Chapter 1 Electric Charges and Fields

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NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields

Electric Charges & Electric Fields is referred to as the first chapter of Physics in 12th Class. It is essential for students going to learn about the physics of static Electric Charges as it provides the foundation for building a mathematical framework with respect to stationary Electric Charges. The first chapter starts with the basic empirical observations from Static Electric Regions and moves forward to establish the mathematical structure of Electromagnetism through Coulomb’s Law, Electric Field and Gauss’s Law. The detailed exploration of these topics is very important to build a strong foundation in Electromagnetism and will help the students to prepare and perform well in the upcoming Board Examinations as well as Competitive Entrance Tests like JEE & NEET.

The NCERT Solutions of Class 12 Physics Chapter 1 (Electric Charges and Fields) has been written logically & simply with a proper logical order so that every student has an understanding of Force & Field Vectors easily. The solutions have been prepared by the ALLEN experts as per the latest NCERT curriculum and are totally aligned with the CBSE Syllabus.

1.0NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields: Key Concepts

Class 12 Physics Chapter 1, Electric Charges and Fields, introduces the basic properties of electric charges and the electric fields produced by them. The chapter covers important concepts that help students understand electrostatic forces, electric fields, electric flux, and Gauss’s law.

Quantisation and Conservation of Charge: Understand that electric charge exists in integral multiples of the elementary charge and that charge cannot be created or destroyed.
Coulomb’s Law: Learn how to calculate the electrostatic force between two point charges and understand how the force depends on the charges and the distance between them.
Electric Field and Field Lines: Study the electric field produced by a charge and understand how electric field lines show the direction and strength of an electric field.
Electric Dipole: Learn about a pair of equal and opposite charges, its electric field, and the torque acting on an electric dipole placed in a uniform electric field.
Electric Flux and Gauss’s Law: Understand electric flux through a surface and learn how Gauss’s law is used to calculate electric fields for symmetrical charge distributions.
Applications of Gauss’s Law: Calculate the electric field due to an infinitely long straight charged wire, a uniformly charged infinite plane sheet, and a uniformly charged thin spherical shell.

2.0NCERT Solutions Class 12 Physics Chapter 1 - Electric Charges and Fields : Detailed Solutions

SOLVED EXAMPLES

If $1 0^{9}$ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body? Sol. In one second $1 0^{9}$ electrons move out of the body. Therefore the charge given out in one second is,

$q = ne, n = 1 0^{9} e^{-} s / sec ., e = 1.6 \times 1 0^{- 19} C q = 1 0^{9} \times 1.6 \times 1 0^{- 19} C q = 1.6 \times 1 0^{- 10} C / sec .$

The time required to accumulate a charge of 1C,

$t = \frac{1 C}{1.6 \times 1 0 ^{- 10} C / sec .} = 6.25 \times 1 0^{9} sec . t = \frac{6.25 \times 1 0 ^{9}}{3600 \times 24 \times 365} = 198 years$

Thus to collect a charge of one coulomb, from a body from which $1 0^{9}$ electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.

How much positive and negative charge is there in a cup of water? Sol. Let us assume that the mass of one cup of water is 250 g. The molecular mass of water is 18 g . Thus, one mole ( $= 6.02 \times 1 0^{23}$ molecules) of water is 18 g. Therefore the number of molecules in one cup of water is,

$= (\frac{250}{18}) \times 6.02 \times 1 0^{23}$

Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. Total positive charge,

$q = (\frac{250}{18}) \times 6.02 \times 1 0^{23} \times 10 \times 1.6 \times 1 0^{- 19} C q = 1.34 \times 1 0^{7} C .$

Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively. (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are $1 \overset{˚}{A} (= 1 0^{- 10} m)$ apart? $(m_{p} = 1.67 \times 1 0^{- 27} kg . m = 9.11 \times 1 0^{- 31} kg)$

Sol.

(a) (i) The electric force between an electron and a proton at a distance r apart is:

$F_{e} = - \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}}$

where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is:

$F_{G} = - G \frac{m _{p} m _{e}}{r ^{2}}$

where $m_{p}$ and $m_{e}$ are the masses of a proton and an electron respectively.

$\frac{F _{e}}{F _{G}} = \frac{e ^{2}}{4 π ε _{0} Gm _{p} m _{e}} = 2.4 \times 1 0^{39}$

(a) (ii) On similar lines. the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is:

$\frac{F _{e}}{F _{G}} = \frac{e ^{2}}{4 π ε _{0} Gm _{p} m _{p}} = 1.3 \times 1 0^{36}$

However. it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

(b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however, the masses of an electron and a proton are different. Thus, the magnitude of force is

$∣ F_{e} ∣ = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}} = \frac{9 \times 1 0 ^{9} Nm ^{2} / C ^{2} \times ( 1.6 \times 1 0 ^{- 19} C ) ^{2}}{( 1 0 ^{- 10} m ) ^{2}} ∣ F_{e} ∣ = 2.3 \times 1 0^{- 8} N$

Using Newton's second law of motion, $F = ma$ , the acceleration that an electron will undergo is, $a_{e} = \frac{∣ F _{e} ∣}{m _{e}}$

$a_{e} = \frac{2.3 \times 1 0 ^{- 8} N}{9.11 \times 1 0 ^{- 31} kg} = 2.5 \times 1 0^{22} m / s^{2}$

Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton. The value for acceleration of the proton is,

$a_{p} = \frac{∣ F _{e} ∣}{m _{p}} = \frac{2.3 \times 1 0 ^{- 8} N}{1.67 \times 1 0 ^{- 27} kg} = 1.4 \times 1 0^{19} m / s^{2}$

A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Figure (a).

