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

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

Edited By Safeer PP | Updated on Sep 14, 2022 09:34 AM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 1 have been prepared by experts to help students understand the basic terms and different elements associated with the practical uses of electricity. NCERT Exemplar Class 12 Physics chapter 1 solutions revolve around field, force and potential surfacing out of charges that are not moving or changing with time i.e. static charges. The overall chapter deals with static charges in different situations which helps to understand their fields, forces, positions and potentials with the help of different examples given.

This Story also Contains
  1. NCERT Exemplar Class 12 Physics Solutions Chapter 1 MCQI
  2. NCERT Exemplar Class 12 Physics Solutions Chapter 1 Very Short Answer
  3. NCERT Exemplar Class 12 Physics Solutions Chapter 1 Short Answer
  4. NCERT Exemplar Class 12 Physics Solutions Chapter 1 Long Answer
  5. Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 1 Electric Charges and Fields
  6. Important Topics, Sub-Topics, and Terms of NCERT Exemplar Class 12 Physics Solutions Chapter 1 Electric Charges and Fields
  7. NCERT Exemplar Class 12 Physics Chapter Wise Links
  8. Important Topics To Cover For Exams From NCERT Exemplar Class 12 Physics Solutions Chapter 1 Electric Charges and Fields

For the purpose of learning and further use, students can avail NCERT Exemplar Class 12 Physics solutions chapter 1 PDF download, that have been prepared by experts after thorough study of the concepts.

Also, read - NCERT Class 11 Physics Syllabus

NCERT Exemplar Class 12 Physics Solutions Chapter 1 MCQI

Question:1

In Fig.1.1, two positive charges q2 and q3 fixed along the y axis, exert a net electric force in the + x direction on a charge q1 fixed along the x axis. If a positive charge Q is added at (x, 0), the force on q1


A. shall increase along the positive x-axis.
B. shall decrease along the positive x-axis.
C. shall point along the negative x-axis.
D. shall increase but the direction changes because of the intersection of Q with q2 and q3.

Answer:

The answer is the option (a)
Explanation: The total force acting on a given charge is the vector sum of individual forces exerted on it by other charges.

Here, the net electrostatic force is in positive x-direction by q2 and q3 on qx. The nature of force between q1 q2 and q1q3 is attractive. Hence qx is negative.
If possible charge Q is placed at (x,0) the nature of force is attractive between q1 and Q while in the positive x-direction. Therefore, net force on qx due to q2, q3 and Q is also in the same direction. Hence, the force in qx should increase with the addition of Q at (x,0).

Question:2

A point positive charge is brought near an isolated conducting sphere (Fig. 1.2). The electric field is best given by

A. Fig (i)
B. Fig (ii)
C. Fig (iii)
D. Fig (iv)

Answer:

The answer is the option (a)
Explanation: The electric field lines are always perpendicular to the conducting surface. There are no lines inside the conducting surface, and the electric field lines are always coming out of positive charge and goes into a negative charge. Therefore (a) is the correct answer.

Question:3

The electric flux through the surface

A. in fig. 1.3 (iv) is the largest
B. in fig. 1.3 (iii) the least
C. in fig 1.3 (ii) is same as (iii) but samaller han (iv)
D. is same for all figure

Answer:

Answer: The answer is the option (d)
Explanation: According to Gauss' theorem, the total electric flux out of an enclosed area is equal to charge enclosed divided by permittivity.
Here all the areas are equal; therefore, the equal flux will pass through all of them. Hence, (d) is correct.

Question:4

Five charges q1, q2, q3, q4, and q5 are fixed at their positions, as shown in Fig. 1.4. S is a Gaussian surface. Gauss's law is given by
\oint_{s} E.ds=\frac{q}{\varepsilon _0}
Which of the following statements is correct?

A. E on the LHS of the above equation will have a contribution from q1, q5 and q3 while q on the RHS will have a contribution from q2 and q4 only.
B. E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q2 and q4 only.
C. E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q1, q3 and q5 only.
D. Both E on the LHS and q on the RHS will have contributions from q2 and q4 only..
Which of the following statements is correct?

Answer:

The answer is the option (b)
Explanation: The E on LHS of the vector sum of all the five charges while the q on RHS is the normal sum of charges enclosed by the surface, therefore, (b) is correct.

Question:5

Figure 1.5 shows electric field lines in which an electric dipole p is placed as shown. Which of the following statements is correct ?

