will the force existing between two point charges change if the medium’s dielectric constant in which they are present increases?

Ans. The dielectric constant of a medium is presented by the following equation

k=FVFM=Force between the charges in vacuumForce between two charges in medium

FM=FVk

According to the expression given above, we can clearly observe that k increase and FM decreases.

5. A charge ‘q’ is positioned at the middle of a cube of side l. What is the electric flux travelling through two opposing faces of the cube? 

Ans.  E= q30

Short Questions and Answers (2 Marks)

Q.1. Why can’t two electric lines of force intersect each other?

Ans. This can be understood through an example. Let’s imagine if they intersect, then at the place of contact, you can draw two tangents from it.

These two tangents are intended to depict two directions of electric field lines, which, at a given spot, is not achievable.

Q.2. Why is the quantization of electric charge when working with macroscopic charge irrelevant?

Ans. When dealing with large-scale and macroscopic charges, the charges used are immense in quantity as compared to the size of the electric charge. That’s why the quantization of electric charge is irrelevant on a macroscopic scale. Therefore, it is ignored, and it is assumed that the electric charge is continuous.

Q.3. Calculate the dipole moment of the dipole when an electric dipole is held at 30° to the uniform electric field of 104 N/C and experiences a torque of 910-26 Nm.

Ans. The details are given as follows.

=30°

E=104 N/C

t=910-26 Nm

P denotes the dipole moment that is required to be calculated.

The torque is presented by =PEsin

P=Esin

P= 910-26104sin 30° = 910-2610-421

P= 1810-30 Cm

Q.4. When a glass rod is brushed with a silk cloth, charges develop on both objects. A similar phenomenon is seen with numerous different pairs of entities. Explain how this occurs with the law of conservation of charge.

Ans. When two objects are rubbed against each other, there is a generation of charges equal in magnitude, but contrary in nature to the bodies that are rubbed. It is also noted that during such an occurrence, charges are produced in pairs. This phenomenon of charging is referred to as charging by friction.

The net charge on the system of both bodies equals zero. This is because the same number of opposite charges in both bodies destroy each other. When a silk cloth is rubbed on a glass rod, opposite-natured charges appear on both those bodies. Thus, this event agrees with the law of conservation of energy. A similar phenomenon can also be observed with numerous different pairs of bodies too.

Q.5. What does the phrase ‘electric charge of a body is quantized’ mean?

Ans: The phrase ‘electric charge of a body is quantized’ says that only an integral (1, 2, 3, 4, …, n) number of electrons may be passed from one body to another. This further implies that charges are not transmitted in fractions. Hence, a body possesses its entire charge only in integral multiples of electric charges.

Q.6. What will the net charge within the box be when the net outwards flux on the box’s surface is 8.0 10³Nm2/C ?

Ans. The information given is as follows

Net outward flux surface of the box 8.0 10³Nm2/C

If a body contains the net charge q, we can indicate the flux by =q0

0=Permittivity of free space=8.854 10-12 N-1 C2 m-2

Hence, the charge q can denote by q=0

⇒q= 8.854 10–12 8.0 10³

⇒q= 7.08 10-8

⇒q= 0.07C

Hence, 0.07C will be the net charge inside the box.

Q.7. An electric field line is a continuous curve, meaning that a field line cannot have unexpected breaks. Why not?

Ans. An electrostatic field line is a constant curve since it is understood that a charge experiences a continuous force when tracked in an electrostatic field. In addition, the field line cannot have unexpected breaks since the charge moves continuously and cannot jump from one place to another.

Q.8. What is an electric dipole moment, and what is its S.I Unit?

Ans. The product of the magnitude of either dipole length or change is called the electric dipole moment of an electric dipole.

S.I. unit of dipole (p) is coulomb metre (Cm).

Short Questions and Answers (2 Marks)

Q.9. A particle having mass m and charge q is discharged from its rest in a uniform electric field of intensity denoted by letter E. Find out the kinetic energy achieved by the particle after travelling a distance.

Ans. In an electric field, E and the electrostatic force is applied to a change can be written as

F = qE          …………………….(1)

We also know Newton’s second law of motion is

F = ma          …………………….(2)

From (1) and (2)

a = qEm           …………………….(3)

The third equation of motion is

v² – u² = 2as

Since the charged particle is at rest

u = 0

v² = 2as       ……………………. (4)

The kinetic energy is given by

KE = 12mv²        ………………….(5)

Substituting (4) in (5), we get,

KE = 12m(2as)=mas   ……………(6)

Substituting (3) in (6) to get,

KE = mqEms

Therefore, the kinetic energy that is achieved by the particle with a charge of q that is moving a distance s in the electric field E is denoted by

∴KE=qEs

Q.10. What will the force between two small spheres that have 210-7 C and 310-7C be, if they are suspended in the air and have 30 cm of distance between them?

