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

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

Edited By Vishal kumar | Updated on Sep 08, 2023 02:29 PM IST | #CBSE Class 12th

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

NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges and Fields: Welcome to the updated NCERT book solutions for the very first chapter of the Class 12 syllabus. On this Careers360 page, you will find comprehensive electric charges and fields class 12 NCERT solutions covering questions from 1.1 to 1.34. Questions 1.1 to 1.23 belong to the exercise section, while questions 1.24 to 1.34 are part of the additional exercise. These class 12 physics ch1 NCERT solutions, crafted by subject experts, are presented in a simple and detailed manner. You can download the PDF for free and use them at your convenience.

Problems in Physics Class 12 Chapter 1 are important from the CBSE class 12 exam point of view. Try solving all the questions of NCERT Class 12 Physics Chapter 1 and if any doubt arises then check the CBSE NCERT solutions for Class 12 Physics Chapter 1 Electric Charges and Fields. Electrostatics is the study of electric charges at rest. The NCERT Class 12th Physics Electric Charges and Fields deal with the charging of a body, properties of charge, Columbus law, electric field, electric flux, Gauss law and application of Gauss law. Two main laws discussed in Physics Class 12 Chapter 1 are Gauss law and Columbus law. NCERT Solutions for Class 12 physics chapter 1 Electric Charges and Fields explains problems related to these topics.

The derivations in Physics Class 12 NCERT Chapter 1 Electric Charges and Fields of NCERT are very important in CBSE board exam point of view. As the second chapter of NCERT solutions for Class 12 is the continuation of the NCERT Class 12 Physics chapter 1 pdf, it is important to remember all points studied in the chapter. These class 12 electric field and charges ncert solutions will help you in a better understanding of the concepts you have studied. While studying NCERT Solutions for Class 12 Physics Chapter 1 formulas can be compared with the chapter Gravitation of Physics NCERT solutions class 11.

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NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields: Exercise Question Answer

Given,

$q_{1}$ = $2 \times 10^{-7}$ C

$q_{2}$ = $3 \times 10^{-7}C$

r = 30 cm = 0.3 m

We know,

Force between two charged particles, $q_{1}$ and $q_{2}$ separated by a distance r.

$F = \frac{1}{4\pi \epsilon _{0}} \frac{q_{1}q_{2}}{r^2}$

$= \frac{1}{4\pi \epsilon _{0}} \frac{2\times10^{-7} \times 3 \times 10^{-7}}{(30\times10^{-2}\ m)^2\ }$

$= (9\times10^9\ N)\times \frac{6\times10^{-14+4}}{900\ m^2\ } = 6\times10^{-3} N$

Since the charges are of the same nature, the force is repulsive.

Given,

$q_{1}$ = $0.4 \mu C$

$q_{2}$ = $-0.8 \mu C$

F = $-0.2N$ (Attractive)

We know,

Force between two charged particles, $q_{1}$ and $q_{2}$ separated by a distance r.

$F = \frac{1}{4\pi \epsilon _{0}} \frac{q_{1}q_{2}}{r^2}$

$\\ -0.2 = 9\times10^9 \times \frac{(0.4\times10^{-6})(-0.8\times10^{-6})}{r^2} \\ \implies r^2 = 9\times10^9\times0.32\times10^{-12}/0.2 \\ \implies r^2 = 9\times0.16\times10^{-2} \\ \implies r = 1.2\times10^{-1} m = 0.12 m =12 cm$

Therefore, the distance between the two charged spheres is 12 cm.

Using Newton's third law, the force exerted by the spheres on each other will be equal in magnitude.

Therefore, the force on the second sphere due to the first = 0.2 N (This force will be attractive since charges are of opposite sign.)

Electrostatic force

$F=\frac{KQ^2}{r^2}$

So the dimension of

$[Ke^2]=[Fr^2]$ ..................(1)

The gravitational force between two bodies of mass M and m is

$F=\frac{GMm}{r^2}$

so dimension of

$[Gm_em_p]=[Fr^2]$ .............(2)

Therefore from (1) and (2)

$[\frac{ke^{2}}{Gm_{e}m_{p}} ]$ is dimensionless

or

Here,

K = $1/4\pi \epsilon _{0}$ , where is the permittivity of space. $[1/ \epsilon _{0}] =[C/V.m] = [Nm^2 C^{-2}]$

e = Electric charge ([e] = [C])

G = Gravitational constant. ([G]= $[Nm^2kg^{-2}]$ )

$m_{e}$ and $m_{p}$ are mass of electron and proton ([ $m_{e}$ ] = [ $m_{p}$ ] = [Kg])

Substituting these units, we get

$[\frac{ke^{2}}{Gm_{e}m_{p}} ] = [\frac{C^2 \times Nm^2C^{-2}}{Nm^2kg^{-2}\times kg\times kg} ] = M^{0}L^{0}T^{0}$

Hence, this ratio is dimensionless.

