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#### Page No 119:

An electron and a proton have equal and opposite charges of magnitude 1.6 × 10−19 C. But it doesn't mean that the electron has 3.2 × 10−19 C​ less charge than the proton.

#### Page No 119:

Yes, there's a lower limit to the electric force between two particles placed at a separation of 1 cm, which is equal to the magnitude of force of repulsion between two electrons placed at a separation of 1 cm.

#### Page No 119:

Electrostatic force follows the inverse square law, $F=\frac{k}{{r}^{2}}$. This means that the force on two particles carrying charges increases on decreasing the distance between them. Therefore, as particle A is slightly displaced towards B, the force on B as well as A will increase.

#### Page No 119:

No, a gravitational field cannot be added vectorially to an electric field.
This is because for electric influence, one or both the bodies should have some net charge and for gravitational influence both the bodies should have some mass. Also, gravitational field is a weak force,while electric field is a strong force.

#### Page No 119:

When a phonograph record is cleaned, it develops a charge on its surface due to rubbing. This charge attracts the neutral dust particles due to induction.

#### Page No 119:

Coulomb's Law states that the force between two charged particle is given by
$F=\frac{{q}_{1}{q}_{2}}{4\pi {\in }_{0}{r}^{2}}$,
where
q1 and q2 are the charges on the charged particles
r = separation between the charged particles
${\in }_{0}$= permittivity of free space

According to the Law of Superposition, the electrostatic forces between two charged particles are unaffected due to the presence of other charges.

#### Page No 119:

4π is the total solid angle. "4π lines of force" is just a way of stating that the field lines extend uniformly in all directions away from the charge.

#### Page No 119:

At the point of intersection, two normals can be drawn. Also, we know that electric field lines are perpendicular to the equipotential surface. This implies that at that point two different directions of the electric field are possible, which is not possible physically. Hence, two equipotential surfaces cannot cut each other.

#### Page No 119:

If a charge is placed at rest in an electric field, its path will be tangential to the lines of force. When the electric field lines are straight lines then the tangent to them will coincide with the electric field lines so the charge will move along them only. When the lines of force are curved, the charge moves along the tangent to them.

#### Page No 119:

The electric lines of force are entering charge q1; so, it is negative. On the other hand, the lines of force are originating from charge q​​​2; so, it is positive. If the lines are drawn in proportion to the charges, then
$\frac{{q}_{1}}{{q}_{2}}=\frac{6}{18}\phantom{\rule{0ex}{0ex}}⇒\frac{{q}_{1}}{{q}_{2}}=\frac{1}{3}$
6 lines are entering q​​​1 and 18 are coming out of q​​​2.

#### Page No 119:

Electrostatic field is a conservative field. Therefore, work done by the electric field does not depend on the path followed by the charge. It only depends on the position of the charge, from which and to which the charge has been moved.

#### Page No 119:

The separation between the two charges forming an electric dipole should be small compared to the distance of a point from the centre of the dipole at which the influence of the dipole field is observed.

#### Page No 119:

The outer electrons of an atom or molecule in a conductor are only weakly bound to it and are free to move throughout the body of the material. On the other hand, in insulators, the electrons are tightly bound to their respective atoms and cannot leave their parent atoms and move through a long distance.

#### Page No 119:

When a charged comb is brought near a small piece of paper, it attracts the piece due to induction. There's a distribution of charges on the paper. When a charged comb is brought near the pieces of paper then an opposite charge is induced on the near end of the pieces of paper so the charged comb attracts the opposite charge on the near end of paper and similar on the farther end. The net charge on the paper remains zero.

#### Page No 119:

(c) EA = EC > EB

The crowding of electric field lines at a point shows the strength of the field at that point. More the crowding of field lines, more will be the field strength. At points A and C, there's equal crowding, whereas at point B, the lines are far apart. Therefore, EA = EC > EB

#### Page No 119:

(d) may increase or decrease

The electric potential energy, E, between the two charges, q1 and q2, separated by the distance, r, is given as

As the distance between the charges is increased, the energy will decrease if both the charges are of similar nature. But if the charges are oppositely charged, the energy will become less negative and, hence, will increase.