Class 12 Physics diagram showing charged spheres A, B, C and D at different separations during electrostatic interaction

The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Figure (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Figure (c). What is the expected repulsion of A on the basis of Coulomb's law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres. Sol. Let the original charge on sphere A be q and that on B be q'. At a distance r between their centres, the magnitude of the electrostatic force on each is given by

$F = \frac{1}{4 π ε _{0}} \times \frac{qq ^{'}}{r ^{2}}$

neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge $\frac{q}{2}$ .

Similarly, after D touches B, the redistributed charge on each is $\frac{q ^{'}}{2}$ . Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is,

$F^{'} = \frac{1}{4 π ε _{0}} \frac{( q /2 ) ( q ^{'} /2 )}{( r /2 ) ^{2}} = \frac{1}{4 π ε _{0}} \frac{( qq ^{'} )}{r ^{2}} = F$

Thus the electrostatic force on A, due to B, remains unaltered.

Consider three charges $q_{1}, q_{2}, q_{3}$ each equal to q at the vertices of an equilateral triangle of side $ℓ$ . What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Figure ?

Class 12 Physics diagram showing electric forces on a charge at the centre of an equilateral triangle

Sol. In the given equilateral triangle ABC of sides of length $ℓ$ , if we draw a perpendicular AD to the side BC, $AD = AC cos 3 0^{\circ} = (3 /2) ℓ$ and the distance AO of the centroid O from A is,

$AO = (\frac{2}{3}) AD = (\frac{1}{3}) ℓ .$

By symmatry $AO = BO = CO$ . Thus, Force $F_{1}$ on Q due to charge q at A,

$F_{1} = \frac{3}{4 π ε _{0}} \frac{Qq}{ℓ ^{2}} along AO$

Force $F_{2}$ on Q due to charge q at B,

$F_{2} = \frac{3}{4 π ε _{0}} \frac{Qq}{ℓ ^{2}} along BO$

Force $F_{3}$ on Q due to charge q at C,

$F_{3} = \frac{3}{4 π ε _{0}} \frac{Qq}{ℓ ^{2}} along CO$

The resultant of forces $F_{2}$ and $F_{3}$ is $\frac{3}{4 π ε _{0}} \frac{Qq}{ℓ ^{2}}$ along OA, by the parallelogram law. Therefo re, the total force on $Q = \frac{3}{4 π ε _{0}} \frac{Qq}{ℓ ^{2}} (\overset{r}{^} - \overset{r}{^}) = 0$ , where $\overset{r}{^}$ is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero.

Consider the charges q, q, and -q placed at the vertices of an equilateral triangle, as shown in Figure. What is the force on each charge?

Electric force vectors on three charges in an equilateral triangle with resultant force shown

Sol. The forces acting on charge q at A due to charges q at B and -q at C are $F_{12}$ along BA and $F_{13}$ along AC respectively, as shown in Figure. By the parallelogram law, the total force $F_{1}$ on the charge q at A is given by $F_{1} = F \overset{r}{^}_{1}$ where $\overset{r}{^}_{1}$ is a unit vector along BC. The force of attraction or repulsion for each pair of charges has the same magnitude $F = \frac{q ^{2}}{4 π ε _{0} ℓ ^{2}}$ . The total force $F_{2}$ on charge q at B is thus $F_{2} = F \overset{r}{^}_{2}$ , where $\overset{r}{^}_{2}$ is a unit vector along AC. Similarly the total force on charge -q at C is $F_{3} = 3 F \overset{n}{^}$ , where $\overset{n}{^}$ is the unit vector along the direction bisecting the $∠ BCA$ . It is interesting to see that the sum of the force on the three charges is zero. i.e. $F_{1} + F_{2} + F_{3} = 0$ The result is not at all surprising. It follows straight from the fact that Coulomb's law is consistent with Newton's third law.

An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude $2.0 \times 1 0^{4} N C^{- 1}$ [Figure(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure (b)]. Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

Electric field between charged plates showing forces acting on positive and negative charges

Sol. In Figure (a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is

$a_{e} = \frac{e E}{m _{e}}$

where $m_{e}$ is the mass of the electron. Starting from rest, the time required by the electron to fall through a distance h is given by

$t_{e} = \frac{2 h}{a _{e}} = \frac{2 hm _{e}}{eE}$

For $e = 1.6 \times 1 0^{- 19} C, m_{e} = 9.11 \times 1 0^{- 31} kg$ .