Putting the value of the constants

$\frac{ke^{2}}{Gm_{e}m_{p}} = \frac{9\times10^9\times(1.6\times10^{-19})^{2}}{6.67\times10^{11}\times9.1\times10^{-31}\times1.67\times10^{-27}}$

$=2.3 \times 10^{40}$

The given ratio is the ratio of electric force $\frac{ke^{2}}{R^2}$ to the gravitational force between an electron and a proton $\frac{Gm_{e}m_{p}}{R^2}$ considering the distance between them is constant!

The given statement "electric charge of a body is quantised" implies that charge on a body can take only integral values. In other words, only the integral number of electrons can be transferred from one body to another and not in fractions.

Therefore, a charged body can only have an integral multiple of the electric charge of an electron.

On a macroscopic level, the amount of charge transferred is very large as compared to the charge of a single electron. Therefore, we tend to ignore the quantisation of electric charge in these cases and considered to be continuous in nature.

When a glass rod is rubbed with a silk cloth, opposite charges appear on both the rod and the cloth.

The phenomenon of charging bodies by rubbing them against each other is known as charging by friction. Here, electrons are transferred from one body to another giving both the bodies an equal but opposite charge. The number of electrons lost by one body (attains positive charge due to loss of negatively charged electrons) is equal to the number of electrons gained by the other body (attains negative charge). Therefore, the net charge of the system is zero. This is in accordance with the law of conservation of charge

Charges at (A, C) and (B, D) are pairwise diametrically opposite and also equal.

Therefore, their force on a point charge at the centre of the square will be equal but opposite in directions.

Now, AC = BD = $\sqrt{2}\times(10\times10^{-2} m) = \sqrt{2}\times0.1 m$

$\therefore$ AO = BO = CO = DO = r = Half of diagonal = $\sqrt{2}\times0.05 m$

Force on point charge at centre due to charges at A and C = $F_{A} = -F_{C} = \frac{k(2\mu C)(1\mu C)}{(r)^2}$

Similarly, force on point charge at centre due to charges at B and D = $F_{B} = -F_{D} = \frac{k(-5\mu C)(1\mu C)}{(r)^2}$

$\therefore$ Net force on point charge = $\\F_{A} + F_{B} + F_{C} + F_{D} \\$

$= -F_{B} +F_{B} - F_{D} + F_{D} = 0$ .

Hence, the charge at the centre experiences no force.

A positive point charge experiences a force in an electrostatic field. Since the charge will experience a continuous force and cannot jump from one point to another, the electric field lines must be continuous.

A tangent drawn at any point on a field line gives the direction of force experienced by a unit positive charge due to the electric field on that point. If two lines intersect at a point, then the tangent drawn there will give two directions of force, which is not possible. Hence two field lines cannot cross each other at any point.

Given, AB = 20 cm

Since, O is the midpoint of the line AB.

AO = OB = 10 cm = 0.1m

The electric field at a point caused by charge q, is given as,

$E = \frac{kq}{r^2}$

Where, q is the charge, r is the distance between the charges and the point O

k = 9x10 9 N m 2 C -2

Now,

Due to charge at A, electric field at O will be $E_{A}$ and in the direction AO.

$E_{A} =\frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2}$

Similarly the electric field at O due to charge at B, also in the direction AO

$E_{B} =\frac{ 9\times10^9 \times (-3\times10^{-6})}{0.1^2}$

Since, both the forces are acting in the same direction, we can add their magnitudes to get the net electric field at O:

E' = $E_{A}$ + $E_{B}$ = 2E (Since their magnitudes are same)

$E' =2\times \frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2} = 5.4\times 10^4 N/C$ along the direction AO.

Let Q = $-1.5 \times10^{-9}$ C

The force experienced by Q when placed at O due to the charges at A and B will be:

$F = Q \times E'$

where 'E' is the net electric field at point O.

$F=1.5 \times 10^{-9} C \times 5.4 \times 10^4 N/C = 8.1\times10^{-3 } N$

Q being negatively charged will be attracted by positive charge at A and repelled by negative charge at B. Hence the direction of force experienced by it will be in the direction of OA.

Given,

$q_{A}=2.5\times 10^{-7}C$ and $q_{B}=-2.5\times 10^{-7}$

The total charge of the system = $q_{A} + q_{B} = 0$

$\therefore$ The system is electrically neutral. (All dipole systems have net charge zero!)