#### Page No 119:

(a) increases

The electric potential energy, E, of a positive charge, q, in a potential, V, is given by E = qV. As the charge is moved from a low-potential region to a high-potential region, i.e. as V is increased, E will increase.

#### Page No 120:

(d) decreases then increases

Let the distance between the points A and B be r.

Let us take a point P at a distance x from A (x < r).

Electric potential V at point P due to two charges of equal magnitude q is given by
$V=\frac{q}{4\pi {\in }_{0}x}+\frac{q}{4\pi {\in }_{0}\left(r-x\right)}\phantom{\rule{0ex}{0ex}}⇒V=\frac{qr}{4\pi {\in }_{0}x\left(r-x\right)}$
Now, differentiating V with respect to x, we get

$\frac{\mathrm{d}V}{\mathrm{d}x}=\frac{-qr\left(r-2x\right)}{4\pi {\in }_{0}{x}^{2}\left(r-x{\right)}^{2}}$
Therefore, x = r/2.
It can be observed that $\frac{\mathrm{d}V}{\mathrm{d}x}<0$ for x < r/2.  Thus, the potential is decreasing first. At
x = r/2, the potential is minimum.
As $\frac{\mathrm{d}V}{\mathrm{d}x}>0$ for x > r/2, the potential is increasing after x = r/2.

#### Page No 120:

(a) A

The potential due to a charge decreases along the direction of electric field. As the electric field is along the positive x-axis, the potential will decrease in this direction. Therefore, the potential is minimum at point (a,0).

#### Page No 120:

(d) may increase slightly or may decrease slightly

If a body is rubbed with another body, it'll either gain some electrons from the other body and become negatively charged or it'll lose some electrons to the other body and become positively charged. Gain of electrons increases the weight of a body slightly and loss of electrons reduces the weight slightly.

#### Page No 120:

(a) is always zero

An electric dipole consists of two equal and opposite charges. When the dipole is placed in an electric field, both its charges experience equal and opposite forces. Therefore, the net resultant force on the dipole is zero. But net torque on the dipole is not zero.

#### Page No 120:

(c) WA = WB = WC

Points A, B and C lie at the same distance from the charge q, i.e. they are lying on an equipotential surface. So, work done in moving a charge from A to B (WAB) or B to C (WBC) is zero.
Hence, work done in bringing a charge from P to A  = WA,
from P to B, WB = WA+WAB = WA
and from to C, WC = WA + WAB + WBC = WA

Hence, WA = WB = WC

#### Page No 120:

(a) zero

The electrostatic field is conservative and the work done by the field is a state function, i.e. it only depends on the initial and final positions of the charge but not on the path followed by it. In completing one revolution, the charge has the same initial and final positions. Therefore, the work done by the field on rotating the charge in one complete revolution is zero.

#### Page No 120:

(a) The total charge of the universe is constant.

According to the principal of conservation of charge, the net amount of positive charge minus the net amount of negative charge in the universe is always constant. Thus, the total charge of the universe is constant. The total positive charge of the universe may increase or decrease, depending on the total increase or decrease in negative charge. This is the principle of conservation of charge that is universal in nature.

#### Page No 120:

(c) may increase if the charge is positive
(d) may decrease if the charge is negative

Electric field is a vector quantity. The electric field at a point due to a number of point charges is the vector sum of electric field due to individual charges. So, when a positive charge is brought into an electric field, the electric field due to the positive charge is added to the electric field already present. Therefore, the electric field increases.
When a negative charge is brought into an electric field, the electric field due to the negative charge is subtracted from the electric field already present. Therefore, the electric field decreases.

#### Page No 120:

None of the above.

