NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields are provided here to help students prepare well for their Class 12 CBSE board exam, as well as competitive exams like JEE and more. Here, in these NCERT Solutions for Class 12 Physics you can find solved answers for Electric Charges and Fields from Class 12, along with important questions from previous year question papers, exemplary problems, MCQs, worksheets and exercises that will help students to prepare for competitive exams as well.

This chapter’s NCERT Solutions for Class 12 Physics will also help you go through the basics and you should expect at least one question in your exam from this chapter. It is easy to prepare Electric Charges and Fields Class 12 notes by studying. Download now the PDF of NCERT Solutions for Class 12 Physics Chapter 1 from the link provided below.

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Class 12 Physics NCERT Solutions Electric Charges and Fields Important Questions


Q 1.1) 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?

Soln: Given,

The Charge on the 1st sphere and 2nd sphere is q1 = 2 x 10-7 C and q2 = 3 x 10-7 C

The distance between two charges is given by r = 30cm = 0.3m

The electrostatic force between the spheres is given by the relation :

F = 14πϵo.q1q2r2 \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{r^{2}}

Here,

ϵo \epsilon _{o} = permittivity of free space and,

14πϵo \frac{1}{4\pi \epsilon _{o}} = 9 x 109 Nm2C-2

Force, F = 9×109×2×107×3×107(0.3)2 \frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}} = 6 x 10‑3 N.

The force between the charges will be repulsive as they have the same nature.

 

Q 1.2) 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?

Soln:

(a) Given,

The charge on 1st sphere (q1 ) and 2nd sphere (q2) is 0.4 µC or 0.4 × 10-6 C and -0.8 × 10-6C respectively.

The electrostatic force on the 1st sphere is given by F = 0.2N.

Electrostatic force between the spheres is given by the relation :

F = 14πϵo.q1q2r2 \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{r^{2}}

Here,

ϵo \epsilon _{o} = permittivity of free space and,

14πϵo \frac{1}{4\pi \epsilon _{o}} = 9 × 109 Nm2C-2

r2 = 14πϵo.q1q2F \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{F}

= 0.4×106×8×106×9×1090.2 \frac{0.4 \times 10^{-6} \times 8 \times 10^{-6} \times 9 \times 10^{9}}{0.2} = 144 x 10-4

r = 144×104 \sqrt{144 \times 10^{-4}} = 12 x 10-2 = 0.12m

Therefore, the distance between the two spheres = 0.12 m

 

(b) Since the spheres have opposite charges, the force on the second sphere due to the first sphere will also be equal to 0.2N.

 

Q 1.3) 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?

Soln.:

The ratio to be determined is given as follows :

ke2Gmemp \frac{ke^{2}}{Gm_{e}m_{p}}

where G is the gravitational constant in N m2 kg-2

me and mp is the masses of electron and proton in kg.

e is the electric charge (unit – C)

k = 14πϵo \frac{1}{4\pi \epsilon _{o}}  (unit – Nm2C-2)

Therefore, the unit of given ratio,

ke2Gmemp \frac{ke^{2}}{Gm_{e}m_{p}} = [Nm2C2][C2][Nm2kg2][kg][kg] \frac{[Nm^{2}C^{-2}][C^{-2}]}{[Nm^{2}kg^{-2}][kg][kg]} = M0L0T0

So, the given ratio is dimensionless.

Given,

e = 1.6 x 10-19 C

G = 6.67 x 10-11 N m2 kg-2

me = 9.1 x 10-31 kg

mp = 1.66 x 10-27 kg

Putting the above values in the given ratio, we get

ke2Gmemp \frac{ke^{2}}{Gm_{e}m_{p}} = 9×109×(1.6×1019)26.67×1011×9.1×1031×1.67×1027 \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}} = 2.3 x 1039

So, the above ratio is the ratio of the electric force to the gravitational force between a proton and an electron when the distance between them is constant.

 

Q 1.4) (i) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(ii) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Soln.:

(i) The ‘electric charge of a body is quantized’ means that only integral (1, 2, …n)  numbers of electrons can be transferred from a body to another.

Charges cannot get transferred in fractions. Hence, the total charge possessed by a body is only in integral multiples of electric charge.

(ii) In the case of large scale or macroscopic charges, the charge which is used over there is comparatively too huge to the magnitude of the electric charge. Hence, on a macroscopic level, the quantization of charge is of no use Therefore, it is ignored and the electric charge is considered to be continuous.

Q 1.5) 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.

Soln.:

When two bodies are rubbed against each other,  a charge is developed on both bodies. These charges are equal but opposite in nature. And this phenomenon of inducing a charge is known as charging by friction. The net charge on both of the bodies is 0 and the reason behind it is that an equal amount of charge repels it. When we rub a glass rod with a silk cloth, charge with opposite magnitude is generated over there. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

 

Q 1.6) Four point charges qA=2μC,  qB=5μC,  qC=2μC,  andqD=5μCq_{A} = 2 \mu C, \; q_{B} = -5 \mu C, \; q_{C} = 2 \mu C, \; and q_{D} = -5 \mu C 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?

Soln.:

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 6

In the above picture, we have shown the square mentioned in the question. Whose side is 10 cm and four charges are placed at the corners of the squares And O is the centre of the square.