$E t_{e} = 2.0 \times 1 0^{4} NC^{- 1}, h = 1.5 \times 1 0^{- 2} m . = 2.9 \times 1 0^{- 9} s$

In figure (b), the field is downward, and the positively charged proton experience a downward force of magnitude eE. The acceleration of the proton is

$a_{p} = \frac{e E}{m _{p}}$

where $m_{p}$ is the mass of the proton; $m_{p} = 1.67 \times 1 0^{- 27} kg$ . The time of fall for the proton is

$t_{p} = \frac{2 h}{a _{p}} = \frac{2 hm _{p}}{eE} = 1.3 \times 1 0^{- 7} s$

Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of 'free fall under gravity' where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field:

$a_{p} = \frac{eE}{m _{p}} = \frac{( 1.6 \times 1 0 ^{- 19} C ) \times ( 2.0 \times 1 0 ^{4} NC ^{- 1} )}{1.67 \times 1 0 ^{- 27} kg} = 1.9 \times 1 0^{12} ms^{- 2}$

which is enormous compared to the value of $g (9.8 m s^{- 2})$ , the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored in this example.

Two point charges $q_{1}$ and $q_{2}$ , of magnitude $+ 1 0^{- 8} C$ and $- 1 0^{- 8} C$ , respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Figure.

Class 12 Physics diagram showing electric field vectors at point C due to two charges

Sol. The electric field vector $E_{1 A}$ at A due to the positive charge $q_{1}$ points towards the right and has a magnitude

$E_{1 A} = \frac{( 9 \times 1 0 ^{9} Nm ^{2} C ^{- 2} ) \times ( 1 0 ^{- 8} C )}{( 0.05 m ) ^{2}} = 3.6 \times 1 0^{4} NC^{- 1}$

The electric field vector $E_{2 A}$ at A due to the negative charge $q_{2}$ points towards the right and has the same magnitude. Hence the magnitude of the total electric field $E_{A}$ at A is

$E_{A} = E_{1 A} + E_{2 A} = 7.2 \times 1 0^{4} NC^{- 1}$

$E_{A}$ is directed towards the right. The electric field vector $E_{1 B}$ at B due to the positive charge $q_{1}$ points towards the left and has magnitude

$E_{1 B} = \frac{( 9 \times 1 0 ^{9} Nm ^{2} C ^{- 2} ) \times ( 1 0 ^{- 8} C )}{( 0.05 m ) ^{2}} = 3.6 \times 1 0^{4} NC^{- 1}$

The electric field vector $E_{2 B}$ at B due to the negative charge $q_{2}$ points towards the right and has a magnitude

$E_{2 B} = \frac{( 9 \times 1 0 ^{9} Nm ^{2} C ^{- 2} ) \times ( 1 0 ^{- 8} C )}{( 0.15 m ) ^{2}} = 4 \times 1 0^{3} NC^{- 1}$

The magnitude of the total electric field at B is

$E_{B} = E_{1 B} - E_{2 B} = - 3.2 \times 1 0^{4} N C^{- 1}$

$E_{B}$ is directed towards the left. The magnitude of each electric field vector at point C, due to charge $q_{1}$ and $q_{2}$ is

$E_{1 C} = E_{2 C} = \frac{( 9 \times 1 0 ^{9} Nm ^{2} C ^{- 2} ) \times ( 1 0 ^{- 8} C )}{( 0.10 m ) ^{2}} = 9 \times 1 0^{3} NC^{- 1}$

The directions in which these two vectors point indicated in figure. The resultant of these two vectors is

$E_{C} = E_{1 C} cos \frac{π}{3} + E_{2 C} cos \frac{π}{3} = 9 \times 1 0^{3} NC^{- 1}$

$E_{C}$ points towards the right.

Two charges $\pm 10 μ C$ are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Figure (a), and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Figure (b).

Electric field along the axial line of an electric dipole with positive and negative charges

(a)

Class 12 Physics diagram showing electric field of a dipole at a point on its perpendicular bisector

(b)

Sol. (a) Field at P due to charge $+ 10 μ C$

$= \frac{1 0 ^{- 5} C}{4 π ( 8.854 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2}} \times \frac{1}{( 15 - 0.25 ) ^{2} \times 1 0 ^{- 4} m ^{2}} = 4.13 \times 1 0^{6} NC^{- 1} along BP$

Field at P due to charge $- 10 μ C$

$= \frac{1 0 ^{- 5} C}{4 π ( 8.854 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2} )} \times \frac{1}{( 15 + 0.25 ) ^{2} \times 1 0 ^{- 4} m ^{2}} = 3.86 \times 1 0^{6} NC^{- 1} along PA$

The resultant electric field at P due to the two charges at A and B is $= 2.7 \times 1 0^{5}$ $NC^{- 1}$ along BP. In this example, the ratio $\frac{OP}{OB}$ is quite large (= 60). Thus we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charge ± q, 2a distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude

$E = \frac{2 p}{4 π ε _{0} r ^{3}} (\frac{r}{a} ≫ 1)$

where $p = 2 aq$ is the magnitude of the dipole moment.