Now, distance between the two charges, d = 15 + 15 = 30 cm = 0.3 m

We know, The electric dipole moment of the system, p = $q_{A}$ x d = $q_{B}$ x d (i.e, the magnitude of charge x distance between the two charges)

$\therefore p = 2.5\times10^{-7} C \times 0.3 m = 7.5\times10^{-8} Cm$

The direction of a dipole is towards the positive charge . Hence, in the positive z-direction.

Given,

Electric dipole moment, p = $4\times 10^{-9} Cm$

$\Theta = 30^{0} \ \therefore sin\Theta = 0.5$

E = $5\times 10^{4}NC^{-1}$

We know, the torque acting on a dipole is given by:

$\tau = p \times E$

$\implies \tau = p Esin\Theta = 4\times10^{-9} \times5\times10^{4}\times0.5 \ Nm$

$\implies \tau =10^{-4}Nm$

Therefore, the magnitude of torque acting on the dipole is $10^{-4}Nm$

Clearly, polyethene being negatively charged implies that it has an excess of electrons(which are negatively charged!). Therefore, electrons were transferred from wool to polyethene.

Given, charge attained by polyethene = $-3\times 10^{-7}$ C

We know, Charge on 1 electron = $-1.6\times10^{-19} C$

Therefore, the number of electrons transferred to attain a charge of $-3\times 10^{-7}$ =

$\frac{-3\times 10^{-7}}{-1.6\times10^{-19} C} = 1.8\times10^{12}$ electrons.

The charge attained by polyethene (and also wool!) is solely due to the transfer of free electrons.

We know, Mass of an electron = $9.1\times10^{-31}\ kg$

The total mass of electron transferred = number of electrons transferred x mass of an electron

= $9.1\times10^{-31} \times 1.8\times10^{12}\ kg = 16.4\times10^{-19} \ kg$

Yes, there is a transfer of mass but negligible.

Since the radii of the spheres A and B are negligible compared to the distance of separation, we consider them as a point object.

Given,

charge on each of the spheres = $6.5\times 10^{-7}C$

and distance between them, r = 50 cm = 0.5 m

We know,

$F = k\frac{q_{1}q_{2}}{r^2}$

Therefore, the mutual force of electrostatic repulsion(since they have the same sign of charge)

$F = 9\times10^9Nm^{2}C^{-2}\times\frac{(6.5\times10^{-7}\ C)^2}{(0.5\ m )^2} = 1.5\times10^{-2} N$

We know, force between two charged particles separated by a distance r is:

$F = k\frac{q_{1}q_{2}}{r^2} = k\frac{q^2}{r^2}$ $(\because q_{1} = q_{2} = q)$

Now if $q\rightarrow 2q\ and\ r\rightarrow r/2$

The new value of force:

$F_{new} =k\frac{(2q)^2}{(r/2)^2} = 16k\frac{q^2}{r^2} = 16F$

Therefore, the force increases 16 times!

$F_{new} =16F = 16\times1.5\times10^{-2} N = 0.24\ N$

When two spheres of the same size are touched, on attaining the equipotential state, the total charge of the system is equally distributed on both of them.

Therefore, (i) When uncharged third sphere C is touched with A, charge left on A = $0.5\times 6.5\times 10^{-7}C$

and charge attained by C = $0.5\times 6.5\times 10^{-7}C$

(ii) Now, charge on B + charge on C = $6.5\times 10^{-7}C$ + $0.5\times 6.5\times 10^{-7}C$ = $1.5\times 6.5\times 10^{-7}C$

When touched, charge left on B = $0.5\times1.5\times 6.5\times 10^{-7}C$

Therefore $q_{A}\rightarrow 0.5\times q_{A}\ and\ q_{B}\rightarrow 0.75\times q_{B}$

Therefore, $F' = 0.5\times0.75 \times F = 0.375 \times 1.5\times10^{-2} N = \boldsymbol{5.7\times10^{-3}\ N}$

Charges 1 and 2 are repelled by the negatively charged plate of the system

Hence 1 and 2 are negatively charged .

Similarly, 3 being repelled by positive plate is positively charged.

(charge to the mass ratio: charge per unit mass)

Since 3 is deflected the most, it has the highest charge to mass ratio .

Given,

$E=3\times 10^{3}\ \widehat{i}\ \frac{N}{C}$

Area of the square = $0.01^2\ m^2$

Since the square is parallel to the yz plane, therefore it's normal is in x-direction.(i.e $\widehat{i}$ direction )

therefore, flux through this surface:

$\phi = E.A$

$\implies \phi = (3\times10^3\ \widehat{i}).(0.01\ \widehat{i}) Nm^2/C = \boldsymbol{30 Nm^2/C}$

Now, Since the normal of the square plane makes a $60^{\circ}$ angle with the x-axis

$cos\Theta = cos(60^{0}) = 0.5$

therefore, flux through this surface:

$\phi = E.A = EAcos\Theta$

$\implies \phi = (3\times10^3)(0.01)(0.5) Nm^2/C = \boldsymbol{15\ Nm^2/C}$

The net flux of the uniform electric field through a cube oriented so that its faces are parallel to the coordinate planes is z ero .