Where,

(Sides) AB = BC = CD = DA = 10 cm

(Diagonals) AC = BD = 102  cm10 \sqrt{2} \; cm

AO = OC = DO = OB = 52  cm5 \sqrt{2} \; cm

At the centre point ‘O’, we have placed a charge of 1  μC1 \; \mu C

In the above case, the repulsive force between the corner A and the centre O is same in magnitude with the repulsive force by the corner C to the centre O, but these forces are opposite in direction. Hence, these forces will cancel each other and from A and C no forces are applied on the centre O. Similarly, from the corner C the attractive force is applying on to the centre O and another force with the same magnitude is applying on the centre O, also these two forces are opposite in direction hence they are also opposing each other.

Therefore, the net force applying to the centre is zero. Because all the forces here are being cancelled by each other.

 

Q 1.7)  (i) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(ii) Explain why two field lines never cross each other at any point.

 

Soln.:

(i) When a charge is placed in an electrostatic field then it experiences a continuous force. Therefore, an electrostatic field line is a continuous curve. And a charge moves continuously and does not jump from on point to the other. So, the field line cannot have a sudden break.

(ii) if two field lines will cross each other at any point then at that point the field intensity will start shooing two directions at the same point which is impossible. Therefore, two field lines can never cross each other.

Q 1.8) Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum.

(i) What is the electric field at the midpoint O of the line AB joining the two charges?

(ii) 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?

Soln.:

(i) The figure given below shows the situation given to us, in which AB is a line and O is the midpoint.

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 8

Distance between two charges, AB = 20 cm

Therefore, AO = OB = 10 cm

Total electric field at the centre is (Point O) = E

Electric field at point O caused by +3  μC+ 3 \; \mu C charge,

E1=14πϵ0.3×106(OA)2=14πϵ0.3×106(10×102)2NC1E _{1} = \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( OA \right )^{2}} = \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( 10 \times 10 ^{-2} \right )^{2}} NC ^{-1}         along OB

Where ϵ0\epsilon _{0} = Permittivity of free space and 14πϵ0=9×109  Nm2C2\frac{1}{ 4 \pi \epsilon _{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}

Therefore,

Electric field at point O caused by 3  μC- 3 \; \mu C charge,

E2=14πϵ0.3×106(OB)2=14πϵ0.3×106(10×102)2NC1E _{2} = \left | \frac {1}{4 \pi \epsilon _{0}}. \frac{- 3 \times 10 ^{-6}}{\left ( OB \right )^{2}} \right | = \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( 10 \times 10 ^{-2} \right )^{2}} NC ^{-1}                 along OB

E1+E2=2×14πϵ0.3×106(10×102)2  NC1∴ E _{1} + E _{2} = 2 \times \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( 10 \times 10 ^{-2} \right )^{2}} \; NC ^{-1}              along OB

[Since the magnitudes of E1  and  E2E _{1} \; and \; E _{2}are equal and in the same direction]

 

E=2×9×109×3×106(10×102)2  NC1=5.4×106NC1∴ E = 2 \times 9 \times 10 ^{9} \times \frac{3 \times 10^{-6}}{\left (10 \times 10^{-2} \right ) ^{2}} \; NC ^{-1} \\ \\ = 5.4 \times 10^{6} NC ^{-1} along OB

Therefore, the electric field at mid – point O is  5.4×106NC15.4 \times 10^{6} NC ^{-1} along OB.

 

(ii) A test charge with a charge potential of 1.5×109  C1.5 \times 10 ^{-9} \; C is placed at mid – point O.

q=1.5×109  Cq = 1.5 \times 10 ^{-9} \; C

Let the force experienced by the test charge be  F

Therefore, F = qE

= 1.5×109×5.4×106=8.1×103  N1.5 \times 10 ^{-9} \times 5.4 \times 10 ^{6} \\ \\ = 8.1 \times 10 ^{-3} \; N

The force is directed along line OA because the negative test charge is attracted towards point A and is repelled by the charge placed at point B. As a result, the force experienced by the test charge is q=8.1×103  Nq = 8.1 \times 10 ^{-3} \; N along OA.

 

Q 1.9) A system has two charges qA=2.5×107  C  and  qB=2.5×107  Cq _{A} = 2.5 \times 10 ^{-7} \;C \; and \; q _{B} = -2.5 \times 10 ^{-7}\;C located at points A :(0, 0, -15 cm) and B (0, 0, + 15 cm), respectively. What is the total charge and electric dipole moment of the system?

Soln.:

The charges which are located at the given points are shown in the co-ordinate system as:

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 9

At point A, total charge amount, qA=2.5×107  Cq _{A} = 2.5 \times 10 ^{-7} \;C

At point B, total charge amount, qB= –2.5×107  Cq _{B} =  – 2.5 \times 10 ^{-7} \;C

Total charge of the system is, qA+qB=2.5×107  C2.5×107  C=0q _{A} + q_{B}= 2.5 \times 10 ^{-7} \;C – 2.5 \times 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=2.5×107×0.3=7.5×108  Cp = q _{A} \times d = q_{B} \times d = 2.5 \times 10 ^{-7} \times 0.3\\ \\ = 7.5 \times 10 ^{-8} \; C m along positive z – axis

Therefore, the electric dipole moment of the system is  7.5×108  C7.5 \times 10 ^{-8} \; C m along positive z−axis.