A. The dipole will not experience any force.
B.The dipole will experience a force towards right.
C.The dipole will experience a force towards left.
D.The dipole will experience a force upwards.

Answer:

The answer is the option (c)
Explanation: Here, the electric field lines are non-uniform, and their strength decreases from left to right. Therefore, the force on -q is more than +q. So, the dipole will experience a force towards the left.

Question:6

A point charge +q, is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of the plane is
A. directed perpendicular to the plane and away from the plane.
B. directed perpendicular to the plane but towards the plane.
C. directed radially away from the point charge.
D. directed radially towards the point charge.

Answer:

The answer is the option (a)
Explanation: The electric field lines are always parallel to the conducting plane; therefore, the lines here will be and directing away from the plane. There are no conducting lines inside the conductor. Hence, (a) is correct.

Question:7

A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed.
A. perpendicular to the diameter
B. parallel to diameter
C. at an angle tilted towards the diameter
D. at an angle tilted away towards the diameter

Answer:

Explanation: In a uniformly positively charged hemisphere, the electric field should be perpendicular on a point away from the center towards the diameter.

NCERT Exemplar Class 12 Physics Solutions Chapter 1 MCQII

Question:8

If \oint _{s} Eds =0over a surface, then

A. The electric field inside the surface and on it is zero.
B. The electric field inside the surface is necessarily uniform.
C. The number of flux lines entering the surface must be equal to the number of flux lines leaving it
D. All charge must necessarily be outside the surface.

Answer:

The correct answers are the options (c,d)
Explanation: \phi _{s}= 0represents that number is electric field lines entering an area is equal to the number of electric field lines leaving. And also, here q=0; therefore the charge must be outside the enclosed area.

Question:9

The Electric field at a point is
A. always continuous.
B. continuous if there is no charge at that point.
C. discontinuous only if there is negative charge at that point.
D. discontinuous if there is a charge at that point.

Answer:

The correct answers are the options (b,d)
Explanation: Electric field lines cannot be defined at a point where a charge is present. While at any point electric field will be continuous if only one charge is present in the medium.

Question:6

If there were only one type of charge in the universe, then
A. \oint_{s}E.ds\neq 0 on any surface
B. \oint_{s}E.ds= 0if the charge is outside the surface.
C. \oint_{s}E.dscould not be defined.

D.\oint_{s}E.ds= \frac{q}{\varepsilon _0} if charge of magnitude q were inside the surface.

Answer:

The correct answers are the options (b,d)
Explanation:\phi _{s}E . dS= \frac{q}{\varepsilon _0} from Gauss' law. Therefore, if \phi _{s}E . dS= 0 then q=0, therefore the net enclosed charge should be 0. And if the charge is outside the surface then also q=0. Therefore, (b,d) are correct.

Question:11

Consider a region inside which there are various types of charges, but the total charge is zero. At points outside the region
A. The electric field is necessarily zero.
B. The electric field is due to the dipole moment of the charge distribution only.
C. The dominant electric field is \alpha \frac{1}{r^3}, for large r, where r is the distance from an origin in this region.
D. The work done to move a charged particle along a closed path, away from the region, will be zero.

Answer:

The correct answers are the options (c,d)
Explanation: There are charges present inside the surface, and their sum is zero. In equation
\phi _{3}=\int E.dS=\frac{q}{\varepsilon _0}
E is the electric field due to all the charges present in and out of the gaussian surface. Therefore, there must be dipoles present inside the surface.
So, at any point outside the region w.r.t the dipoles inside the region the dominant electric field \alpha \frac{1}{r^3} for large r.
Also, the electric field is conservative, and work done to move a charge on a closed path away from the region will be zero. Here c and d are correct.

Question:13

A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring (Fig. 1.7). Then


A. If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.
B. If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.
C. If q < 0, it will perform SHM for small displacement along the axis.
D. q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0.

Answer:

Answer: The correct answers are the options (a, b, d)
Explanation: As the positive charge is uniformly distributed over the ring, the charge at the centre will be zero. So, no force is exerted on charge at the centre. If the charge is pushed away from the centre, it will experience a force pushing back towards the centre, and if the charge is negative, then it will experience a force towards the ring circumference. Also, the negative charge will be in an unstable equilibrium. So, options a, b, and d are correct.

Question:12

Refer to the arrangement of charges in Fig. 1.6 and a Gaussian surface of radius R with Q at the centre. Then

A. total flux through the surface of the sphere is \frac{-Q}{\varepsilon _0}

B. field on the surface of the sphere is \frac{-Q}{4\pi \varepsilon _0R^{2}}.
C. flux through the surface of sphere due to 5Q is zero.
D. field on the surface of sphere due to –2Q is same everywhere.