The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from -q to q). Here $p = 1 0^{- 5} C \times 5 \times 1 0^{- 3} m = 5 \times 1 0^{- 8} Cm$ Therefore,

$E = \frac{2 \times 5 \times 1 0 ^{- 8} Cm}{4 π ( 8.854 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2} )} \times \frac{1}{( 15 ) ^{3} \times 1 0 ^{- 6} m ^{3}} = 2.6 \times 1 0^{5} NC^{- 1}$

along the dipole moment direction AB, which is close to the result obtained earlier.

(b) field at Q due to charge $+ 10 μ C$ at B

$= \frac{1 0 ^{- 5} C}{4 π ( 8.854 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2} )} \times \frac{1}{[ 1 5 ^{2} + ( 0.25 ) ^{2} ] \times 1 0 ^{- 4} m ^{2}} = 3.99 \times 1 0^{6} NC^{- 1} along QA$

Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is

$= 2 \times \frac{0.25}{1 5 ^{2} + ( 0.25 ) ^{2}} \times 3.99 \times 1 0^{6} NC^{- 1}$

along $BA = 1.33 \times 1 0^{5} NC^{- 1}$ along BA. As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole:

$E = \frac{p}{4 π ε _{0} r ^{3}} (\frac{r}{a} ≫ 1) = \frac{5 \times 1 0 ^{- 8} Cm}{4 π ( 8.854 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2} )} \times \frac{1}{( 15 ) ^{3} \times 1 0 ^{- 6} m ^{3}} = 1.33 \times 1 0^{5} NC^{- 1}$

The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again, the result agrees with that obtained before.

The electric field components in Figure are $E_{x} = α x^{1/2}, E_{y} = E_{z} = 0$ , in which $α = 800 NC^{- 1} m^{- 1/2}$ . Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that $a = 0.1 m$ .

Gaussian surface diagram showing left and right normal vectors across a cubic region for electric flux

Sol. (a) Since the electric field has only an x component, for faces perpendicular to x direction, the angle between $E$ and $Δ S$ is $\pm \frac{π}{2}$ . Therefore, the flux $ϕ = E .Δ S$ is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is

$E_{L} = α x^{1/2} = α a^{1/2} (x = a at the left face) .$

The magnitude of electric field at the right face is

$E_{R} = α x^{1/2} = α (2 a)^{1/2} (x = 2 a at the right face) .$

The corresponding fluxes are

$ϕ_{L} = E_{L} \cdot Δ S = Δ SE_{L} \cdot \overset{n}{^}_{L} = E_{L} Δ S cos θ = - E_{L} Δ S, since θ = 18 0^{\circ} = - E_{L} a^{2} ϕ_{R} = E_{R} \cdot Δ S = E_{R} Δ S cos θ = E_{R} Δ S$

$ϕ_{R} = E_{R} a^{2}$ , since $θ = 0^{o}$ Net flux through the cube.

$= ϕ_{R} + ϕ_{L} = E_{R} a^{2} - E_{L} a^{2} = a^{2} (E_{R} - E_{L}) = α a^{2} [(2 a)^{1/2} - a^{1/2}] = α a^{5/2} (2 - 1) = 800 (0.1)^{5/2} (2 - 1) = 1.05 Nm^{2} C^{- 1}$

(b) We can use Gauss's law to find the total charge q inside the cube. We have

$ϕ q = \frac{q}{ε _{0}} . Therefore, = 1.05 \times 8.854 \times 1 0^{- 12} C = 9.27 \times 1 0^{- 12} C .$

An electric field is uniform, and in the positive x-direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that $E = 200 \hat{i} N / C$ for $x > 0$ and $E = - 200 \hat{i} N / C$ for $x < 0$ . A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at $x = + 10 cm$ and the other is at $x = - 10 cm$ (Figure). (a) What is the net outward flux through each flat face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder? Sol. (a) We can see from the figure that on the left face $E$ and $Δ S$ are parallel. Therefore, the outward flux is

$ϕ_{L} = E \cdot Δ S = - 200 \hat{i} \cdot Δ S = + 200Δ S, since \hat{i} \cdot Δ S = - Δ S = + 200 \times π (0.05)^{2} = + 1.57 Nm^{2} C^{- 1}$

On the right face $E$ and $Δ S$ are parallel and therefore

$ϕ_{R} = E \cdot Δ S = + 1.57 Nm^{2} C^{- 1} .$

(b) For any point on the side of the cylinder $E$ is perpendicular to $Δ S$ and hence $E .Δ S = 0$ . Therefore, then flux out of the side of the cylinder is zero. (c) Net outward flux through the cylinder

$ϕ = 1.57 + 1.57 + 0 = 3.14 Nm^{2} C^{- 1}$

Electric flux through a cylindrical Gaussian surface in a uniform electric field

(d) The net charge within the cylinder can be found by using Gauss's law which gives

$q = ε_{0} ϕ = 3.14 \times 8.854 \times 1 0^{- 12} C = 2.78 \times 1 0^{- 11} C$

An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?