Electric field,
$E=\frac{-dV}{dr}$, where V = electric potential
For E = 0,  V should be constant.
So, when E = 0,  it is not necessary that V should be 0.
Hence, none of the above signifies the correct relation.

#### Page No 120:

(b) may be equal to 20 Vcm−1
(c) may be greater than 20 Vcm−1

Change in the electric potential, dV = 40 V
Change in length, $∆r$ = −1−1 = −2 cm
Electric field,
$E=\frac{-dV}{dr}$

This is the value of the electric field along the x axis.
Electric field is maximum along the direction in which the potential decreases at the maximum rate. But here, direction in which the potential decreases at the maximum rate may or may not be along the x-axis. From the given information,the direction of maximum decrease in potential cannot be found out accurately. So, E can be greater than 20 V/cm in the direction of maximum decrease in potential.
So, the electric field at the origin may be equal to or greater than 20 Vcm−1.

#### Page No 120:

(b) Potential difference between two points
(d) Change in potential energy of a two-charge system

Potential and potential energy depend on the choice of a reference point of zero potential or zero potential energy. But the difference of potential and energy does not depend on the choice of the reference point. Hence, the correct options are (b) and (d).

#### Page No 120:

(d) The torque on the dipole due to the field may be zero.
Torque acting on a dipole placed in an electric field,

where $\theta$ is the angle between the force F and the arm of the couple.
So, for $\theta$=0 or pi ,the torque will be zero. Hence, the amount of torque acting on the dipole depends on the orientation of the dipole in the given electric field.

#### Page No 120:

(b) The magnitudes of the forces will be equal.

We know:
$\stackrel{\to }{F}=q\stackrel{\to }{E}$

For an electron and a proton, the value of q will be same, but the sign will be opposite.
Hence, they will experience a force that will be equal in magnitude but opposite in direction.
Now,
$\stackrel{\to }{F}=q\stackrel{\to }{E}=m\stackrel{\to }{a}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{a}=\frac{q\stackrel{\to }{E}}{m}$

As the electron and proton have different values of mass m, they will have different magnitudes of acceleration. Also, they will differ in direction due to the opposite signs of q

#### Page No 120:

(c) it is proportional to r2

Given:
E$\propto$r  and V = 0 at r =0
$⇒$E = kr
Also, E = $\frac{-dV}{dr}$

Using the condition, V = 0 at r = 0, we get  C =0.
Therefore,

#### Page No 121:

By Coulomb's Law,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$
$⇒{\epsilon }_{0}=\frac{1}{4\mathrm{\pi }F}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$
Using [F] = [MLT−2]
[r] = [M0L1T0]
[q] = [M0L0T1A1], we get
0] = [M−1L−3T4A2]

#### Page No 121:

Given:
q1 = q2 = q = 1.0 C
Distance between the charges, r = 2 km = 2 × 103 m
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

Let my mass, m, be 50 kg.
Weight of my body, W = mg
⇒ W =
50 × 10 N = 500 N
Now,

So, the force between the charges is 4.5 times the weight of my body.

#### Page No 121:

Given:
Magnitude of charges, q1 = q2 = 1 C
Electrostatic force between them, F = Weight of a 50 kg person = mg = 50 × 9.8 = 490 N
Let the required distance be r.
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}⇒490=\frac{9×{10}^{9}×1×1}{{r}^{2}}\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\frac{9×{10}^{9}}{490}\phantom{\rule{0ex}{0ex}}$

#### Page No 121:

Let the magnitude of each charge be q.
Separation between them, = 1 m
Force between them, F = 50 × 9.8 = 490 N
By Coulomb's Law force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

#### Page No 121:

We know:
Charge on a proton, q = 1.6 × 10−19 C
Given, separation between the charges, r = 1015 m
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

#### Page No 121:

Given:

Let the third charge, q, be placed at a distance of x cm from charge q1, as shown in the figure.