 

Q 1.10) An electric dipole with dipole moment 4×109   Cm4 \times 10^{-9}\;  C\, m is aligned at 30° with the direction of a uniform electric field of magnitude 5×104  NC15 \times 10^{4}\; N C^{-1}. Calculate the magnitude of the torque acting on the dipole.

Soln.:

Electric dipole moment, p=4×109  Cmp = 4 \times 10^{-9}\;C\,m

Angle made by p with a uniform electric field, θ=30\theta = 30^{\circ}

Electric field, E=5×104  NC1E = 5 \times 10^{4}\; NC^{-1}

Torque acting on the dipole is given by the relation,

τ=pEsinθ=4×109×5×104×sin30=20×105×12=104  Nm\tau = pE \sin \theta\\ \\ = 4 \times 10^{-9} \times 5 \times 10^{4} \times \sin 30 = 20 \times 10 ^{-5} \times \frac {1}{2} = 10^{-4}\;Nm

Therefore, the magnitude of the torque acting on the dipole is 104  Nm10^{-4}\;Nm

 

Q 1.11) A polythene piece rubbed with wool is found to have a negative charge of 3×107  C3 \times 10^{-7}\; C.

(i) Estimate the number of electrons transferred (from which to which?)

(ii) Is there a transfer of mass from wool to polythene?

 

Soln.:

(i) Since the wool is positively charged and the polythene is negatively charged, so we can say that few amounts of electrons are transferred from wool to polythene.

Charge on the polythene, q = 3×107  C3 \times 10^{-7}\; C.

Amount of charge on an electron, e=1.6×1019  Ce = -1.6 \times 10^{-19}\; C

Let number of electrons transferred from wool to polythene be n

So, by using the given equation we can calculate the value of n,

q = ne

n=qe=3×1071.6×1019=1.87×1012\Rightarrow n = \frac{q}{e} = \frac{-3 \times 10 ^{-7}}{-1.6 \times 10 ^{-19}} = 1.87 \times 10 ^{12}

 

Therefore, the number of electrons transferred from wool to polythene is 1.87×10121.87 \times 10 ^{12}

 

(ii) Yes,

Mass is also transferred as an electron is transferred from wool to polythene and an electron particle have some mass.

Mass of an electron, me=9.1×1031kgm_{e} = 9.1 \times 10 ^{-31} kg

Total mass transferred , m = me×n=9.1×1031×1.87×1012=1.701×1018  kgm_{e} \times n \\ \\ = 9.1 \times 10 ^{-31} \times 1.87 \times 10^{12} \\ \\ = 1.701 \times 10^{-18} \; kg

Here, the mass transferred is too low that it can be neglected.

 

Q 1.12) (i) 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 6.5×107  C6.5 \times 10^{-7} \; C each?
The radii of A and B are negligible compared to the distance of separation.

(ii) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

 

Soln.:

(i) Charge on sphere A, qA=6.5×107  Cq_{A} = 6.5 \times 10^{-7} \; C

Charge on sphere B, qB=6.5×107  Cq_{B} = 6.5 \times 10^{-7} \; C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres, F=14πϵ0.qAqBr2F = \frac{1}{4 \pi \epsilon _{0}}. \frac{q _{A} q_{B}}{r ^{2}}

Where ϵ0\epsilon _{0} = Permittivity of free space and 14πϵ0=9×109  Nm2C2\frac{1}{ 4 \pi \epsilon _{0}} = 9 \times 10^{9} \; Nm^{2} C^{2}

Therefore,

F=9×109×(6.5×107)2(0.5)2=1.52×102  NF = \frac{9 \times 10 ^{9} \times \left ( 6.5 \times 10^{-7} \right )^{2}}{\left ( 0.5 \right )^{2}}\\ \\ = 1.52 \times 10^{-2}\; N

Therefore, the force between the two spheres is 1.52×102  N1.52 \times 10^{-2}\; N

 

(ii) After doubling the charge,

Charge on sphere A, qA=1.3×106  Cq_{A} = 1.3 \times 10^{-6} \; C

Charge on sphere B, qB=1.3×106  Cq_{B} = 1.3 \times 10^{-6} \; C

The distance between the spheres is halved.

r=0.52=0.25  m∴ r = \frac {0.5}{2} = 0.25 \; m

Force of repulsion between the two spheres,

F=14πϵ0.qAqBr2=9×109×1.3×106×1.3×106(0.25)2=16×1.52×102=0.243  NF = \frac{1}{4 \pi \epsilon _{0}}. \frac{q _{A} q_{B}}{r ^{2}} = \frac{9 \times 10 ^{9} \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6} }{\left ( 0.25 \right )^{2}}\\ \\ = 16 \times 1.52 \times 10^{-2} \\ \\ = 0.243 \; N

Therefore, the force between the two spheres is 0.243 N.

 

Q 1.13) Suppose the spheres A and B in Exercise 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?

 Soln.:

Distance between the spheres, A and B, r = 0.5 m

Initially, the charge on each of sphere q=6.5×107  Cq = 6.5\times 10^{-7} \; C

When the sphere A is touched with An uncharged sphere C, then half of the charge will be transferred to the sphere C. Hence the charge on both the spheres A and C will be q/2.