Answer:

The correct answers are the options (a,c)
Explanation: 5Q charge lies outside the Gaussian surface, hence making no contribution to the electric flux. So, a and c are correct options.

NCERT Exemplar Class 12 Physics Solutions Chapter 1 Very Short Answer

Question:14

An arbitrary surface encloses a dipole. What is the electric flux through this surface?

Answer:

According to Gauss' law

\phi _s=\int E . dS =\frac{q}{\varepsilon _0}

where q= enclosed charge. So, +q and -q will cancel each other in a dipole and the sum of the charges will be zero.

Question:15

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

Answer:

Q is kept at the center of the spherical cavity. –Q charge is induced at the inner surface and +Q on the outer surface.
The surface density is denoted by σ.
a) The surface charge density on the inner surface is σ1= charge/ Area =
\frac{-Q}{4\pi R_{2}^{1}}
b) The surface charge density on the other surface is σ2=
\frac{+Q}{4\pi R_{2}^{2}}

Question:16

The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?

Answer:

Answer: The electric field inside a conductor is zero because the atoms are of order of am angstrom. While an excess charge can cause electric field to generate inside a conductor.

Question:17

If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.

Answer:

If the total charge is zero enclosed by a surface, it does not necessarily mean that the electric field is also zero enclosed by the surface. The electric field can be perpendicular to the surface. While conversely, it does imply that the electric field is zero on the surface.

Question:18

Sketch the electric field lines for a uniformly charged hollow cylinder shown in Fig 1.8.

Answer:

The side view and the top view of the electric field lines for a uniformly charged hollow cylinder are given below:

Question:19

What will be the total flux through the faces of the cube (Fig. 1.9) with side of length a if a charge q is placed at
(a) A: a corner of the cube.
(b) B: mid-point of an edge of the cube.
(c) C: centre of a face of the cube.
(d) D: mid-point of B and C.

Answer:

a) The charge is equally distributed amongst the 8 cubes of a divided cube with sides 2a when the charge is placed at the center of the large cube. Therefore, the total flux through the faces of the cube= \frac{q}{8\varepsilon _0}

b) If the charge is placed at B, it is equally distributed amongst the 4 cubes. Hence, the total flux through the four faces is = \frac{q}{4\varepsilon _0}


c) If the charge is placed at C, then the charge is share by 2 cubes equally, therefore the total flux will be= \frac{q}{2\varepsilon _0}

d) If the charge is placed at D, it is distributed amongst two cubes and therefore, the flux through them will be= \frac{q}{2\varepsilon _0}

NCERT Exemplar Class 12 Physics Solutions Chapter 1 Short Answer

Question:20

A paisa coin is made up of Al-Mg alloy and weighs 0.75g. It has a square shape and its diagonal measures 17 mm. It is electrically neutral and contains equal amounts of positive and negative charges. Treating the paisa coins made up of only Al, find the magnitude of equal number of positive and negative charges. What conclusion do you draw from this magnitude?

Answer:

1 molar mass M of Al
= 6.023 \times 10^{23}\: \: atoms

A 1 paisa coin containing Al atoms

= N = \frac{ 6.023 \times 10^{23}}{26.9815 \times 0.75 }= 1.76 \times 10^{22}
13 electrons and 13 protons are present in Al as it has a charge number of 13.
The magnitude of negative and positive charges = 3.48 \times 10^{4} C = 34.8 kC

Question:21

Consider a coin of Example 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude 34.8 kC. Suppose that these equal charges were concentrated in two point charges separated by (i) 1 cm , (ii) 100 m ( \sim length of a long building), and (iii) 106 m (radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?

Answer:

a)\: \: F1=\frac{\left | q \right |^{2}}{4\pi \varepsilon _0{r_{1}}^{2}}= 1.09\times 10^{23}N

b)\: \: F2=\frac{\left | q \right |^{2}}{4\pi \varepsilon _0{r_{2}}^{2}}= 1.09\times 10^{15}N

c)\: \: F3=\frac{\left | q \right |^{2}}{4\pi \varepsilon _0{r_{3}}^{2}}= 1.09\times 10^{7}N

Question:22

Fig. 1.10 represents a crystal unit of cesium chloride, CsCl. The cesium atoms, represented by open circles are situated at the corners of a cube of side 0.40nm, whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron.