This is because the number of lines entering the cube is the same as the number of lines leaving the cube.

Alternatively,

using Gauss’s law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$ times the total charge enclosed by S.

i.e. $\phi = q/\epsilon _{0}$

where, q = net charge enclosed and $\epsilon _{0}$ = permittivity of free space (constant)

Since there is no charge enclosed in the cube, hence $\phi = 0$ .

Using Gauss’s law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$ times the total charge enclosed by S.

i.e. $\phi = q/\epsilon _{0}$

where, q = net charge enclosed and $\epsilon _{0}$ = permittivity of free space (constant)

Given, $\phi = 8.0 \times10^3\ Nm^2/C$

$\therefore q = \phi \times\epsilon _{0} = (8.0\times10^3 \times8.85\times10^{-12})\ C$

$\implies q = 70 \times10^{-9} C = \boldsymbol{0.07 \mu C}$

This is the net charge inside the box.

Using Gauss's law, we know that $\phi = q/\epsilon _{0}$

Since flux is zero, q = 0, but this q is the net charge enclosed by the surface.

Hence, we can conclusively say that net charge is zero, but we cannot conclude that there are no charges inside the box.

Let us assume that the charge is at the centre of the cube with edge 10 cm.

Using Gauss's law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$ times the total charge enclosed by S.

i.e. $\phi = q/\epsilon _{0}$

where, q = net charge enclosed and $\epsilon _{0}$ = permittivity of free space (constant)

Therefore, flux through the cube: $\phi = (10\times10^{-6} C)/\epsilon _{0}$

Due to symmetry, we can conclude that the flux through each side of the cube, $\phi'$ , will be equal.

$\therefore \phi' = \frac{\phi}{6} = \frac{10^{-5}}{6\times\epsilon _{0}} = \frac{10^{-5}}{6\times\ 8.85\times10^{-12}} = \boldsymbol{1.9 \times 10^5\ Nm^2C^{-1}}$

Given,

q = net charge inside the cube = $2.0\mu C$

Using Gauss’s law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$ times the total charge enclosed by S.

i.e. $\phi = q/\epsilon _{0}$

where, q = net charge enclosed and $\epsilon _{0}$ = permittivity of free space (constant)

$\therefore \phi = 2.0\times10^{-6}/8.85\times10^{-12}\ Nm^2C^{-1} = \boldsymbol{2.2\times10^5\ Nm^2C^{-1}}$

(Note: Using Gauss's formula, we see that the electric flux through the cube is independent of the position of the charge and dimension of the cube! )

Q 1.20 (a) A point charge causes an electric flux of $-1.0\times 10^{3}\frac{Nm^{2}}{C}$ to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

Given,

$\phi = -1.0\times 10^{3}\frac{Nm^{2}}{C}$

Using Gauss’s law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$ times the total charge enclosed by S.

i.e. $\phi = q/\epsilon _{0}$

where, q = net charge enclosed and $\epsilon _{0}$ = permittivity of free space (constant)

Therefore, flux does not depend on the radius of the sphere but only on the net charge enclosed. Hence, the flux remains the same although the radius is doubled.

$\phi' = -10^{3}\frac{Nm^{2}}{C}$

Given,

$\phi = -1.0\times 10^{3}\frac{Nm^{2}}{C}$

Using Gauss’s law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$ times the total charge enclosed by S.

i.e. $\phi = q/\epsilon _{0}$

where, q = net charge enclosed and $\epsilon _{0}$ = permittivity of free space (constant)

$\therefore q = -10^{3}Nm^2C^{-1} \times 8.85\times10^{-12}N^{-1}m^{-2}C^{2} = -8.8\times10^{-9}\ C \\ = \boldsymbol{-8.8\ nC}$

Q 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 $1.5 \times 10^{3}\frac{N}{C}$ and points radially inward, what is the net charge on the sphere?

We know, for determining the electric field at r>R for a conducting sphere, the sphere can be considered as a point charge located at its centre.

Also, electric field intensity at a point P, located at a distance r, due to net charge q is given by,

$E = k\frac{q}{r^2}$

Given, r = 20 cm = 0.2 m (From the centre, not from the surface!)