After that when sphere C with charge q/2 is brought in touch with sphere B with charge q, then charge on each of the sphere will be divided in two equal parts, is.

12(q+q2)=3q4\frac{1}{2}\left ( q + \frac{q}{2} \right ) = \frac{3q}{4}

Hence, charge on each of the spheres, C and B , is 3q4 \frac{3q}{4}

Force of repulsion between sphere A and B is:

F=14πϵ0.qAqBr2=14πϵ0.q2×3q4r2=14πϵ0.3q28r2=9×109×3×(6.5×107)28×(0.5)2=5.703×103  N F = \frac{1}{4 \pi \epsilon _{0}}. \frac{q _{A} q_{B}}{r ^{2}} = \frac{1}{4 \pi \epsilon _{0}}. \frac{\frac{q}{2} \times \frac{3q}{4}}{r^{2}} \\ \\ = \frac{1}{4 \pi \epsilon _{0}}. \frac{3q^{2}}{8r^{2}} \\ \\ = \frac{9 \times 10^{9} \times 3 \times \left ( 6.5 \times 10^{-7} \right )^{2}}{8 \times \left ( 0.5 \right )^{2}} = 5.703 \times 10^{-3} \; N

Therefore, the force of attraction between the two spheres is 5.703×103  N5.703 \times 10^{-3} \; N

 

Q 1.14) The figure below 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?

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 14

Soln.:

We can see here that the particles 1 and 2 are moving in the direction of the positive charge and we know that the opposite charge attracts each other and the same charge repels each other. So, here we can say that the charged particles 1 and 2 which are going towards the charged particle are negatively charged. And the particle 3 is being attracted towards the negative charge. So, the particle 3 will be the positively charged particle.

The EMF or charge to mass ratio is directly proportional to the amount of depletion or depletion at a given velocity. Here we can see that the particle 3 is depleting more as compared to the other two. Therefore, it will have a higher charge to mass ratio.

 

Question 1.15)

Consider a uniform electric field E = 3 × 10 3 î N/C.

(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the y z – plane?

(b) What is the flux through the same square if the normal to its plane makes a 60 ° angle with the x-axis?

Soln. :

(a) Electric field intensity, E = 3 × 10 3 î  N / C

Magnitude of electric field intensity, | E | = 3 × 10 3 N / C

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

Area of the square, A = s 2 = 0.01 m 2

The plane of the square is parallel to the y – z plane. Hence, angle between the unit vector normal to the plane and electric field, θ = 0 °

Flux ( φ ) through the plane is given by the relation,

φ = | E |A cos θ

φ  = 3 × 10 3 × 0.01 × cos 0 °

φ = 30 N m 2 /C

(b) Plane makes an angle of 60 ° with the x – axis.

Hence, θ = 60°

Flux, φ = | E |A cos θ

Flux, φ  = 3 × 10 3 × 0.01 × cos 60°

Flux, φ  = 30 x 0.5

Flux, φ  = 15 N m 2 /C

 

Question 1.16)

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?

Soln.:

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.

 

Question 1.17)

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 × 10 3 N m 2 /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?

Soln.:

(a) Net outward flux through the surface of the box, φ = 8.0 × 10 3 N m 2 /C

For a body containing net charge q, flux is given by the relation,

ϕ=qε0\phi = \frac{ q }{ \varepsilon _{ 0 }}

 

ε0=permitivityoffreespace\varepsilon _{ 0 } = permitivity of free space

 

ε0=8.854×1012N1C2m2\varepsilon _{ 0 } = 8.854 \times 10 ^{ – 12 } N^{ – 1 } C ^{ 2 } m ^{ – 2 }

 

q=ε0ϕq = \varepsilon _{ 0 } \phi

= 8.854 x 10 – 12 x 8.0 x 10 3 C = 7.08 x 10 – 8 C = 0.07 μC\mu C

Thus, the total charge inside the box is 0.07 μC\mu C

(b) No

Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

 

Question 1.18)

A point charge + 10 μC is at 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 )

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 18

 Soln.:

The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.

ϕTotal=qϵ0\phi _{ Total } = \frac{ q }{ \epsilon _{0}}

Hence, electric flux through one face of the cube i.e., through the square is

ϕ=ϕTotal6=16.qϵ0\phi = \frac{\phi _{ Total}}{6} = \frac{ 1 }{ 6 }. \frac{ q }{ \epsilon _{ 0 }}

Here,

ϵ0{ \epsilon _{ 0 }} = permittivity of free space = 8.854 x 10 – 12 N – 1 C 2 m – 2

q=10μC=10×106Cq = 10 \mu C = 10 \times 10 ^{ – 6 } C

Therefore,

ϕ=16.10×1068.854×1012\phi = \frac{ 1 }{ 6 } . \frac{ 10 \times 10 ^{ – 6 }}{ 8.854 \times 10 ^{ – 12 }}

 

ϕ\phi = 1.88 x 10 5 N m 2 C – 1

Therefore, electric flux through the square is 1.88 × 10 5 N m 2 C – 1

 

Question 1.19)

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?