Gaussian surface enclosing a charged sphere with electric field measured at an external point P

Sol. The charge distribution for this model of the atom is as shown in Figure. The total negative charge in the uniform spherical charge distribution of radius R must be -Z e, since the atom (nucleus of charge Z e + negative charge) is neutral. This immediately gives us the negative charge density $ρ$ , since we must have

$\frac{4 π R ^{3}}{3} ρ = 0 - Ze$

or $ρ = - \frac{3 Ze}{4 π R ^{3}}$ To find the electric field $E (r)$ at a point P which is a distance r away from the nucleus, we use Gauss's law. Because of the spherical symmetry of the charge distribution, the

magnitude of the electric field $E (r)$ depends only on the radial distance, no matter what the direction of r. Its direction is along (or opposite to) the radius vector r from the origin to the point P. The obvious Gaussian surface is a spherical surface centred at the nucleus. We consider two situations, namely, $r < R$ and $r > R$ .

(i) $r < R$ : The electric flux $ϕ$ enclosed by the spherical surface is

$ϕ = E (r) \times 4 π r^{2}$

where E (r) is the magnitude of the electric field at r. This is because the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there, and has the same magnitude at all points on the surface. The charge q enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius r,

$i.e. q = Z e + \frac{4 π r ^{3}}{3} ρ$

Substituting for the charge density $ρ$ obtained earlier, we have

$q = Z e - Z e \frac{r ^{3}}{R ^{3}}$

Gauss's law then gives,

$E (r) = \frac{Ze}{4 π ε _{0}} \frac{1}{r ^{2}} - \frac{r}{R ^{3}}; r < R$

The electric field is directed radially outward. (ii) $r > R$ : In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. Thus, from Gauss's law.

$E (r) \times 4 π r^{2} = 0 or E (r) = 0; r > R$

At $r = R$ , both cases give the same result; $E = 0$ .

EXERCISE QUESTION WITH SOLUTIONS

What is the force between two small charged spheres having charges of $2 \times 1 0^{- 7} C$ and $3 \times 1 0^{- 7} C$ placed 30 cm apart in air? Sol. $q_{1} = 2 \times 1 0^{- 7} C$

$q_{2} = 3 \times 1 0^{- 7} C r = 30 cm = 0.3 m$

Coulomb's law diagram showing repulsive forces between two point charges separated by 0.3 m

Electrostatic force between the spheres is given by the relation

$F = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{r ^{2}}$

Where, $ε_{0} =$ Permittivity of free space and

$\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} Nm^{2} C^{- 2}$

Therefore, force

$F = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 7} \times 3 \times 1 0 ^{- 7}}{( 0.3 ) ^{2}} = 6 \times 1 0^{- 3} N$

Hence, force between the two small charged spheres is $6 \times 1 0^{- 3} N$ . The charges are of same nature. Hence, force between them will be repulsive.

The electrostatic force on a small sphere of charge $0.4 μ C$ due to another small sphere of charge $- 0.8 μ C$ in air is 0.2 N . (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Sol. (a) Electrostatic force on the first sphere,

$F = 0.2 N$

Charge on this sphere,

$q_{1} = 0.4 μ C = 0.4 \times 1 0^{- 6} C$

Charge on the second sphere,

$q_{2} = - 0.8 μ C = - 0.8 \times 1 0^{- 6} C$

Electrostatic force between the spheres is given by the relation $F = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{r ^{2}}$ therefore, $r^{2} = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{F}$

$= F_{OA} = - F_{OC}, F_{OB} = - F_{OD} = 1.44 \times 1 0^{- 4} \Rightarrow r = 144 \times 1 0^{- 4} = 12 \times 1 0^{- 2} = 0.12 m = 12 cm$

The distance between the two spheres is 12 cm.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.

Check that the ratio $\frac{ke ^{2}}{Gm _{e} m _{p}}$ is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify? Sol. The given ratio is $\frac{ke ^{2}}{Gm _{e} m _{p}}$ . Where, G = Gravitational constant, Its unit is $Nm^{2} kg^{- 2}$ $m_{e}$ and $m_{p} =$ Masses of electron and proton respectively and their unit is kg. $e =$ electronic charge. Its unit is C. $k = \frac{1}{4 π ε _{0}}$ and its unit is $Nm^{2} C^{- 2}$ . Therefore, unit of the given ratio

$\frac{ke ^{2}}{Gm _{e} m _{p}} = \frac{[ Nm ^{2} C ^{- 2} ] [ C ^{2} ]}{[ Nm ^{2} kg ^{- 2} ] [ kg ] [ kg ]} = M^{0} L^{0} T^{0}$

Hence, The given ratio is dimensionless.