$\\ \therefore 1.5\times10^3 = 9\times10^9\times \frac{q}{0.2^2} \\ \implies q = \frac{1.5\times0.04}{9}\times10^{-6} = 6.67\times10^{-9}\ C$

Therefore, charge on the conducting sphere is $- 6.67\ nC$ (since flux is inwards)

Given,

Surface charge density = $80.0\mu Cm^{-2}$

Diameter of sphere = 2.4 m $\therefore$ radius of sphere, r = 1.2 m

The charge on the sphere, Q= surface charge density x surface area of the sphere

$= (80\times10^{-6})\times(4\pi r^2) = 320\times22/7\times(1.2)^2 = \boldsymbol{1.45\times10^{-3}\ C}$

Q 1.22 (b) A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of $80.0 \frac{\mu C}{m^{2}}$ . (b) What is the total electric flux leaving the surface of the sphere?

Using Gauss's law, we know that :

$\phi = q/ \epsilon_{0}$

$\implies \phi = 1.45\times10^{-3}/ 8.85\times10^{-12} = 1.6\times10^{8}\ Nm^2C^{-1}$

Given,

$\lambda = 9\times 10^{4}\frac{N}{C}$

d = 2 cm = 0.02 m

We know, For an infinite line charge having linear charge density $\lambda$ , the electric field at a distance d is:

$E = k\lambda / d$

$\therefore 9\times10^4 = 9\times10^{9}\lambda / 0.02$

ï¿¼

The linear charge density is $10 \mu C/cm$ .

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density $\sigma$ = $\sigma/2\epsilon_{0}$ .

(To note: It's independent of distance from the plate!)

In the region outside the first plate,

since both plates have the same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.

(E due to positive plate away from it and E due to negative plate towards it!)

Hence, the electric field in the outer region of the first plate is zero.

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density $\sigma$ = $\sigma/2\epsilon_{0}$ .

(To note: It's independent of distance from the plate and same everywhere!)

In the region outside the second plate,

Since both plates have the same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.

(E due to positive plate away from it and E due to negative plate towards it!)

Hence, the electric field in the outer region of the second plate is zero.

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density $\sigma$ = $\sigma/\epsilon_{0}$ .

(To note: It's independent of distance from the plate!)

Let A and B be the two plates such that:

$\sigma_{A}= 17\times10^{-22} Cm^{-2}$ = $\sigma$

$\sigma_{B}= -17\times10^{-22} Cm^{-2}$ = - $\sigma$

Therefore,

The electric field between the plates, E = $E_{A} + E_{B}$ = $\sigma_{A}/2\epsilon_{0} + (-\sigma_{B}/2\epsilon_{0})$

$= \sigma/2\epsilon_{0} = 17\times10^{-22}/8.85\times10^{-12} = 1.92\times10^{-10 } NC^{-1}$

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields: Additional Exercises Question

The force due to the electric field is balancing the weight of the oil droplet.

weight of the oil drop = density x volume of the droplet x g = $\rho \times \frac{4}{3}\pi r^3 \times g$

Force due to the electric field = E x q

charge on the droplet, q = No. of excess electrons x charge of an electron = $12\times q_{e} = 12\times 1.6\times10^{-19} = 1.92\times10^{-18} C$

Balancing forces:

$\rho \times \frac{4}{3}\pi r^3 \times g = E\times q$

Putting known and calculated values:

$\\ r^3 =\frac{3}{4\pi} E\times q / \rho\times g = \frac{3}{4}\times\frac{7}{22}\times(2.55\times10^4)\times 1.92\times10^{-18} / (1.26\times10^3 \times10 ) \\ = 0.927\times10^{-18} m^3$

$r = 0.975\times10^{-6} m = 9.75\times10^{-4} mm$

(a) Wrong, because field lines must be normal to a conductor.

(b) Wrong, because field lines can only start from a positive charge. It cannot start from a negative charge,

(c) Right;

(d) Wrong, because field lines cannot intersect each other,

(e) Wrong, because electrostatic field lines cannot form closed loops.

Force on a charge F=qE

but here E is varying along the Z direction.

Force can be written as,

$F=q\frac{dE}{dz}dz=P\frac{dE}{dz}=10^{-7}\times10^5=10^{-2}N$

Torque experienced =0 since both dipole and electric field are in the Z direction. The angle between dipole and the electric field is 180 degrees

$\tau = \mathbf{P}\times \mathbf{E}=PEsin\180=0$

We know that the electric field inside a conductor is zero.

Using Gauss' law, if we draw any imaginary closed surface inside the solid, net charge must be zero.(Since E= 0 inside)

Hence there cannot be any charge inside the conductor and therefore, all charge must appear on the outer surface of the conductor.

We know, electric field inside a conductor is zero.

Now, imagine a Gaussian surface just outside the cavity inside the conductor. Since, E=0 (Using Gauss' law), hence net charge must be zero inside the surface. Therefore, -q charge is induced on the inner side of the cavity(facing conductor B).

Now consider a Gaussian surface just outside the conductor A. The net electric field must be due to charge Q and q. Hence, q charge is induced on the outer surface of conductor A. Therefore, the net charge on the outer surface of A is Q + q.