Soln.:

Net electric flux ( φ Net ) through the cubic surface is given by

ϕnet=qϵ0\phi _{ net } = \frac{ q }{ \epsilon _{ 0 }}

Here,

ϵ0{ \epsilon _{ 0 }} = permittivity of free space = 8.854 × 10 – 12 N – 1 C 2 m – 2

q=total  charge  contained  in  the  cube  given  =  2.0μ  C  =  2×106Cq = total \;charge\; contained \;in \;the\; cube\; given \;= \;2.0 \mu\; C \;=\; 2 \times 10 ^{ – 6 } C

Therefore,

ϕnet=2×1068.854×1012\phi _{ net } = \frac{ 2 \times 10 ^{ – 6 }}{ 8.854 \times 10 ^{ – 12 }}

 

ϕnet=2.26×105Nm2C1\phi _{ net } = 2.26 \times 10 ^{ 5 } N m ^{ 2 } C ^{ – 1}

The total electric flux through the surface of the cube given is , ϕnet=2.26×105Nm2C1\phi _{ net } = 2.26 \times 10 ^{ 5 } N m ^{ 2 } C ^{ – 1}

 

Question 1.20)

A point charge causes an electric flux of – 1.0 × 10 3 N m 2 /C to pass through a spherical Gaussian surface of 10.0 cm radius centred 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?

Soln. :

(a) Electric flux, Φ = – 1.0 × 10 3 N m 2 /C

Radius of the Gaussian surface, r = 10.0 cm

Electric flux penetrating through a surface depends on the net charge enclosed inside the body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., – 10 3 N m 2 /C.

(b) Electric flux is given by the relation

ϕTotal=qϵ0\phi _{ Total } = \frac{ q }{ \epsilon _{0}}

Here,

ϵ0{ \epsilon _{ 0 }} = permittivity of free space = 8.854 x 10 – 12 N – 1 C 2 m – 2

q = total charge contained enclosed by the spherical surface = φ ε 0

q = – 1.0 x 10 3 x 8.854 x 10 – 12

q = – 8.854 x 10 – 9 C

q = – 8.854 n C

Therefore, the value of the point charge is – 8.854 n C.

 

Question 1.21)

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 × 10 3 N/C and points radially inward, what is the net charge on the sphere?

 Soln. :

Electric field intensity ( E ) at a distance ( d ) from the centre of a sphere containing net charge q is given by the relation,

E=14πϵ0.qd2E = \frac{ 1 }{ 4\pi \epsilon _{ 0 }} . \frac{ q }{ d ^{ 2 }}

Where, q = Net charge = 1.5 × 10 3 N/C

d = Distance from the centre = 20 cm = 0.2 m

ε 0 = Permittivity of free space and 14πϵ0=9×109Nm2C2\frac{ 1 }{ 4 \pi \epsilon _{ 0 }} = 9 \times 10 ^{ 9 } N m ^{ 2 } C ^{ – 2 }

Therefore, the net charge on the sphere is 6.67 n C.

 

Question 1.22)

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC /m 2

(a) Find the charge on the sphere.

(b) What is the total electric flux leaving the surface of the sphere?

Soln. :

(a) The diameter of the sphere, d = 2.4 m

The radius of the sphere, r = 1.2 m

Surface charge density, σ = 80.0 μC /m 2 = 80 × 10 – 6 C/m 2

The total charge on the surface of the sphere can be calculated as follows,

Q = Charge density × Surface area

= σ × 4πr 2

= 80 × 10 – 6 × 4 × 3.14 × ( 1.2 ) 2

= 1.447 × 10 – 3 C

Therefore, the charge on the sphere is 1.447 × 10 – 3 C.

(b) Total electric flux ( φ Total  ) leaving out the surface of a sphere containing net charge Q is given by the relation,

ϕTotal=Qϵ0\phi _{ Total } = \frac{ Q }{ \epsilon _{0}}

Here,

ϵ0{ \epsilon _{ 0 }} = permittivity of free space = 8.854 x 10 – 12 N – 1 C 2 m – 2

Electric flux = 1.63 x 10 8 N C – 1 m 2

Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10 8 N C – 1 m 2

 

Question 1.23)

An infinite line charge produces a field of 9 × 10 4 N/C at a distance of 2 cm. Calculate the linear charge density.

Soln. :

Electric field produced by the infinite line charges at a distance d having linear charge density λ is given by the relation,

E=λ2πϵ0dE = \frac{ \lambda }{ 2 \pi \epsilon _{ 0 } d }

 

λ=2πϵ0dE\lambda = 2 \pi \epsilon _{ 0 } d E

Here,

d = 2 cm = 0.02 m

E = 9 x 10 4 N / C

ε 0 = Permittivity of free space and 14πϵ0=9×109Nm2C2\frac{ 1 }{ 4 \pi \epsilon _{ 0 }} = 9 \times 10 ^{ 9 } N m ^{ 2 } C ^{ – 2 } λ=0.02×9×1042×9×109λ=10μC/m\\ \lambda = \frac{0.02\times 9\times 10^{4} }{2\times 9\times 10^{9} } \\ \lambda = 10 \mu C/m

Question 1.24)

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 /m 2

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?

Soln. :

The situation given in the question can be depicted through the following figure.