Ans. The charge of the 1st and 2nd spheres is 210-7 C and 310-7C

The distance r is = 30 cm = 0.3m

The electrostatic force that exists between the spheres can be denoted as

F= 140.q1q2r2

Here,

0= permittivity of free space and 140= 9109 Nm² C-2

Force, F=9109210-7310-70.32= 610-3 N

A repulsive force will exist between the charges as they are of similar nature.

Q.11. A sphere S1having radius R1 surrounds a charge Q. Another concentric sphere S2 of radius R2 (R2﹥R1) with no additional charge between S1 and S2, find the ratio of electric flux between S1 and S2.

Ans. We can recall that according to Gauss’s law, the expression for electric flux going through a surface q is

= Q0

Here, 0 denotes the permittivity of the medium.

The electric flux that is going through sphere S1 is denoted by

s1= Q0       …………….(1)

We know that there is no additional charge exists between the two spheres. The flux existing between S2 is given by

s2= Q0       ……………..(2)

s1s2=Q0Q0

∴s1s2=11

Hence the required ratio will be 1:1

Q.12. If a polythene piece is rubbed with wool and results in a negative charge of 3 10-7, answer the following.

• Calculate how many electrons will be moved, noting which item they will come from and go to.

• Will there be a transfer of mass from wool to polythene?

Ans. (a) When a piece of polythene is rubbed against the wool, a certain number of electrons are passed on from wool to polythene. So, this results in wool becoming positively charged, while losing electrons and polythene becomes negatively charged when it gains the electrons.

Charge on the polyethene piece,

q = 310-7

And the charge of electrons,

e =-1.610-19 C

Now let’s imagine n to be the number of electrons that are transferred from wool to polythene, after that, using the property of quantization of charge, we can say

n = -310-7-1.610-19

∴n = 1.871012

Hence, 1.871012 will be the number of electrons that are transferred from wool to polythene.

(b) Yes, there will be a transfer of electrons from wool to polythene, and mass will also be transferred.

For instance, m is the mass being passed on in the given case, and me is the mass of the electron then,

m = men

m = 9.110-311.851012

∴m = 1.70610-18

Therefore, we can observe that only a small amount of mass is passed on from wool to polythene.

Q.13. Imagine there is a uniform electric field E = 3103 îN/C. Find out the following.

• The flux of this electric field is through a square of side 10 cm, the plane of which is parallel to the y-z plane.

• The flux through the same square if the 60° is formed by the normal plane with the x-axis.

Ans. (a) The information given is as follows.

Electric field intensity, E = 33103 îN/C

The magnitude of the electric field intensity, E = 3103 N/C

Side of the square, a=10cm = 0.1m

Area of the square, A = a² = 0.01m²

Since the square’s plane is parallel to the y-z plane, the normal plane would move in the x direction. So, the angle between the normal plane and the electric field would be, = 0°

When we substitute the values, we get,

= 310³0.01cos 0°

∴ = 3-Nm²/c

The net flux through the above-mentioned surface will be = 3-Nm²/c

(b) When the angle is 60° with the x-axis, the flox through the given surface will be,

= 310³0.01cos 60°

= 3012

∴ = 15Nm²/C

So the flux here would be 15Nm²/C.

Q.14. The distance between a point charge +10C is 5 cm above the square’s centre, having a side of 10cm, as given in the figure below. What will the magnitude of electric flux through the square be? (Hint: Imagine one face of a cube to be a square with an edge of 10cm.)

Ans. Imagine one face of a cube to be a square having 10cm with a charge q in its centre. Now, as explained by Gauss’s theorem for a cube, total electric flux is through its every six faces.

total =  q0

The electric flux passing through one cube’s face can be given by, = total6

= 16q0

The permittivity of free space, 0 = 8.85410-12N-1C²m-2

The net charge enclosed, q = 10C = 1010-6C

If we substitute the values in the question

= 161010-6C8.854105Nm2C-1

∴ = 1.88105Nm2C-1

Hence, the electric flux through the square will be 1.88105Nm2C-1.

Q.15. A spherical Gaussian surface with a 10 cm radius that is centred on a point charge experiences an electric flux of  -0.1103Nm2/C. Answer the following questions.

• How much flux is capable of passing through the surface if the radius of the Gaussian surface is doubled?

• What will the value of the point charge be?

Ans (a). Electric flux caused by the point charge is -0.1103Nm2/C.