We know that the electric field inside a conductor is zero.

Therefore, a possible way to shield from the strong electrostatic fields in its environment is to enclose the instrument fully by a metallic surface.

Let the surface area of the sphere be S.

And assume that the hole is filled. For a point B, just above the hole, considering a gaussian surface passing through B, we have

$\oint E.dS = q/\epsilon_{0}$

Now, since the electric field is always perpendicular to the surface of the conductor.

$\\ \therefore \oint E.dS = E.S= q/\epsilon_{0} = (\sigma.S)/\epsilon_{0} \\ \implies E = \sigma/\epsilon_{0}$

Using Superposition principle, $E =E_{1} + E_{2}$ ,

where $E_{1}$ is due to the hole and $E_{2}$ is due to the rest of the conductor. (both pointing outwards, i.e away from the centre)

Again, for a point A, just below the hole, Electric field will be zero because of electrostatic shielding.

Using the superposition principle, this will be due to $E_{1}$ pointing inwards(towards the centre) and due to $E_{2}$ (Pointing away from the centre)

$0 =E_{1}-E_{2}$ $\implies E_{1}=E_{2}$

Using this relation, we get:

$E =E_{1} + E_{2} = 2E_{1} \implies E_{1} = E/2 = \sigma/2\epsilon_{0}$

Since this is pointing outwards,

$\boldsymbol{E_{1}} = \sigma/2\epsilon_{0}\ \boldsymbol{\widehat{n}}$ is the electric field in the hole.

(Trick: 1. Assume the hole to be filled.

2. Consider 2 points just above and below the hole.

3. Electric fields at these points will be due to the hole and rest of the conductor. Use superposition principle.)

[Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Let AB be a long thin wire of uniform linear charge density λ.

Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.

The charge on a small length dx on the line AB is q which is given as q = λdx.

So, according to Coulomb’s law, the electric field at P due to this length dx is

$dE'=\frac{1}{4\pi \epsilon}\frac{\lambda dx}{(PC)^2}$

But

This electric field at P can be resolved into two components as dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only dEcosθ component.

So, the net electric field at P due to dx is

dE' = dE cosθ

………………….(1)

In ΔPOC,

⇒ x = h tan θ

Differentiating both sides w.r.t. θ,

⇒ dx = h sec 2 θ dθ …………………….(2)

Also, h 2 + x 2 = h 2 + h 2 tan 2 θ

⇒ h 2 + x 2 = h 2 (1+ tan 2 θ)

⇒ h 2 + x 2 = h 2 sec 2 θ ………….(3)

(Using the trigonometric identity, 1+ tan 2 θ = sec 2 θ)

Using equations (2) and (3) in equation (1),

,

The wire extends from to since it is very long.

Integrating both sides,

This is the net electric field due to a long wire with linear charge density λ at a distance h from it.

In the question linear charge density =E

therefor

$E'=\frac{E}{2\pi \epsilon h}$

Given, a proton and a neutron consist of three quarks each.

And, ‘up’ quark is of charge + $\left ( \frac{2}{3} \right )$ e, and the ‘down’ quark of charge $\left (- \frac{1}{3} \right )$ e

Let the number of 'up' quarks be n. Therefore, the number of 'down' quarks is (3-n).

$\therefore$ The net charge = $\\ n\times\frac{2}{3}e + (3-n)\frac{-1}{3}e = (n-1)e$

Now, a proton has a charge +1e

$\therefore$ $(n-1)e = +1e \implies n =2$

Proton will have 2 u and 1 d, i.e, uud

Similarly, the neutron has a charge 0

$\therefore$ $(n-1)e = 0 \implies n =1$

Neutron will have 1 u and 2 d, i.e, udd

For equilibrium to be stable, there must be a restoring force and hence all the field lines should be directed inwards towards the null point. This implies that there is a net inward flux of the electric field. But this violates Gauss's law, which states that the flux of electric field through a surface not enclosing any charge must be zero. Hence, the equilibrium of the test charge cannot be stable.

Therefore, the equilibrium is necessarily unstable.

Two charges of same magnitude and sign are placed at a certain distance apart. The mid-point of the line joining these charges will have E =0.

When a test charged is displaced along the line towards the charges, it experiences a restoring force(which is the condition for stable equilibrium). But if the test charge is displaced along the normal of the line, the net electrostatic force pushes it away from the starting point. Hence, the equilibrium is unstable.

Let s be the vertical deflection, t be the time taken by the particle to travel between the plates

$\therefore$ $s = ut + \frac{1}{2}at^2$

Here , u =0 , since initially there was no vertical component of velocity.

The particle in the elecric field will experience a constant force (Since, Electric field is constant.)