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 24

Parallel plates A and B are placed close to each other as shown in the figure. The region between plates A and B is labelled as II, and outer region of plate A is labelled as I, outer region of plate B is labelled as III.

The charge density of plate A can be calculated as,

σ = 17.0 × 10 – 22 C/m 2

Similarly, the charge density of plate B can be calculated as,

σ = – 17.0 × 10 – 22 C/m 2

In regions, I and III, electric field E is zero. This is because the charge is not enclosed by the respective plates.

Electric field E in region II is given by the relation,

E=σϵ0E = \frac{\sigma }{ \epsilon _{ 0 }}

here,

ε 0 = Permittivity of free space = 8.854 × 10 – 12 N – 1C 2 m – 2

E=17.0×10228.854×1012E = \frac{ 17.0 \times 10 ^{ – 22 }}{ 8.854 \times 10 ^{ – 12 }} = 1.92 x 10 – 10 N/C

Therefore, electric field between the plates is 1.92 x 10 – 10 N/C

Question 1.25)

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10NC–1 (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).

Soln :

Excess electron on an oil drop, n = 12

Electric field intensity, E= 2.55 × 10NC–1

Density of oil, ρ = 1.26 g cm–3

Acceleration due to gravity, g = 9.81 m s–2

Charge of an electron, e = 1.60 × 10–19 C

Radius of the oil drop = r

Force due to the electric field will be equal to the weight of the oil drop (W)

F = W

Eq = mg

Here,

q = ne = net charge on the oil drop

m = mass of the oil drop

= volume of the oil drop x density of the oil

= (4/3)πr3 x ρ

Ene= (4/3)πr3 x ρ x g

r=3Ene4πρg3r = \sqrt[3]{\frac{3Ene}{4\pi \rho g }}

 

r=3×2.55×104×12×1.6×1.6×10194×3.14×1.26×103×9.813r = \sqrt[3]{\frac{3\times 2.55\times 10^{4}\times 12\times 1.6\times 1.6 \times 10^{-19}}{4\times 3.14\times 1.26\times 10^{3}\times 9.81 }}

 

=946.09×10213= \sqrt[3]{946.09\times 10^{-21}}

= 9.82 x 10-7 m

The radius of the oil drop is 9.82 x 10-7 m

Question 1.26)

Which among the curves shown in Fig. cannot possibly represent electrostatic field lines?

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 26

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 26 (d and e)

Soln:

a) The field lines are not normal to the surface of the conductor. Therefore, it does not represent electrostatic field lines.

(b) In this figure, the field lines are emerging from the negative charge and terminating at the positive charge. This is not possible. Therefore, it does not represent electrostatic field lines.

(c) The electric field is emerging from the positive charge and it repels each other. Therefore, the field lines shown represent electrostatic field lines.

(d) The field lines shown intersect each other. Therefore, the field lines shown does not represent electrostatic field lines.

(e) The closed loops are formed in the area between the field lines. Therefore, it does not represent electrostatic field lines.

Question 1.27)

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?

Soln:

Total dipole moment of the system, p=q×dl=−10−7 cm

The rate of increase of the magnitude of the electric field along the positive z-direction = 105 NC–1 per metre.

The force experienced by the system is given as

F=qE

F=q(dE/dl) x dl
= p x (dE/dl)
=- 10−7 ×10−5

=−10−2N

The force experienced by the system is – 10−2 N. This is in the negative z-direction and opposite to the direction of the electric field. Therefore, the angle between the dipole moment and the electric field is 180o.
Torque (τ) =pEsin1800=0

Therefore, the torque experienced by the system is zero.

Question 1.28)

(a) A conductor A with a cavity as shown in Fig. (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. (b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 28

Soln:

(a) A Gaussian surface is considered within the conductor that encloses the cavity. Within the conductor, the electric field intensity (E) will be zero.

The charge inside the conductor is q and the permittivity of free space is ∈

According to Gauss’s law,

Flux, ϕ=E.ds = q/∈
Electric field intensity, E=0

q/∈= 0

The permittivity of the free space, ∈≠ 0

Therefore, the charge inside the conductor, q=0

Therefore, the entire charge appears on the outer surface of the conductor.

(b) A conductor B is kept in the cavity and it is insulated from A. Hence a charge – q will be induced in the inner surface of conductor A, this in turn will induce a charge +q on the outer surface of A.  Therefore, the total charge on the outer surface of A will be equal to Q+q.

(c) The sensitive instrument can be shielded from the electrostatic field by keeping it fully enclosed inside a metallic surface.

Question 1.29)

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is σ2ϵ0n^\frac{\sigma }{2\epsilon _{0}}\hat{n}, where nˆ is the unit vector in the outward normal direction and σ is the surface charge density near the hole

Soln:

Considering that the hole is filled up, then the electric field intensity at a point close to the surface of the conductor is given by Gauss’s law,

Flux, ϕ=E.ds=qϵ0\phi =E.ds=\frac{q}{\epsilon _{0}}

q = σ x ds

here, σ is the surface charge density

ds is the area of the Gaussian surface

Eds= (σ x ds)/∈0

E = σ /∈

=  σϵ0n^\frac{\sigma }{\epsilon _{0}}\hat{n}

Let E1 be the electric field due to the hole and E2 is the electric field due to the rest of the conductor. E1 and E2 are opposite in direction and equal in magnitude since the total electric field inside the conductor is zero.