$e = 1.6 \times 1 0^{- 19} C G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2} m_{e} = 9.1 \times 1 0^{- 31} kg_{m}, m_{p} = 1.67 \times 1 0^{- 27} kg$

Hence, the numerical value of the given ratio is

$\frac{ke ^{2}}{Gm _{e} m _{p}} = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{6.67 \times 1 0 ^{- 11} \times 9.1 \times 1 0 ^{- 31} \times 1.67 \times 1 0 ^{- 27}} \approx 2.4 \times 1 0^{39}$

This is the ratio of electric force to the gravitational force between a proton and an electron. keeping distance between them constant.

(a) Explain the meaning of the statement 'electric charge of a body is quantised'. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges? Sol. (a) Electric charge of a body is quantised i.e. charge on any body can be added or removed in the form of integral multiple of e (charge of electron). (b) We ignore quantisation of electric charge at macroscopic level because at macroscopic level the grainyness of charge is not visible, also the charge is not countable just like while breathing we feel air as continuous but we know very well that it is made up of atoms and molecules of different gases.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pair of bodies. Explain how this observation is consistent with the law of conservation of charge. Sol. As no net charge is created or destroyed in isolation, only equal and opposite charges are created or neutralised, so this observation is consistent with the law of conservation of charge.
Four point charges $q_{A} = 2 μ C, q_{B} = - 5 μ C$ , $q_{C} = 2 μ C$ , and $q_{D} = - 5 μ C$ are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of $1 μ C$ Placed at the centre of the square? Sol. The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

Electric force vectors on a central charge due to four corner charges of a square

$F_{O A} = \frac{k ( 2 μ C ) ( 1 μ C )}{A O ^{2}}, F_{OB} = \frac{k ( 5 μ C ) ( 1 μ C )}{B O ^{2}}, F_{OC} = \frac{k ( 2 μ C ) ( 1 μ C )}{C O ^{2}}, F_{O D} = \frac{k ( 5 μ C ) ( 1 μ C )}{D O ^{2}}$

In a square $AO = BO = CO = DO$ So, $F_{OA} = - F_{OC}, F_{OB} = - F_{OD}$ These forces are same in magnitude but opposite in direction, so resultant force will be equal to zero. $F_{R} = F_{OA} + F_{OC} + F_{OB} + F_{OD} = 0$

$F_{R} = 0$

(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why? (b) Explain why two field lines never cross each other at any point ? Sol. (a) An electric field line represent electric field in a given region, as electrostatic field is long range and extended up to infinity so the line representing it is also a continuous curve either terminates on negatively charged body or extended up to infinity. (b) Two electric field lines can never cross each other because if they cross, there will be two directions of the resultant electric field at the point of intersection (say A); which is impossible.

Electric field lines 1 and 2 extending from point A, with electric field vectors E1 and E2 shown in their respective directions.

Two point charges $q_{A} = 3 μ C$ and $q_{B} = - 3 μ C$ are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude $1.5 \times 1 0^{- 9} C$ is placed at this point, what is the force experienced by the test charge ? Sol. (a)

Electric field at midpoint O between +3 μC and −3 μC charges, with electric field components E_A and E_B directed to the right.

Electric field at point ' O ' is

$E = E_{A} + E_{B} = 2 E_{A} (E_{A} = E_{B} in magnitude) = \frac{2 kq _{A}}{r ^{2}} = \frac{2 \times 9 \times 1 0 ^{9} \times 3 \times 1 0 ^{- 6}}{( 10 \times 1 0 ^{- 2} ) ^{2}} = 5.4 \times 1 0^{6} N / C along AO$

(b) Force on test charge

$F = qE = 1.5 \times 1 0^{- 9} \times 5.4 \times 1 0^{6} = 8.1 \times 1 0^{- 3} N along OA$

A system has two charges $q_{A} = 2.5 \times 1 0^{- 7} C$ and $q_{B} = - 2.5 \times 1 0^{- 7} C$ located at points A: (0, 0, -15 cm) and B : (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system? Sol. Total charge $= 2.5 \times 1 0^{- 7} - 2.5 \times 1 0^{- 7} = 0$ Dipole moment

$p = (qd) (- \hat{k}) = - (2.5 \times 1 0^{- 7} \times 30 \times 1 0^{- 2}) \hat{k} p = - 7.5 \times 1 0^{- 8} \hat{k} C - m$

An electric dipole with dipole moment $4 \times 1 0^{- 9} C - m$ is aligned at $3 0^{\circ}$ with the direction of a uniform electric field of magnitude $5 \times 1 0^{4} NC^{- 1}$ . Calculate the magnitude of the torque acting on the dipole. Sol. $τ = pE sin θ = 4 \times 1 0^{- 9} \times 5 \times 1 0^{4} sin 3 0^{\circ}$

$= 4 \times 1 0^{- 9} \times 5 \times 1 0^{4} \times \frac{1}{2} = 1 0^{- 4} N - m$

A polythene piece rubbed with wool is found to have a negative charge of $3 \times 1 0^{- 7} C$ . (a) Estimate the number of electrons transferred (from which to which ?) (b) Is there a transfer of mass from wool to polythene ? Sol. (a) $∵ Q = Ne$