By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{Q}_{1}{Q}_{2}}{{r}^{2}}$
Force on charge q due to q1,
$F=\frac{9×{10}^{9}×2.0×{10}^{-6}×q}{{x}^{2}}$
Force on charge q due to q2,
$F\text{'}=\frac{9×{10}^{9}×{10}^{-6}×q}{{\left(10-x\right)}^{2}}\phantom{\rule{0ex}{0ex}}$
According to the question,
$F-F\text{'}=0$

So, the third charge should be placed at a distance of 5.9 cm from q1.

#### Page No 121:

Given:

Since both the charges are opposite in nature, the third charge cannot be placed between them. Let the third charge, q, be placed at a distance of x cm from charge q1, as shown in the figure.

By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{Q}_{1}{Q}_{2}}{{r}^{2}}$
So, force on charge q due to q1,
$F=\frac{9×{10}^{9}×2.0×{10}^{-6}×q}{{x}^{2}}$
Force on charge q due to q2,

#### Page No 121:

We know that minimum charge on a body (q) = charge on an electron
q = 1.6 × 10−19 C
Given:
Separation between the charges, r = 1 cm = 10-2 m
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

#### Page No 121:

Molecular mass of water= 18 g
Number of molecules in 18 g of H2O = Avogadro's number
= 6.023 × 1023
Number of electrons in 1 molecule of H2O = (2 × 1) + 8 = 10
Number of electrons in 6.023 × 1023 molecules of H2O = 6.023 × 1024
That is, number of electrons in 18 g of H2O = 6.023 × 1024
So, number of electrons in 100 g of H2O = $\frac{6.023×{10}^{24}}{18}×100$
= 3.34 × 1025
∴ Total charge = 3.34 × 1025 × 1.6 × 10−19
= 5.34 × 106 C

#### Page No 121:

Molecular mass of water= 18 g
So, number of atoms in 18 g of H2O = Avogadro's number
= 6.023 × 1023
Number of electrons in 1 atom of H2O = (2 × 1) + 8 = 10
Number of electrons in 6.023 × 1023 atoms of H2O= 6.023 × 1024
That is, number of electrons in 18 g of H2O = 6.023 × 1024
So, number of electrons in 100 g of H2O = $\frac{6.023×{10}^{24}}{18}×100$
= 3.34 × 1025
Total charge = 3.34 × 1025 × (−1.6 × 10−19)
=− 5.34 × 106 C
So total charge of electrons in 100 gm of water, q1 = −5.34 × 106 C
Similarly, total charge of protons in 100 gm of water, q2 = +5.34 × 106 C
Given, r = 10 cm = 0.1 m
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

This force will be attractive in nature.
Result shows that the electrostatic force is much stronger than the gravitational force between any us and earth(weight=gravitational force between us and earth).

#### Page No 121:

Given, radius of the sphere, R = 6.9 fermi
So, the largest separation between two protons = 2R = 13.8 fermi
Charge on a proton, q =
By Coulomb's Law, force of repulsion,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

Inside the nucleus, another short-range attractive force (nuclear force) acts on the protons. That's why these protons do not fly apart due to the Coulombian repulsion.

#### Page No 121:

Given:
Force of attraction between the spheres, F = 0.1 N
Separation between the spheres, r = 10−2 m
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$
Let the no. of electrons transferred from one sphere to the other be n. Then,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{\left(nq\right)}^{2}}{{r}^{2}}$
$⇒0.1=\frac{9×{10}^{-9}×{n}^{2}×{\left(1.6×{10}^{-19}\right)}^{2}}{{10}^{-4}}\phantom{\rule{0ex}{0ex}}⇒{n}^{2}=\frac{0.1×{10}^{-4}}{9×{10}^{9}×1.6×1.6×{10}^{-38}}\phantom{\rule{0ex}{0ex}}⇒n=2×{10}^{11}$

#### Page No 121:

Let the given separation between the ions, i.e. 2.75 × 10−8 cm, be the separation between the transferred electron and the sodium nucleus.