F = ma = -qE (Using Newton's Second Law, F = ma)

$\therefore$ a = -qE/m (-ve sign implies here in downward direction)

Again, t = Distance covered/ Speed = $L/ v_{x}$

(In x-direction, since there is no force, hence component of velocity in x-direction remains constant = $v_{x}$ .

And, the distance covered in x-direction = length of the plate = L)

Putting these values in our deflection equation,

$\\ s = ut + \frac{1}{2}at^2 = (0)(L/v_{x}) + \frac{1}{2}(-qE/m)(L/v_{x})^2 \\ \implies s =\frac{-qEL^2}{2mv_{x}^2}$

(S is -ve, which implies it deflect in downwards direction.)

$\therefore$ The vertical deflection of the particle at the far edge of the plate is $\frac{qEL^{2}}{2mv_{x}^{2}}$ .

This motion is similar to the motion of a projectile in a gravitational field, which is also a constant force. The force acting on the particle in the gravitational field is mg whiles in this case, it is qE. The trajectory will be the same in both cases.

$\therefore$ The vertical deflection of the particle at the far edge of the plate is $s=\frac{qEL^{2}}{2mv_{x}^{2}}$

given s= 0.5cm=0.005cm

calculate for L from the above equation

$L=\sqrt{\frac{2smv_x^2}{qE}}=\sqrt{\frac{2\times 0.005\times 9.1\times 10^{-31}(2\times10^6)^2 }{1.6\times 10^{-19}\times9.1\times10^2}}$

L=1.6 cm

class 12 physics chapter 1 ncert solutions is very important for board exams, JEE, and NEET. It covers basic concepts like electric charges, Coulomb's law, and electric fields. Understanding these basics is essential for success in board exams, where questions test these fundamentals. In competitive exams like JEE and NEET, a strong grasp of class 12 physics chapter 1 exercise solutions is crucial for the physics section. Moreover, this chapter's concepts are interconnected with later chapters in Class 12, forming the groundwork for more advanced physics topics. So, mastering electric charges and fields class 12 is not only exam-important but also essential for building a strong physics foundation.

NCERT Solutions for Class 12 Physics- Chapter Wise

 Electric Charges and Fields Current Electricity Moving Charges and Magnetism Magnetism and Matter Alternating Current Electromagnetic Waves Ray Optics and Optical Instruments Wave Optics Solutions Dual nature of radiation and matter Atoms Nuclei Semiconductor Electronics Materials Devices and Simple Circuits
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Electric charges and fields class 12 NCERT solutions: Important Formulas and Diagrams

In class12 physics ch1 ncert solutions, you'll find important formulas and diagrams that aid in understanding electrostatic concepts. These resources are valuable for exam preparation and building a strong foundation in electrostatics.

• Coulomb's Law

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Where:

K is the proportionality, Constant and Q1 and Q2 are two-point charges.

The vector form of Coulomb's Law:

• Electric Field

Electric Field Intensity(E):

Where:

F is the force experienced by qo. The SI unit of E is V/m.

Electric Field Due To A Point Charge:

Electric Flux :

>> Flux is a scalar quantity. The SI unit of flux is volt-meter or newton-meter.

Electric Dipole:

Electric Dipole Moment:

Where: 2l is the dipole length.

>> The electric dipole moment is a vector quantity and the S.I unit is coulomb-meter (C-m).

• Gauss's Law

The total electric flux through a closed surface enclosing a charge q is equal to times the total charge q enclosed by the surface.

Topics Covered in Class 12 NCERT Chapter Electric Charges and Fields:

The main topics covered in the ch 1 Physics class 12 are:

Charge and its properties

This NCERT Class 12 topic discuss the concept of charge with practical examples of charging through friction, conduction and induction. Also, the concepts of properties like conservation, additivity and quantisation of charge are discussed in the chapter electric charges and fields. Questions based on this are present in electric charges and fields Class 12 solutions.

Coulomb's law and its applications

Coulomb's law discusses the relation between the force exerted between two charges separated by a distance. For example, questions 1 and 12 of NCERT solutions for Class 12 chapter 1 uses this law to solve it.

• Electric field: electric field and properties of electric field lines are discussed in this topic

• Electric dipole: The concepts of dipole and field due to dipole at the axial, equatorial and general point and torque due to electric dipole are discussed.

• Continuous charge distribution: Linear charge, surface charge and volume charge distributions are covered in this class 12 NCERT Physics chapter 1 topic

• Electric flux, Gauss's law and its applications: The concepts of electric flux, the relation between electric charge and flux and application of Gauss law to different charge configurations are discussed in class 12 NCERT.

All the topics mentioned in Physics Class 12 chapter 1 are important and students are advised to go through all the concepts mentioned in the topics. Questions from all the above topics are covered in the electric charges and fields NCERT solutions.