(i) What is the net electric field on the Cl atom due to eight Cs atoms?
(ii) Suppose that the Cs atom at the corner A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?

Answer:

a) As the chlorine atoms are distributed equally from all the eight corners of the cube from the centre, the net electric field at the centre is therefore zero.
b) The net vector sum of the electric field due the charge A is E_A + E_{seven \: \: charges} = 0

Question:23

Two charges q and –3q are placed fixed on x-axis separated by distance 'd'. Where a third charge 2q should be placed such that it will not experience any force?

Answer:


At point P,

\frac{q}{4\pi \varepsilon _0}x^{2}= \frac{3q}{4\pi \varepsilon _0}\: \: \left ( {x+d} \right )^{2}

\left ( {x+d} \right )^{2}=3x^{2}
On further solving the equation, we will get the distance between the 2q and q charges as
\frac{d}{2}(1+\sqrt{3})

Question:24

Fig. 1.11 shows the electric field lines around three-point charges A, B and C.

(a) Which charges are positive?
(b) Which charge has the largest magnitude? Why?
(c) In which region or regions of the picture could the electric field be zero? Justify your answer.
(i) near A, (ii) near B, (iii) near C, (iv) nowhere.

Answer:

i) A and C are positive charges as the electric field lines are protruding out of these two charges.
ii) Charge C has the largest magnitude as it has the greatest number of electric field lines coming out.
iii) The neutral region is in between the charges A and C. Charges C is a stronger charge; therefore the neutral point naturally lies near to A (charge A is a weak charge).

Question:25

Five charges, q each are placed at the corners of a regular pentagon of side 'a' (Fig. 1.12).


(a)
(i) What will be the electric field at O, the center of the pentagon?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by –q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its corners?

Answer:

i) The net charge at the centre will be zero as all the charges are equidistant from the centre and they will cancel out each other.
ii) It can be written as that the electric field at O due to A and all other charges is zero. The equation would be as follows:
\vec{E}_A+\vec{E}_{f\! our\: charges}=0
Hence \vec{E}_{f\! our\: charges}=-\vec{E}_A\: \: or \left | \vec{E}_{f\! our\: charges} \right |= \left |\vec{E}_A \right |
When charge q is removed from A, net electric field at the centre due to
remaining charges \left | \vec{E}_{f\! our\: charges} \right |= \left |\vec{E}_A \right |=\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\: \: along\: \: OA

(iii) If charge q at A is replaced by -q, then electric field due to this negative charge
\vec{E_{-q}}=\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\: along \: OA
Hence net electric field at the centre
\vec{E_{net}}=\vec{E_{-q}}+\vec{E}_{force \: charge}=\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\: +\frac{1}{4\pi \varepsilon _0}\frac{q}{r^{2}}\:
\vec{E}_{net}=\frac{1}{4\pi \varepsilon _0}\frac{2q}{r^{2}}\: \: along \: OA
b) If pentagon is replaced by a polygon and charge Q at each corner, then also they all will cancel out each other, and the net electric field at O will be zero.

NCERT Exemplar Class 12 Physics Solutions Chapter 1 Long Answer

Question:26

In 1959, Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a net charge. Suppose that the universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.

(a) Find the critical value of y such that expansion may start.
(b) Show that the velocity of expansion is proportional to the distance from the centre.

Answer:

a) Let the radius of the Universe be R and assume that the hydrogen atoms are uniformly distributed. The expansion will only occur if the Coulomb repulsion is larger than the gravitational attraction at a distance R
The Hydrogen atom contains one proton and one electron charge on each hydrogen atom
e_H=e_P+e=-(1+y)e+e=-ye=\left | ye \right |
Let E be electric field intensity at distance R, on the surface of the sphere, then according to Gauss' theorem,
\\\oint \vec{E}.d\vec{S}= \frac{q_{enclosed}}{\varepsilon _0}\\ \\ \Rightarrow E(4\pi R^2)=\frac{4}{3}\frac{\pi R^3 N\left | ye \right |}{\varepsilon _0}\\ \\ \Rightarrow E=\frac{1}{3}\frac{ N\left | ye \right |R}{\varepsilon _0}.........1

Question:31

Total charge –Q is uniformly spread along the length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.

(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.