E1=E2\left | E_{1} \right |=\left | E_{2} \right | ——-(1)

The electric field E = σ /∈is due to the superposition of E1 and E2.

E = E1 + E2

From equation (1) we know E1 = E2

Therefore, E = 2E1

E1 = E/2

Therefore, the electric field in the hole

E1σ2ϵ0n^\frac{\sigma }{2\epsilon _{0}}\hat{n}

Question 1.30)

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Soln:

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 30

Consider a long thin wire of uniform linear charge density λ. A point P is taken at a distance r from the midpoint C of the wire. The electric field at a point P due to the wire is E.

Let E be the electric field at point P due to the wire

Consider a small length element dx on the wire with centre O, such that OC = x

Let q be the charge on this element, q = λdx

The electric field at A due to the element is

dE=14πϵ0λdxOP2dE = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda dx}{OP^{2}}

OP2 = r2 + x2

dE=14πϵ0λdx(r2+x2)dE = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda dx}{(r^{2}+x^{2})}

The electric field (dE) is resolved into two rectangular components dEcosθ and dEsinθ . dEsinθ is the parallel component and dEcosθ is the perpendicular component. When the whole wire is considered, the parallel component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point P.

dE=14πϵ0λdxcosθ(r2+x2)dE = \frac{1}{4\pi \epsilon _{0}}\frac{\lambda dx cos\theta }{(r^{2}+x^{2})} ——(1)

In the ΔOCP, x = r tanθ —–(2)

Differentiating equation (1) we get

dx = rsec2θdθ

r2 + x2 = r2 + r2 tan2θ

= r2 (1+ tan2θ)  = r2 sec2θ

Substituting in (1)

dE=λrsec2θdθ4πϵ0r2sec2θcosθdE= \frac{\lambda rsec^{2}\theta d\theta }{4\pi \epsilon_{0}r^{2}sec^{2}\theta }cos\theta

 

dE=λ4πϵ0rcosθdεθdE= \frac{\lambda }{4\pi \epsilon_{0}r }cos\theta d\varepsilon \theta

The wire is of infinite length. Therefore, A and B are infinite distance apart. Therefore, θ varies from -π/2 to +π/2. The electric field intensity at the point P due to the whole wire is

E=π/2π/2λ4πϵ0rcosθdθE = \int_{-\pi /2}^{\pi /2}\frac{\lambda }{4\pi \epsilon _{0}r}cos\theta d\theta

λ4π0r[sinθ]π/2π/2=λ2π0r\frac{\lambda}{4 \pi \in_{0} r}[\sin \theta]_{-\pi / 2}^{\pi / 2}=\frac{\lambda}{2 \pi \in_{0} r}

Question 1.31)

It is now established 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.

Soln:

The net charge of a proton is +e. The value +e can be got when p=(23+2313)ep = \left ( \frac{2}{3}+\frac{2}{3}-\frac{1}{3} \right )e

Therefore, the proton has two up quark and one down quark

The net charge of a neutron is 0. This can be got when n=(231313)en = \left ( \frac{2}{3}-\frac{1}{3}-\frac{1}{3} \right )e = 0

Therefore, the neutron has one up quark and two down quark.

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.

Soln:

(a) Let us consider that the equilibrium is stable. Then if the test charge is displaced in any direction it will experience a restoring force towards the null-point. This means that all the field lines near the null point will be directed towards the null point. That is, there is a net inward flux of electric field through a closed surface around the null-point. But according to Gauss’s law, the flux of electric field through a surface, not enclosing any charge, must be zero. Hence, equilibrium cannot be stable.
(b) The null-point is the mid-point of the line joining the two charges. If the test charge is displaced along the line from the null-point there is a restoring force. If the test charge is displaced normal to the line the net force takes it away from the null-point. Hence for the stability of equilibrium needs restoring force in all directions.

Question 1.33)

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 the figure in question 1.14). The length of the plate is L and a 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 vx2). Compare this motion with the motion of a projectile in a gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Soln:

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Question 33

Mass of the particle = m

Charge on a particle  =−q

Velocity of the particle =vx

Length of the plates =L

Electric field between the plates =E

Mechanical force, F = Mass (m) × Acceleration (a)

a = F/m

Electric force, F=qE

Therefore, acceleration, a = qE/m ——–(1)

Time required for the particle to field of length L is given by,
t= distance/time

= Length of the plate/Velocity of the particle = L/vx
​In the vertical direction, initial velocity, u=0

The vertical deflection of the particle can be got from the equation

s=ut+(1/2)at2

s=0+12(qEm)(Lvx)2s = 0 + \frac{1}{2}\left ( \frac{qE}{m} \right )\left ( \frac{L}{v_{x}} \right )^{2}

 

s=12(qEm)(Lvx)2s = \frac{1}{2}\left ( \frac{qE}{m} \right )\left ( \frac{L}{v_{x}} \right )^{2}

This is similar to the motion of horizontal projectiles under gravity.

Question 1.34)

Suppose that the particle in Exercise 1.33 is an electron projected with velocity v= 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.)