$∴ N = \frac{Q}{e} = \frac{3 \times 1 0 ^{- 7}}{1.6 \times 1 0 ^{- 19}} = 1.9 \times 1 0^{12}$

Number of electrons transferred from wool to polythene $= 1.9 \times 1 0^{12}$ (b) Mass gained $= Nm_{e}$

$= 1.9 \times 1 0^{12} \times 9.1 \times 1 0^{- 31} = 17.3 \times 1 0^{- 19} kg$

(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 1 0^{- 7} C$ ? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved? Sol. (a) Charge on sphere A, $q_{A} = 6.5 \times 1 0^{- 7} C$ Charge on sphere $B, q_{B} = 6.5 \times 1 0^{- 7} C$ Distance between the spheres,

$r = 50 cm = 0.5 m$

Force of repulsion between the two spheres.

$F = \frac{1}{4 π ε _{0}} \cdot \frac{q _{A} q _{B}}{r ^{2}}$

Therefore,

$F = \frac{9 \times 1 0 ^{9} \times ( 6.5 \times 1 0 ^{- 7} ) ^{2}}{( 0.5 ) ^{2}} = 1.52 \times 1 0^{- 2} N$

Therefore, the force between the two spheres is $1.52 \times 1 0^{- 2} N$ . (b) If the charge on each sphere is doubled and the distance between them is halved, the force of repulsion, F' would become 16 times (as $F \propto q_{A} q_{B}$ and $F \propto 1/ r^{2}$ ). Clearly

$F^{'} = 16 F = 16 (1.52 \times 1 0^{- 2} N) = 0.243 N$

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio ?

Sol. Since particle (1) and (2) are deflected opposite to E so particle (1) and (2) are negatively charged and (3) is positively charged. Using, Newton's second equation of motion

$y = u_{y} t + \frac{1}{2} a_{y} t^{2} u_{y} = 0, F_{y} = m a_{y} = QE a_{y} = \frac{QE}{m} y = 0 + \frac{1}{2} (\frac{QE}{m}) t^{2} \Rightarrow y = \frac{1}{2} (\frac{QE}{m}) t^{2} \Rightarrow y \propto (\frac{Q}{m})$

The displacement, $y \propto (\frac{Q}{m})$ therefore particle (3) has highest charge to mass ratio.

Consider a uniform electric field $E = 3 \times 1 0^{3} \hat{i}$ N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis ? Sol. (a) Electric flux through the square,

$ϕ = E \cdot d A = (3 \times 1 0^{3} \hat{i}) \cdot (\frac{10}{100} \times \frac{10}{100} \hat{i}) = 30 Nm^{2} C^{- 1}$

(b) Again, $ϕ = E . d A = E dA cos θ$

Electric flux diagram showing a surface area vector at a 60° angle to the electric field, used to calculate electric flux as 15 N m² C⁻¹.

$= (3 \times 1 0^{3}) (\frac{10}{100} \times \frac{10}{100}) (cos 6 0^{\circ}) = (3 \times 1 0^{3}) (1 0^{- 2}) (\frac{1}{2}) = 15 Nm^{2} C^{- 1} .$

What is the net flux of the uniform electric field of Exercise 14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? Sol. All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines emerging out of the cube. As a result, net flux through the cube is zero.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 1 0^{3} Nm^{2} / C$ . (a) What is the net charge inside the box ? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box ? Why or Why not ? Sol. (a) Using $ϕ = \frac{q}{ε _{0}}$ , we get $q = ϕ ε_{0}$

$= (8 \times 1 0^{3}) (8.85 \times 1 0^{- 12}) = 70.8 \times 1 0^{- 9} C = 0.07 μ C$

(b) No, it can not be said to because there may be equal number of positive and negative elementary charges inside the box. It can only be said that net charge inside the box is zero.

A point charge $+ 10 μ C$ is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. What is the magnitude of the electric flux through the square ? (Hint : Think of the square as one face of a cube with edge 10 cm.)

Parallelogram diagram showing a +10 μC point charge positioned 5 cm above the centre of a square of side 10 cm. I prefer this response

Sol. In this situation, square may be consider as one out of six faces of cube of side 10 cm. According to Gauss theorem, total flux through 6 faces of cube is,

$ϕ_{cube} = \frac{q}{ε _{0}}$

∴ Flux through square,

$= \frac{1}{6} (\frac{q}{ε _{0}}) = \frac{4 π \times 9 \times 1 0 ^{9} \times 10 \times 1 0 ^{- 6}}{6} = 6 π \times 1 0^{4} Nm^{2} C^{- 1} = 1.88 \times 1 0^{5} Nm^{2} C^{- 1}$

A point charge of $2.0 μ C$ is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface ? Sol. Electric Flux, $ϕ = \frac{q}{ε _{0}} = \frac{2 \times 1 0 ^{- 6}}{8.85 \times 1 0 ^{- 12}}$