Answer:

Let the charge q is displaced slightly by z\left ( z< < R \right ) along the axis of ring. Let force on the charge q will be towards O. The motion of charge q will be simple harmonic, it the force on charge q must be proportion to z and is directed towards O.
Electric field at axis of the ring at a distance z from the centre of ring
E=\frac{1}{4\pi \varepsilon _0}\frac{Q_z}{\left ( R^2+Z^2 \right )^{3/2}};towords O
the net force on the charge F_{net}=qE
F_{net}=\frac{1}{4\pi \varepsilon _0}\frac{qQz}{\left ( R^2+Z^2 \right )^{3/2}};
\Rightarrow F_{net}=\frac{1}{4\pi \varepsilon _0}\frac{qQz}{R^3\left ( \frac{1+z^2}{R^2} \right )^{3/2}}
As z<<R\: then,
\Rightarrow F_{net}=\frac{1}{4\pi \varepsilon _0}\frac{qQz}{R^3\left ( \frac{1+z^2}{R^2} \right )^{3/2}} or \vec{F}_{net}=-K\vec{z}
Where K=\frac{Qq}{4\pi \varepsilon _{0}R^{3}}=constant
Clearly, the force on q is proportional to the negative of its displacement. Therefore the motion of q is simple harmonic.
\\\omega \sqrt{\frac{K}{m}}\: and\: T=\frac{2\pi}{\omega }=2\pi\sqrt{\frac{m}{K}}\\T=2\pi\sqrt{\frac{m4\pi \varepsilon _0\: R^3}{Qq}} \\T=2\pi\sqrt{\frac{4\pi \varepsilon _0\: m\: R^3}{Qq}} \\

Question:30

Two charges –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x<<d) perpendicular to the line joining the two fixed charged as shown in Fig. 1.14. Show that q will perform simple harmonic oscillation of time period.

Answer:

Let the charge q is displaced slightly by x(x<<d)perpendicular to the line joining the two fixed charges. The net force on the charge q will be towards O.The motion of charge -q to be simple harmonic if the force on charge q must be proportional to its distance from the centre O and is directed towards O.
The net force on the charge F_{net}= 2F \cos\theta
Here F=\frac{1}{4\pi \varepsilon _0}\frac{q(q)}{r^2}= \frac{1}{4\pi \varepsilon _0}\frac{q^2}{\left ( d^2+x^2 \right )}
And\cos \theta =\frac{x}{\sqrt{x^2+d^2}}
Hence, F_{net}=2 \left [ \frac{1}{4\pi \varepsilon _0}\frac{q^2}{\left ( d^2+x^2 \right )} \right ]\left [ \frac{x}{\sqrt{x^2+d^2}} \right ]
= \frac{1}{2\pi \varepsilon _0}\frac{q^2x}{\left ( d^2+x^2 \right )^{3/2}}=\frac{1}{2\pi \varepsilon _0}\frac{q^2x}{d^3\left ( 1+\frac{x^2}{d^2} \right )^{3/2}}
As x<<d, then F_{net}=\frac{1}{2 \pi \varepsilon _0}\frac{q^2x}{d^3}\: or \: F_{net}=Kx
i.e. force on charge q is proportional to its displacement from the centre O and it is directed towards O.
Hence, the motion of charge q would be simple harmonic, where
\omega = \sqrt{\frac{K}{m}}\\ and \: T=\frac{2 \pi }{\omega}=2 \pi\sqrt{\frac{m}{K}}
\Rightarrow T= 2 \pi \sqrt{\frac{m.4 \pi \varepsilon _0d^3}{2q^2}}=\left [ \frac{8 \pi^3 \varepsilon _0md^3}{q^2} \right ]^{1/2}

Question:29

There is another useful system of units, besides the SI/mks system, called the cgs (centimetre-gram-second) system. In this system, Columb's law is given by

F=\frac{Qq}{r^2}

where the distance r is measured in cm (= 10^{-2} m), F in dynes (=10^{-5} N) and the charges in electrostatic units (es units), where 1es unit of charge

\frac{1}{3}\times10^{-9}C
The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 \times 10^8 \: m/s. An approximate value of c then is c = [3] \times 10^8 m/s.
(i) Show that the coloumb law in cgs units yields
1 esu of charge = 1 (dyne)1/2 cm.
Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L.
(ii) Write 1 esu of charge = x\: C, where x is a dimensionless number. Show that this gives

\frac{1}{4\pi\epsilon_0}=[3]^2\times10^{9}\frac{Nm^2}{C^2}

Answer:


(i) Acording to relation,
F =\frac{Qq}{r^2}= 1dyne= \frac{\left [ esu\; of\: chaarge \right ]^2}{\left [ 1cm \right ]^{2}}
So, 1 esu of charge
=\left ( 1dyne \right )^{1/2}\times 1 cm
F^{1/2}.L=\left [ MLT^{2} \right ]^{1/2}\left [ L \right ]= \left [ M^{1/2}L^{3/2}T^{1} \right ]
\Rightarrow Hence, [1 esu of charge] =M^{1/2}L^{3/2}T^{1}
Thus the charge in C.G.S. unit (in esu) is represented in terms of fractional power 1/2 of M and 3/2 of L.
(ii) If two charge each of magnitude 1 esu separated by 1 cm, Coulomb force on the charge is 1 dyne = 10^{-5}N.
Let 1 esu of charge =x C, where x is a dimensionless number. We can consider a situation, two charges of magnitude x C separated by 10^{-2}m.
The force between the charge
F=\frac{1}{4 \pi \varepsilon _0}\frac{x^{2}}{\left ( 10^{-2} \right )^{2}}= 1dyne=10^{-5}N
\therefore \frac{1}{4 \pi \varepsilon _0}= \frac{10^{-9}}{x^{2}}\frac{Nm^{2}}{C^{2}}
Taking,
\\x=\frac{1}{\left | 3 \right |\times 10 ^{9}},\\ We \: get\: \: \frac{1}{4\pi \varepsilon _0}=10^{-9}\times \left | 3 \right |^{2}\times 10^{18} \\ \frac{1}{4\pi \varepsilon _0}=9\times 10^{9}\frac{Nm^{2}}{C^{2}}

Question:28

Two fixed, identical conducting plates (\alpha and \beta ), each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (\gamma), free to move is located on the other side of the plate with charge q at a distance d (Fig 1.13). The third plate is released and collides with the plate β. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst (\beta \: and\: \gamma )


(a) Find the electric field acting on the plate \gamma before the collision.
(b) Find the charges on β and \gamma after the collision.
(c) Find the velocity of the plate \gamma after the collision and at a distance d from the plate β.

Answer:

(a) Net electric field field at plate \gamma before collision is vector sum of electric field at plate \gammadue to plate \alpha\: and\: \beta.
The electric field at plate \gamma due to plate \alpha is \vec{E}_1= \frac{Q}{S(2\varepsilon _0)}(-\hat{i}),

The electric field at plate \gamma due to plate \beta is \vec{E}_2= \frac{q}{S(2\varepsilon _0)}(\hat{i}),
Hence the net electric field at plate \gamma before collision is
\\ \vec{E}=\vec{E}_1+\vec{E}_2= \frac{q-Q}{S(2\varepsilon _0)}(\hat{i}),\\ \\ \vec{E}=\vec{E}_1+\vec{E}_2= \frac{Q-q}{S(2\varepsilon _0)}(\hat{i}),
Or \frac{Q-q}{S(2\varepsilon _0)} to the left, if Q>q

(b) During collision, plates \alpha\: and\: \beta are in contact with each other, hence their potential becomes the same.
Suppose charge on plate \beta \: is\: q_1\: and charge on plate \gamma \: is\: q_2\:. at any point O , in between the two plates, the electric field must be zero.
Electric field at O due to \alpha,
\vec{E}_\alpha = \frac{Q}{S(2\varepsilon _0)}\left (- \hat{i} \right )
Electric field at O due to \beta,
\vec{E}_2 = \frac{q_1}{S(2\varepsilon _0)}\left ( \hat{i} \right )
Electric field at O due to \gamma,
\vec{E}_\gamma = \frac{q_2}{S(2\varepsilon _0)}\left (- \hat{i} \right )
As the electric field at O is zero, therefore
\frac{Q+q_2}{S(2\varepsilon _0)}=\frac{q_1}{S(2\varepsilon _0)}\\ \\ \because Q+q_2=q_1
As there is no loss of charge on collision,
Q+q=q_1+q_2
On solving Equ. (1) and (2) We get
q_1=(Q+q/2)=Charge on plate \beta
q_1=(q/2)=Charge on plate \gamma