Soln:

Velocity of the electron, v= 2.0 × 106 m s–1

Separation between the plates, d = 0.5 cm = 0.5 x 10-2 m

Electric field between the plates = 9.1 × 102 N/C

s=12(qEm)(Lvx)2s = \frac{1}{2}\left ( \frac{qE}{m} \right )\left ( \frac{L}{v_{x}} \right )^{2}

The distance at which the electron strike the upper plate is

L=2dmvx2qEL = \sqrt{\frac{2dmv_{x}^{2}}{qE}}

 

L=2×0.005×9.1×1031×(2×106)21.6×1019×9.1×102L = \sqrt{\frac{2\times 0.005\times 9.1\times 10^{-31}\times (2\times 10^{6})^{2}}{1.6\times 10^{-19}\times 9.1\times 10^{2}}}

= 0.364×101914.56×1017\sqrt{\frac{0.364 \times 10^{-19}}{14.56 \times 10^{-17}}}

= 2.5×104\sqrt{2.5 \times 10^{-4}}

= 1.6 cm

The electron strikes the upper plate after 1.6 cm

 

Our NCERT Solutions for Class 12 Physics deals with questions regarding the linear charge density of an infinite line charge and many other questions regarding the electric field. This chapter is one of the most important chapters for Class 12 students as competitive exams might ask you questions from this chapter too.

Class 12 Physics NCERT Solutions for Electric Charges

In NCERT Solutions for chapter Electric Charges and Fields, we will be solving questions on how to calculate forces between two charged particles, which are kept at a certain distance from each other. We will be explaining the reason behind the charge appearing in materials due to rubbing against each other just like when you rub a glass rod with a silk cloth. Do you know why we ignore quantization of charge when it comes to large scale charges? Find it out here.

Subtopics of Class 12 Physics Chapter 1 Electric Charges and Fields

  1. Introduction
  2. Electric Charge
  3. Conductors And Insulators
  4. Charging By Induction
  5. Basic Properties Of Electric Charge
    1. Additivity of Charges
    2. Charge is Conserved
    3. Quantisation of Charge
  6. Coulomb’s Law
  7. Forces Between Multiple Charges
  8. Electric Field
    1. Electric field due to a system of charges
    2. The physical significance of the electric field
  9. Electric Field Lines
  10. Electric Flux
  11. Electric Dipole
    1. The field of an electric dipole
    2. The physical significance of dipoles
    3. Dipole in a Uniform External Field
  12. Continuous Charge Distribution
  13. Gauss’s Law
  14. Applications Of Gauss’s Law
  15. Field due to an infinitely long straight uniformly charged wire
    1. Field due to a uniformly charged infinite plane sheet
    2. Field due to a uniformly charged thin spherical shell

One question shows you how much charge applied to the centre point of a square where four charges are kept in four edges.

Can two field lines cross each other? Find out here in this chapter. An electrostatic field is a continuous curve, find the reason why. We will solve questions on finding the total charge and dipole moment of the system along with torque acting on the dipole and electric field of a dipole. When you rub wool with polythene, do you think the mass was transferred from wool to polythene, find out the answer below?

What will happen when the charge on the two spheres which are kept nearby is doubled and the distance is reduced, we will be showing you the answer in one of the questions below. We will be facing questions on the flux of the field through a square, cube and even a surface. We will even be seeing net charge inside a box and what will happen if the Gaussian surface is doubled and what is the amount of flux passing through the surface.

Class 12 Physics is tough for many students and they struggle to get a good score in the Class 12 Physics examination. In order to score good marks students are advised to follow the NCERT Solutions provided here. It will build much-needed confidence in you as it gives knowledge of the types of questions appearing in the examination.

Why Opt for BYJU’S?

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Frequently Asked Questions on NCERT Solutions for Class 12 Physics Chapter 1

Is the NCERT Solutions for Class 12 Physics Chapter 1 sufficient for the exam preparation?

The NCERT Solutions for Class 12 Physics Chapter 1 are designed by the faculty at BYJU’S to help students ace the exam without fear. The fundamental concepts are explained in the most systematic way to improve confidence among students. Each and every minute detail is covered in the NCERT Solutions to help students with their exam preparation. The solutions are available in both online and offline mode, which can be used by the students based on their requirements.

Explain Coulomb’s law covered in the Chapter 1 of NCERT Solutions for Class 12 Physics.

According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges. This concept is briefly explained in the NCERT Solutions for Class 12 Physics Chapter 1 curated by the experts at BYJU’S. The solutions are elaborated in a simple language to make it easier for the students while learning.

What are the main topics covered in Chapter 1 of NCERT Solutions for Class 12 Physics?

The main topics covered in Chapter 1 of NCERT Solutions for Class 12 Physics are listed here:
1. Introduction
2. Electric Charge
3. Conductors And Insulators
4. Charging By Induction
5. Basic Properties Of Electric Charge
6. Coulomb’s Law
7. Forces Between Multiple Charges
8. Electric Field
9. Electric Field Lines
10. Electric Flux
11. Electric Dipole
12. Continuous Charge Distribution
13. Gauss’s Law
14. Applications Of Gauss’s Law
15. Field due to an infinitely long straight uniformly charged wire

 

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