Net charge on thechlorine ion, q1 = $-1.6×{10}^{-19}$ C
Net charge on the sodium ion, q2 = $1.6×{10}^{-19}$ C
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

#### Page No 121:

We know that the mass of a proton, m = 1.67 × 10−27 kg
Charge on a proton, q = 1.6 × 10−19 C
Gravitational constant,
Electrostatic force,

Gravitational force,

#### Page No 121:

(a) Given, nuclear force of attraction, $F=C\frac{{e}^{-Kr}}{{r}^{2}}$
Here eKr is just a pure number, i.e. a dimensionless quantity. So,
$\left[C\right]=\left[F\right]×\left[{r}^{2}\right]\phantom{\rule{0ex}{0ex}}\left[C\right]=\left[{\mathrm{MLT}}^{-2}\right]×\left[{\mathrm{L}}^{2}\right]$
$\left[\mathrm{C}\right]=\left[{\mathrm{ML}}^{3}{\mathrm{T}}^{-2}\right]$

(b)By Coulomb's Law, electric force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{e}^{2}}{{r}^{2}}$
Taking , we get
$F=\frac{9×{10}^{9}×{\left(1.6×{10}^{-19}\right)}^{2}}{{\left(5×{10}^{-15}\right)}^{2}}$
And nuclear force, F = Ce−kr/r2
Taking r = 5 × 10-15 m and k = 1 fermi−1, we get
$F=\frac{C×{10}^{-5}}{{\left(5×{10}^{-15}\right)}^{2}}\phantom{\rule{0ex}{0ex}}$
Comparing both the forces, we get

#### Page No 121:

Since all the charges are of equal magnitude, the force on the charge at A due to the charges at B and C will be of equal magnitude. (As shown in the figure)

That is, ${F}_{\mathrm{BA}}={F}_{\mathrm{CA}}=F\left(\mathrm{say}\right)$

The horizontal components of force cancel each other and the net force on the charge at A,

Given: r = 5 cm =0.05 m
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$
$F\text{'}=\frac{2×9×{10}^{9}×{\left(2×{10}^{-6}\right)}^{2}×\mathrm{sin}60°}{{\left(0.05\right)}^{2}}\phantom{\rule{0ex}{0ex}}F\text{'}=\frac{2×9×{10}^{9}×{\left(2×{10}^{-6}\right)}^{2}×\frac{\sqrt{3}}{2}}{{\left(0.05\right)}^{2}}$
F' = 24.9 N

#### Page No 121:

Given,
Magnitude of the charges,
Side of the square,
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

So, force on the charge at A due to the charge at B,

Force on the charge at A due to the charge at C,

Force on the charge at A due to the charge at D,
${\stackrel{\to }{F}}_{\mathrm{DA}}={\stackrel{\to }{F}}_{\mathrm{BA}}$

The resultant force at A, F'= ${\stackrel{\to }{F}}_{\mathrm{BA}}+{\stackrel{\mathit{\to }}{\mathit{F}}}_{\mathrm{CA}}+{\stackrel{\mathit{\to }}{\mathit{F}}}_{\mathrm{DA}}$
The resultant force of will be $\sqrt{2}{F}_{\mathrm{BA}}$ in the direction of ${\stackrel{\mathit{\to }}{F}}_{\mathrm{CA}}$. Hence, the resultant force,

#### Page No 121:

Given:
Separation between the two charges, r = 0.53 Å = 0.53 × 10−10 m
${q}_{1}={q}_{2}=e$
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

#### Page No 121:

Given:
Separation between the two charges, r = 0.53 Å = 0.53 × 10−10 m
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$
Here, ${q}_{1}={q}_{2}=e$

Now, mass of an electron, Me = 9.12 × 1031 kg
The necessary centripetal force is provided by the Coulombian force.