(c) Let the velocity be \nu at the distance d from plate \beta after the collision.
If m is the mass of the plate \gamma, then the gain in K.E. over the round trip must be equal to the work done by the electric field.
After the collision, electric field at plate \gamma is
\vec{E}_2= \frac{Q}{2\varepsilon _0S}(-\hat{i}),+\frac{\left ( Q+q/2 \right )}{2\varepsilon _0S}\hat{i}=\frac{\left ( q/2 \right )}{2\varepsilon _0S}\hat{i}
Just before collision, electric field at plate \gamma is
\vec{E}_1= \frac{Q-q}{2\varepsilon _0S}\: \hat{i}.
If F1 is force on plate before collision., then
F_1=\vec{E}_1Q= \frac{\left ( Q-q \right )Q}{2\varepsilon _0S}\: \hat{i}\: and\: F_2=\vec{E}_2\frac{q}{2}= \frac{\left ( \frac{q}{2} \right )^{2}}{2\varepsilon _0S}\: \hat{i}
Total work done by the electric field is the round trip movement of plate \gamma,
W=(F_1+F_2)d
=\frac{\left [ \left ( Q-q \right )Q+\left ( q/2 \right )^{2} \right ]d}{2\varepsilon _0S}=\frac{\left ( Q-q/2 \right )^{2}d}{2\varepsilon _0S}
If m is the mass of plate \gamma, the KE gained by the plate =
\frac{1}{2}m\nu ^2
According to work energy principle,
\frac{1}{2}m\nu ^2=W \Rightarrow \frac{1}{2}m\nu ^2=\frac{\left ( Q-q/2 \right )^{2}d}{2\varepsilon _0S}
\Rightarrow \nu =\left ( Q-q/2 \right )\left ( \frac{d}{m\varepsilon _0S} \right )^{1/2}

Question:27

Consider a sphere of radius R with charge density distributed as

\rho (r) = kr for r \leq R\\ = 0\: f\! or \: r >R .
a) Find the electric field at all points r.
(b) Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero? Assume that the introduction of the proton does not alter the negative charge distribution.

Answer:

a) The charge density distribution expression in the sphere tells that the electric field is radial.

Let us consider a sphere S of radius R and two hypothetic spheres of radius r< R and r> R.
Let us first consider for point r< R, electric field intensity will be given by,
\oint \vec{E}.d\vec{S}= \frac{1}{\varepsilon _0}\int \rho \: dV
Here dV =4\pi r^2\:dr
\Rightarrow \oint \vec{E}.d\vec{S}= \frac{1}{\varepsilon _0}4 \pi K \int_{0}^{r}r^3dr\left ( \because \rho \left ( r \right )=Kr \right )
\Rightarrow \left ( E \right )4 \pi r^2= \frac{4 \pi K}{\varepsilon _0}\frac{r^4}{4}
We get
E= \frac{1}{4\varepsilon _0}Kr^2
As charge density is positive, it means the direction of E is radially outwards.

Now consider points r>R , electric field intensity will be given by
\oint \vec{E}.d\vec{S}= \frac{1}{\varepsilon _0}\int \rho dV
\Rightarrow E(4 \pi r^2)= \frac{4 \pi K}{\varepsilon _0}\int_{0}^{R}r^3dr=\frac{4 \pi K}{\varepsilon _0}\frac{R^4}{4}
Which gives E=\frac{K}{4 \varepsilon _0}\frac{R^4}{r^2}
Here also the charge density is again positive. So, the direction of E is radially outward.
(b) The two proton must be placed symmetrically on the opposite sides of the centre along a diameter. This can be shown by
the figure given below. Charge on the sphere,
\\q=\int_{0}^{R}\rho dV= \int_{0}^{R}\left ( Kr \right )4 \pi r^2dr\\ \\ q=4 \pi K\frac{R^4}{4}2e\\ \\\because K=\frac{2e}{\pi R^4}


The proton 1 and 2 are embedded at distance r from the centre of the sphere as shown, then attractive force on proton 1 due to charge
distribution is
F_1=eE=\frac{-eKr^2}{4\varepsilon _0}
and repulsive force on proton 1 due to proton 2 is
F_2=\frac{e^2}{4 \pi\varepsilon _0 (2r)^2}
Net force on proton 1, F=F_1+F_2
F=\frac{-eKr^2}{4\varepsilon _0}+\frac{e^2}{16 \pi \varepsilon _0r^2}
Thus, net force on proton 1 will be zero, when
\frac{er^22e}{4\varepsilon _0 \pi R^4}= \frac{e^2}{16 \pi \varepsilon _0r^2}
\Rightarrow r^4= \frac{R^4}{8}
Hence the protons must be at a distance r=\frac{R}{(8)^{1/4}} from the centre.

The NCERT Exemplar solutions for Class 12 Physics chapter 1 provided here are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams.

Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 1 Electric Charges and Fields

  • Introduction
  • Electric Charge
  • Conductors and Insulators
  • Charging by Induction
  • Basic Properties of Electric Charge
  • Additivity of charges
  • Charge is conserved
  • Quantisation of charge
  • Coulomb’s Law
  • Forces between Multiple Charges
  • Electric Field
  • Electric field due to a system of charges
  • Physical significance of electric field
  • Electric Field Lines
  • Electric Flux
  • Electric Dipole
  • The field of an electric dipole
  • Physical significance of dipoles
  • Dipole in a Uniform External Field
  • Continuous Charge Distribution
  • Gauss’s Law
  • Applications of Gauss’s Law
  • Field due to an infinitely long straight uniformly charged wire
  • Field due to a uniformly charged infinite plane sheet
  • Field due to a uniformly charged thin spherical shell

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Important Topics, Sub-Topics, and Terms of NCERT Exemplar Class 12 Physics Solutions Chapter 1 Electric Charges and Fields

  • Students will be able to collect information and basic knowledge about electric charges, its movement and the field formed due to these charges.
  • This will also enlighten students about the electrostatic field formed by different charges, electric field formed by different distribution of charges and its mathematical analysis and implication.
  • NCERT Exemplar Class 12 Physics chapter 1 solutions also provide a logical explanation for different situations that we encounter in real lives such as lightning during thunderstorms, hearing a crackle while changing synthetic cloths, experiencing a current from a plastic chair and so on.
  • Class 12 NCERT Exemplar Physics solutions chapter 1 also provides an interpretation of basic qualities of electric charges and different materials such as conductors and insulators which develops the basic knowledge and makes them aware about the uses, and misuses of same in various devices and how they are used in devices we use around us.
  • Students will be able to relate this concept of charging of bodies to daily routines in real life.
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NCERT Exemplar Class 12 Physics Chapter Wise Links

Important Topics To Cover For Exams From NCERT Exemplar Class 12 Physics Solutions Chapter 1 Electric Charges and Fields

It is important that students cover some key topics from this chapter for exams. Here are a few of them:

· NCERT Exemplar Class 12 Physics solutions chapter 1 enlightens about the different properties of electric charges, and how there are positive and negative charges in different things as well as the human body that attract and repel each other and how these charges can be transferred to different devices through different methods.

· The chapter also covers a variety of numericals that help to calculate the force between two point charges as well as forces between multiple charges with the application of Coulomb’s law.

· The presence of electric field around any charge and its electric flux is also explained in the chapter and its application with the help of Gauss’s law.

NCERT Exemplar Class 12 Solutions

Also, check the NCERT solutions of questions given in book

Also read NCERT Solution subject wise

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Frequently Asked Question (FAQs)

1. How many questions are solved in this chapter?

 We have solved all 31 questions from the main exercise that is mentioned in the NCERT book in NCERT exemplar Class 12 physics solutions chapter 1.

2. Are these NCERT Exemplar Class 12 Physics Solutions Chapter 1 helpful?

Yes, these solutions are highly helpful to clear both theoretical understandings of the chapter, and also get an idea of how to solve numerical questions in an exam.

3. What are the topics covered in NCERT Exemplar Class 12 Physics ?

Here are the topics covered in NCERT Exemplar Class 12 Physics

  • Introduction
  • Electric Charge
  • Conductors and Insulators
  • Charging by Induction
  • Basic Properties of Electric Charge
  • Additivity of charges
  • Charge is conserved
  • Quantisation of charge
  • Coulomb’s Law
  • Forces between Multiple Charges
  • Electric Field
  • Electric field due to a system of charges
  • Physical significance of electric field
  • Electric Field Lines
  • Electric Flux
  • Electric Dipole
  • The field of an electric dipole
  • Physical significance of dipoles
  • Dipole in a Uniform External Field
  • Continuous Charge Distribution
  • Gauss’s Law
  • Applications of Gauss’s Law
  • Field due to an infinitely long straight uniformly charged wire
  • Field due to a uniformly charged infinite plane sheet
  • Field due to a uniformly charged thin spherical shell

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

https://www.careers360.com/courses/graphic-designing-course

Thank you

Hope this information helps you.

Read Complete Answer
Sajal Trivedi 29 January,2024
i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer
Upasana Barman 24 August,2022
i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Read Complete Answer
Mayankjha.student2007 23 August,2022
if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer
Nuthalapati Vyshnavi 21 August,2022
I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer
Muskan Dhalwal 17 August,2022
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

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\; K\;
Option 3)

zero\;

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K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

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be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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