$= 2.25 \times 1 0^{5} Nm^{2} / C$

A point charge causes an electric flux of $- 1.0 \times 1 0^{3} Nm^{2} / C$ to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ? (b) What is the value of the point charge ? Sol. (a) The electric flux from new Gaussian surface remains same i.e. $- 1 \times 1 0^{3} Nm^{2} C^{- 1}$ because electric flux depends only on the charges enclosed in the gaussian surface and it is independent of the size of the gaussian surface. (b) Using $ϕ = \frac{q}{ε _{0}}$ , we get

$q = ε_{0} ϕ = (8.85 \times 1 0^{- 12}) (- 1 \times 1 0^{3}) i.e. q = - 8.85 \times 1 0^{- 9} C$

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is $1.5 \times 1 0^{3} N / C$ and points radially inward, what is the net charge on the sphere ? Sol. Using $E = \frac{9 \times 1 0 ^{9} ( q )}{r ^{2}}$ , we get

$q = \frac{Er ^{2}}{9 \times 1 0 ^{9}} = \frac{( 1.5 \times 1 0 ^{3} ) ( 20 \times 1 0 ^{- 2} ) ^{2}}{9 \times 1 0 ^{9}} = \frac{( 1.5 \times 1 0 ^{3} ) ( 0.2 ) ^{2}}{9 \times 1 0 ^{9}} = 6.67 nC$

Nature of charge will be negative because electric field points radially inward.

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of $80.0 μ C / m^{2}$ . (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? Sol. (a) $d = 2.4 m, r = 1.2 m$ ,

$σ = 80.0 μ C / m^{2} = 80 \times 1 0^{- 6} C / m^{2}$

charge on the sphere,

$q = σ A = σ \times 4 π r^{2} = 80 \times 1 0^{- 6} \times 4 \times 3.14 \times (1.2)^{2} = 1.447 \times 1 0^{- 3} C$

Therefore, the charge on the sphere is $1.447 \times 1 0^{- 3} C$ . (b) Total electric flux $(ϕ_{Total})$ leaving out the surface of a sphere containing net charge Q is given by the relation,

$ϕ_{total} = \frac{Q}{ε _{0}}$

$ϕ_{total} = \frac{1.447 \times 1 0 ^{- 3}}{8.854 \times 1 0 ^{- 12}} = 1.63 \times 1 0^{8} NC^{- 1} m^{2}$

therefore, the total electric flux leaving the surface of the sphere is $1.63 \times 1 0^{8} NC^{- 1} m^{2}$ .

An infinite line charge produces a field of $9 \times 1 0^{4} N / C$ at a distance of 2 cm . Calculate the linear charge density. Sol. Electric field produced by the infinite line charge at a distance 'd' having linear charge density $λ$ is given by the relation,

$E = \frac{λ}{2 π ε _{0} d} \Rightarrow E = \frac{2 k λ}{d} \Rightarrow λ = \frac{E d}{2 k}$

Where, $d = 2 cm = 0.02 m$

$E = 9 \times 1 0^{4} N / C$

therefore,

$λ = \frac{0.02 \times 9 \times 1 0 ^{4}}{2 \times 9 \times 1 0 ^{9}} = 1 0^{- 7} C / m = 0.1 μ C / m$

Therefore, the linear charge density is $0.1 μ C / m$ .

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 1 0^{- 22}$ $C / m^{2}$ . What is E : (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates ?

Sol. (a) $E_{A} = E_{P} - E_{Q} = \frac{σ}{2 ε _{0}} - \frac{σ}{2 ε _{0}}$

$= (\frac{17 \times 1 0 ^{- 22} - 17 \times 1 0 ^{- 22}}{2 ε _{0}}) = 0$

i.e. Electric field is zero.

Electric field calculation showing equal and opposite fields from two charged plates, resulting in zero electric field at point A.

(b) Similarly, electric field at a point B is zero $(E_{B} = 0)$ . (c) Electric field at C is given by

$E_{C} = E_{P} + E_{Q} = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{2 σ}{2 ε _{0}} = \frac{2 \times 17 \times 1 0 ^{- 22}}{2 \times 8.854 \times 1 0 ^{- 12}} = 1.92 \times 1 0^{- 10} NC^{- 1}$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Access NCERT Solutions for Class 12 Physics chapter-wise with exercise answers, important formulas, and clear step-by-step solutions to help students learn and practise.

Chapter Number	Chapterwise NCERT Solutions
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 1

Step-by-Step Derivations: Electric field derivations are explained step by step, making them easier to understand, learn, and revise for exams.
Correct Vector Use: The solutions show the correct use of vector signs and directions, helping students write and solve vector-based questions correctly.
Helpful for Competitive Exams: Detailed solutions help students understand important concepts and practise questions for CBSE board exams, JEE, NEET, and other entrance exams.
Correct Sign Conventions: The solutions use the correct signs for charge and electric flux, helping students avoid mistakes while solving numerical problems.
CBSE Exam Format: Answers are presented with the required formulas, steps, and explanations to help students understand how to write solutions in CBSE examinations.

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NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields

1.0NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 1 - Electric Charges and Fields : Detailed Solutions

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EXERCISE QUESTION WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 1

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