# NCERT solutions for class 12 Physics chapter 1 Electric Charges and Fields

NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields: Electrostatics is the study of electric charges at rest. The chapter 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. Solutions of NCERT class 12 physics chapter 1 electric charges and fields explain problems related to these topics. Try solving all the questions of NCERT and if any doubt arises then check the CBSE NCERT solutions for class 12 physics chapter 1 electric charges and fields. The derivations in NCERT chapter 1 electric charges and fields are very important in CBSE board exam point of view. Also, the steps in derivations are important as it will help in NCERT solutions for class 12 physics chapter 1 electric charges and fields. As the second chapter of NCERT class 12 is the continuation of the first chapter, so it is important to remember all points studied in the chapter. The NCERT solutions will help you in better understanding of the concepts you have studied.

While studying NCERT solutions for class 12 physics chapter 1, electric charges and fields you can compare the formulas of this chapter with the chapter gravitation of class 11 NCERT.

## Electrostatics and gravitation some comparisons

 Electrostatics Gravity The force between two charges separated by a distance r $F=\frac{Kq_1q_2}{r^2}$ The force between two masses separated by a distance r $F=\frac{Gm_1m_2}{r^2}$ Field strength $E=\frac{F}{q}$ Field strength $G=\frac{F}{m}$ Field strength $E=\frac{Kq}{r^2}$ Field Strength $g=\frac{Gm}{r^2}$

NCERT solutions for class 12 physics chapter 1 electric charges and fields exercises:

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 = 9x109 N m2 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 zero.

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 :

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

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

## Q 1.23 An infinite line charge produces a field of $9\times 10^{4}\frac{N}{C}$ at a distance of 2 cm. Calculate the linear charge density.

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:

$E = k\lambda / d$

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

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:

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 sec2θ dθ …………………….(2)

Also, h2 + x2 = h2 + h2tan2θ

⇒ h2 + x2 = h2(1+ tan2θ)

⇒ h2 + x2 = h2 sec2θ ………….(3)

(Using the trigonometric identity, 1+ tan2θ = sec2θ)

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

## NCERT solutions for class 12 physics- chapter wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

Subject wise solutions-

## Importance of solutions of NCERT for class 12 physics chapter 1 electric charges and fields in board exams:

The physics paper for CBSE board exam is for 70 marks. In 2019 CBSE board exam 9 marks question were asked from the session electrostatics that includes the first two chapters of NCERT class 12 physics. Practising problems from the NCERT solutions for class 12 physics chapter 1 electric charges and fields help students to score well in the final exam.

• Q.

### Can I get good marks in physics class 12 by solving the NCERT questions?

Yes,  by solving the NCERT questions you can get a good per cent of marks in physics. After studying all NCERT concepts and then by solving NCERT questions you can score more than 80% mark in Physics. Along with this if you are going through previous year board exam papers you and revising concepts well then you can score more than 90%.

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