Importance of NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields in Board Exams:

The physics paper for the CBSE board exam is for 70 marks. In 2019 CBSE board exam 9 marks questions were asked from class 12 chapter 1 physics that includes the first two chapters, and this makes solving NCERT class 12 physics chapter 1 quite important. Practicing problems from the NCERT class 12 Physics solutions chapter 1 help students to score well in the final exam. Students can also use NCERT class 12 Physics chapter 1 solutions pdf download button to read chapter 1 physics class 12 NCERT solutions offline.

Key Features of Class 12 Electric Field and Charges NCERT Solutions

1. Comprehensive Coverage: This chapter covers fundamental concepts related to electric charges, Coulomb's law, electric fields, and their applications.

2. Exercise and Additional Exercise Questions: It includes a wide range of exercises and additional exercise questions to practice and assess understanding.

3. Detailed Solutions: Class12 physics ch1 ncert solutions provided in the NCERT textbook and supplementary materials offer step-by-step explanations for better comprehension.

4. Foundation for Advanced Topics: The concepts introduced in this class 12 physics chapter 1 ncert solutions serve as the building blocks for more complex topics in electromagnetism and modern physics.

5. Preparation for Exams: A strong grasp of this electric charges and fields class 12 is essential for success in board exams, JEE, and NEET.

6. Interconnected Concepts: ncert solution class 12 physics chapter 1 establishes connections to later chapters in the Class 12 physics syllabus like the gravitation chapter, emphasizing its significance.

Also Check NCERT Books and NCERT Syllabus here:

NCERT Exemplar Class 12 Solutions

 NCERT Exemplar Class 12 Chemistry Solutions NCERT Exemplar Class 12 Mathematics Solutions NCERT Exemplar Class 12 Biology Solutions NCERT Exemplar Class 12 Physics Solutions

Subject Wise Solutions

Excluded Content:

Certain portions of the chapter have been omitted or removed. These exclusions include:

1. Activity involving paper strips and the creation of an electroscope from Section 1.2 on Electric Charge.
2. The concept related to earthing in Section 1.3 on Conductors and Insulators.
3. The topic of Charging by Induction from Section 1.4.
4. Exercises numbered 1.13 and the range of exercises from 1.25 to 1.34.

These exclusions from class 12 electric charges and fields ncert solutions should be noted by students, and they should refer to their course materials for a precise understanding of what is covered in their syllabus.

1. What is the weightage of the chapter electric fields and charges for CBSE board exam

From the Class 12 Physics Chapter 1 NCERT solutions electric charges and fields 3 to 5 marks questions can be expected for CBSE bord exam. Questions based on the NCERT exercise questions can be expected for the exam. So practising the NCERT solutions for Class 12 Physics chapter 1 is important. The questions may be theory based, numerical or derivations.

2. Where can I find complete solutions of NCERT class 12 Physics

The detailed NCERT Class 12 Physics solutions chapter 1 is provided by careers 360. All the exercise and additional exercise questions of NCERT chapter Electric charges and Fields are solved with necessary explanations.

3. Whether the unit electrostatics is helpful in higher studies?

Yes, it is helpful in engineering streams and for research in the science field. The concepts studied in electrostatics will be useful in electrical and electronics-related branches.

4. What are the important topics of physics NCERT class 12 chapter 1

The two important laws discussed in NCERT solutions for Class 12 Physics Chapter 1  are important. That is the Coulombs law and the Gauss law. The derivations using Gauss laws and numerical related to both laws are important.

5. What is the weightage of electric charges and fields in JEE Mains

In JEE mains around 3.3% of questions are asked from the Physics class 12 chapter 1 questions and answers. That is 1 to 2 questions can be expected.

6. What is the weightge of NCERT chapter electric charges and fields for neet exam

For NEET exam 4% questions can be expected from the NCERT physics chapter 1 Electric Charges and Fields.

7. What are the main topics to be covered for NCERT Class 12 Physis Chapter 1 solutions
• Electric Charge
• Conductors And Insulators
• Charging By Induction
• Basic Properties Of Electric Charge
• Coulomb’s Law
• Electric Field
• Electric Field Lines
• Electric Flux
• Electric Dipole
• Continuous Charge Distribution
• Gauss’s Law And Applications

8. Can I prepare for the CBSE Exam with the NCERT Solutions for Class 12 Physics Chapter 1?

The experts at Careers360 created the ncert solution for class 12 physics chapter 1 to assist students in acing the board test with confidence. To boost pupils' confidence, the key principles are presented in the most organised manner possible. NCERT Solutions address every little nuance to aid students in their board exam preparation.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

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) Option 2) Option 3) Option 4)

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

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) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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

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 Option 4) be a function of the molecular mass of the substance.

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.

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

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