The Gaussian surface’s radius that encloses the point charge is

r = 10.0 cm.

The net charge contained by the surface, as determined by Gauss’ law, determines how much electric flux pierces through the surface.

It is unaffected by the size of the hypothetical surface that is expected to encapsulate this charge.

The flux travelling through the surface, or 103Nm² / C, remains the same if the radius of the Gaussian surface is doubled.

(b) The electric flux is denoted by the relation,

total= q0

Where,

q = net charge encircled by the spherical surface

Permittivity of free space, 0= 8.85410-12C²m-2

q= 0

If we substitute the value,

q = -1.010³8.85410-12 =-8.85410-9C

∴ = -8.854nC

Hence, -8.854nC will be the value of the point charge.

Q.16. The surface charge density of a uniformly charged conduction sphere having a diameter of 2.4m is 80.0C/m². Answer the following questions.

• Find the charge that the sphere would have.

• What will the total electron flux leaving the surface of the sphere be?

Ans. (a) The diameter of the sphere is

d = 2.4m

The radius of the sphere is

r = 1.2m

The surface charge density,

= 80.0C/m² = 8010-6 C/m²

The total charge that will be present on the surface of the sphere will be

Q = Charge densitySurface area

Q = 4r²+ 8010-6 43.14(1.2)²

∴Q = 1.44710-3C

Hence, the charge present on the surface will be 1.44710-3C.

(b) As given by the Gauss’s law, the total electric flux (total) leaving the surface that contains net charge denoted by Q is

total = Q0    …………………(1)

The permittivity of free space is

0 = 8.85410-12N-1C²m-2

The charge of the sphere will be

Q = 1.44710-3C

When we substitute these in (1), we get,

total = 1.44710-38.85410-12

∴total = 1.6310-8NC-1m²

Hence, the total electric flux that will be leaving the sphere’s surface will be 1.6310-8NC-1m².

Short Questions and Answers (5 Marks) 

Q.1. Answer the following questions.

(a) Define equipotential surface. Demonstrate that an equipotential surface has an electric field that is always perpendicular to it.

(b) Find an expression for the potential at a location along a short electric dipole’s axial line.

Ans (a).  A surface with uniform potential is known as an equipotential surface.

We know that

dW = Fdx

dW = (qE)dx

(F = (qE), force on the test charge)

Since there is no work done when moving a test charge across an equipotential surface,

O = (qE)dx

Edx = O

Edx

(b) Consider an electric dipole with a dipole length of 2a and a point P on the axial line where O, the dipole’s centre, is such that OP = r.

V=VPA+VPB

V=K(-q)(r+a)+K(+q)(r-a)

V=Kq1r-a-!r+a

V=Kqr+a-r+a(r-a)(r+a)

V = Kq2ar2-a2              (∴P = 2aq)

V = KPr2-a2

If a dipole has a short length, “a” can be neglected

We get,

V+ = KPr2

Q.2. Check whether the ratio Ke2Gmemp will be dimensionless. You can take the help of the physical constant’s table and check the value of this ratio. What is the significance of the ratio?

Ans. The given ratio is Ke2Gmemp.

Here, G is the gravitational constant and is denoted by Nm²kg-2.

me  is the mass of the electron, and mp is the mass of a proton. kge is the electric charge and is denoted by C. permittivity of free space is, 0 and its unit is Nm²C-2.

Hence the unit of the ratio Ke2Gmemp is

= (Nm2C-2)(C2)(Nm²kg-2)(kg)(kg)

The dimension can be given as = (M0L0T0)

We know,

e = 1.610-19C

G = 6.6710-11Nm²kg-2

me= 9.110-31kg

mp = 1.6610-27 kg,

Therefore, the numerical value of the ratio will be

Ke2Gmemp = 9109(1.610-19)26.6710-119.110-311.6610-27≈2.31039

This ratio illustrates the relationship between the gravitational and electric forces acting on a proton and an electron while maintaining their relative distance from one another.

Q.3. Four charges named qA=2C, qB=-5C, qC=2C, and qD=-5C are situated in the corners of a 10 cm square ABCD. What is the force acting on a 1C charge in the square’s centre?

Ans. Four charges are positioned at the corners of the square in the given figure, which has four sides that are each 10 cm long. The square’s centre is O.

AB, BC, CD, and AD are the sides of the square. The diagonals of the square of length 102cm, AC and BD, are each 10 cm long.

AO, OB, OC, OD, are the length of 52cm.

The charge of the amount 1C is located at the centre of the square.

The sides of the square are AB, BC, CD, and AD. The diagonals of the square with a length of 10 cm each are AC and BD.

When compared to the repulsion between the charges placed at C and O, the force between the charges positioned at A and O is equally strong, but is directed in the other direction. They will therefore balance each other out.

Similar to the attraction force between the charges placed at D and O, there is an attraction force between charges placed at B and O that is similar in magnitude but moves in the opposite direction. As a result, their forces cancel one another out.

As a result, the 1C charge located at centre O is not subject to any net force from the four charges placed at the corners of the square.

Q.4. A system having two charges qA=2.510-7 and qB=-2.510-7C is positioned at A(0,0,-15) and B(0,0,15). What will the electric dipole moment and total charge of the system be?

Ans.

The value of charge at A, qA = 2.510-7

The value of charge at B, qB = -2.510-7C

Net amount of charge, qnet = qA+qB

qnet = +2.510-7 -2.510-7C

qnet = 0

The distance between two charges at A and B,

d = 15+15 = 30cm

d = 0.3m

The electric dipole moment of the system can be represented by

P = qAd = qB0.3

P = 7.510-8 Cm along the z-axis.

Hence, the electric dipole moment of the system will be 7.510-8, and it will be along the positive z-axis.

Q.5. (a) The centres of two insulated charged copper spheres, A and B are separated by a distance of 50 cm. If the charge on each is 6.510-7, what is the electrostatic repulsion’s mutual force? The separation distance is much greater than the radii of A and B.

(b) If each sphere is charged twice as much as above and their distance is halved, what is the force of repulsion?

Ans. (a) Charge on the sphere, A,qA = charge on sphere B,qB = 6.510-7

The distance between the sphere, r = 50cm = 0.5m

The force of repulsion between the two spheres,

F = qAqB40r2

Here,

0 = free space permittivity

140 = 9109 Nm²C-2

∴F = 9109(6.510-7)2(0.5)2

= 1.5210-2N

Hence, the force that would exist between the two spheres is

1.5210-2N

(b)

When the charge is doubled, charge on the sphere

A,qA= charge on the sphere

B,qB= 26.510-7C=1.310-6C.

The distance that would remain between the spheres will be halved.

∴r = 0.52 = 0.25m

Force of repulsion that would exist between the two spheres

F = qAqB40r2

= 91091.310-51.310-5(0.25)2

= 161.5210-2

= 0.243 Ns

Hence, 0.243 Ns will be the force that would exist between the spheres.

Q.6. The entire electric field in a particular area of space is along the z-axis. Though it is not constant, the strength of the electric field grows uniformly along the positive z-direction at a rate of 105 NC-1 per metre. What torque and force does a system with a total dipole moment of 10-7 Cm in the negative z-direction experience?

Ans. The dipole moment of the system, p = qdl = -10-7

The rate at which the electric field per unit increases

dedl = 10-5 NC-1

Force (f) that the system experiences can be described by the following relation,

F = qE

F = qdEDldl

= pdEDl

= -10-710-5

= -10-2N

Hence, -10-2N will be the force in the negative z-direction or the opposite to the direction of the electric field. Therefore, the angle that would exist between the dipole moment and the electric field is 180°.

Torque(T) can be described by the relation,

(T) = pE sin180° = 0

Hence, the torque that the system will experience will be zero.

Q.7. (a) Take a random electrostatic field arrangement as an example. A small test charge is positioned at the configuration’s null point or the location where E = 0.

Show that the test charge’s equilibrium is inherently unstable.

(b) Verify this result using the basic arrangement of two charges with the same magnitude and sign spaced apart.

Ans. (a) Let’s start by assuming that the little test charge positioned at the null point of

the setup is stable and in equilibrium.

In order for the equilibrium to remain stable, the test charge must only move very slightly.

The charge will return to the null point in any direction because there will be surrounding it are tremendous restoring powers.

This shows even more strongly that all electric lines of force acting near the null point are toward the specified null point and inward.

However, according to Gauss’s law, the net electric flow through a chargeless surface enclosed is equal to zero. This fact defies the presumption that we had commenced with. In light of this, it may be said that the balance of the test charge must be unstable.

(b) The null point occurs to be at the midpoint of the line connecting these two charges when we take into account this configuration setup with two charges of the same sign and magnitude arranged at a specific distance apart.

According to the presumption made before, when the test charge is positioned at this midpoint, it will encounter substantial restoring forces as it tries to move itself.

However, because there is no restoring force along the normal to the line taken into account, the test charge is pulled off when it tries to displace in a direction normal to the line connecting the two charges.

The assumption is further challenged in this situation, since stable equilibrium prioritizes restoring force in all directions.

Important Question of Physics Class 12 Chapter 1

Important Questions for Class 12 Physics Chapter 1 Electric Charges and Fields

When you practice these Class 12 Physics Chapter 1 Important Questions, you will learn many things that will help you enhance your fundamental knowledge and be able to obtain good marks in the exams. Class 12 Physics Chapter 1 Important Questions are a lifesaver for those students who have a weak foundation of these concepts. Some of the knowledge that students learn during studying physics class 12 ch 1 key questions are as follows.

Electric Charge

The term ‘electricity’ is derived from the Greek word elektron, which means amber. The electric and magnetic forces existing inside matter, atoms, and molecules are employed to establish their properties. There are just two forms of the entity, which are known as electric charge.

An experiment was carried out to examine the electric charge. This experiment indicated that there exist two kinds of electrification, which are: (i) like charges repel from one another and (ii) unlike charges seem to attract one another. There’s a property which separates these two sorts of charges known as the polarity of charge.

Electric charge is recognized as the physical feature of matter which allows it to experience a force whenever it gets put on an electromagnetic field. Electric charges are separated into two sorts which are positive and negative. The negatively charged matter is called an electron, whereas a positively charged matter is called a proton.

Properties of Electric Charge

Some fundamental features of the electric charge are as follows.

Additive: This feature of electric charge implies that the total charge of a body can represent the complete sum of all the individual charges that appear to be working on the system.

Additive: This feature of electric charge implies that the total charge of a body can represent the complete sum of all the individual charges that appear to be working on the system. Quantization: This feature of electric charge implies that the total charge of a body can reflect the integral multiple of a basic quantum of charge.

Conservation: This feature of electric charge implies that the total or the full charge of a given system remains unchanged with time. In simple words, when an item gets charged because of some friction, a transition of charge from one item to another occurs. A charge is something that can’t be destroyed or created.

Electric Field

An electric field is described as an electric force present in one unit of charge. The field’s direction is believed to be identical to the direction of the force and assumed to exert a positive test charge. When the electric field is radially outward, then it is regarded as a positive charge, and when the electric field is radially inward, then it is regarded as a negative charge. Volt per metre is the SI unit of electric fields.

We may describe the electric field as a physical field which can be found encircling each electric charge and responsible for exerting force on all the other charges existing on the field. Electric fields are considered to be produced from time-varying or electric magnetic fields. Both electric fields and magnetic fields are viewed as representations of electromagnetic force, which is among the four fundamental forces that are present in the environment or nature.

Properties of Electric Field

Given below are some properties of an electric field.

• The lines present in an electric field never intersect with one another.
• The surface charge is perpendicular to the lines in the electric field.
• When the lines are close to each other, the field is strong, but when the lines are free and are not near together, the field is weak.
• The total lines in the field are exactly proportional to the amount of the charge.
• The lines present in the electric field seem to originate from a positive charge and finish with a negative charge.
• When it comes to a single charge, they seem to start and stop at infinity.
• The line curves of the electric field are continuously in a charge-free zone.

These are the basic theories of the chapter ‘Electric Charges and Fields’. This chapter contains more advanced concepts and theories that students will learn when they go through this chapter. It is advisable for students to practise the Class 12 Physics Chapter 1 Important Questions regularly to learn the advanced concepts so that they will be able to achieve good marks in the exams.

Class 12 Physics Chapter 1 Important Questions

Some questions that may come up in your Physics exams are as follows.

• What do you mean by the electric charge? Explain the team using an experiment.
• What do you mean by the term conductors?
• What do you mean by the term insulators?
• Explain the difference between the conductors and insulators.
• What is Coulomb’s law? Explain it with some examples and experiments.
• What is an electric dipole?

Benefits of Important Questions of Physics Class 12 Chapter 1

These Class 12 Physics Chapter 1 Important Questions will be advantageous for students who are weak in science. These notes will act as a guide for every pupil. Some of the advantages of referring to crucial questions for Class 12 Physics Chapter 1 are as follows.

• The questions which are provided are selected while considering the format and the CBSE syllabus as prescribed because students can lose a lot of marks if those norms are not followed.
• These questions are selected by subject experts.
• The questions are prepared while keeping in mind the intellectual aptitude of students so that they can understand every question.
• The questions that are mentioned in these notes are most likely to appear in the final exams, therefore prepping the students better.

Conclusion

It would be beneficial for students to refer to these Class 12 Physics Chapter 1 Important Questions so that they can understand the concepts better and would be able to achieve good marks.

These CBSE notes include the important questions mentioned in NCERT books. After going through these CBSE revision notes, students will learn all about electric charges and fields and will gain confidence to solve CBSE past years’ question papers.