#### Page No 121:

By Coulomb's Law, force (F) on charge q due to one charge,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}q}{{r}^{2}}$
So, net force due to ten charges,

#### Page No 121:

Given:
Magnitude of the charges, q =  2.0 × 10−8 C
Separation between the charges, r = 1 m
The tension in the string will be same as the electrostatic force between the charged particles.
So, $T=F$
By Coulomb's Law,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}^{2}}{{r}^{2}}$

#### Page No 121:

Given:
Magnitude of the charges, q =  2.0 × 10−7 C
Separation between the charges,
Length of the string,
Mass of the balls, m = 100 g = 0.1 kg

(a) By Coulomb's Law, the electric force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}^{2}}{{r}^{2}}$

(b) The components of the resultant force along it is zero because mg balances Tcosθ, as shown in the figure. So,
F = Tsinθ

(c) Tension in the string,
Tsinθ = F     ..(1)
Tcosθ = mg    ...(2)
Squaring equations (1) and (2) and adding, we get

#### Page No 121:

Let the tension in the string be T and the force of attraction between the two balls be F.
From the free-body diagram of the balls, we get
Tcosθ = mg    ...(1)
Tsinθ = F    ...(2)
From ∆ABC,
$\mathrm{sin}\theta =\frac{1}{20}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\sqrt{1-{\left(\frac{1}{20}\right)}^{2}}$
From equation (1) and (2),

#### Page No 121:

Tcosθ = mg   ...(i)
Tsinθ = Fe   ...(ii)
Here,  ${F}_{e}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{{q}^{2}}{{r}^{2}}$
$\mathrm{tan}\theta =\frac{2}{39.9}$
Dividing equation (ii) by (i), we get

#### Page No 122:

Let the tension in the string be T and force of attraction between the two balls be F.
From the free-body diagram of the ball, we get
Tcos θ = mg    ...(1)
Tsin θ = F    ...(2)

Here, separation between the two charges,  $r=2l\mathrm{sin}\theta$
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}^{2}}{{r}^{2}}=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}×\frac{{q}^{2}}{{\left(2l\mathrm{sin}\theta \right)}^{2}}$
Substituting the value of F in equation (3), we get
$m=\frac{{q}^{2}\mathrm{cot}\theta }{16\mathrm{\pi }{\epsilon }_{0}g{l}^{2}{\mathrm{sin}}^{2}\theta }$

#### Page No 122:

Given:
Mass of the bob, m = 100 g = 0.1 kg
So, tension in the string, T = mg
⇒ T = 0.1 × 9.8 = 0.98 N
For the tension to be zero, the repelling force (Fe) on the bob = T
Magnitude of the charge placed below the bob, q =  2.0 × 10−4 C
Separation between the charges, r = 0.1 m
When the electrostatic force between the bob and the particle is balanced by the tension in the string then the string will become loose.
Let the required charge on the bob be q' .
$⇒{F}_{e}=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{qq\text{'}}{{r}^{2}}=T$

#### Page No 122:

For equilibrium, ${\stackrel{\to }{\mathrm{F}}}_{AC}+{\stackrel{\to }{\mathrm{F}}}_{CB}=0$

Let the charge at point c be θ.

For a charge at rest,

#### Page No 122:

Let the extension in the string be x.
Given:
Magnitude of the charges, q =  2.0 × 10−8 C
Separation between the charges, r = (0.1+ x) m
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}^{2}}{{r}^{2}}$
The spring force due to extension x,
$F=Kx$
For equilibrium,
Electrostatic force = Spring force

Yes, the assumpton is justified. As two similar charges are present at the ends of the spring so they exert repulsive force on each other.Due to the repulsive force between the charges,an extension x is produced in the spring. Springs are made up of elastic material.When a spring is extended then a restoring force acts on it which is always proportioanal to the extension produced and directed opposite to the direction of applied force.The restoring force depends on the elasticity of the material. When the extension is small then only the restoring force is proportinal to the extension.If the extension s camparable to the natural length of the spring then the restoring force will depend on higher powers of the extension produced.

#### Page No 122:

Given:
Magnitude of charge of particle A, q1 =  2.0 × 10−5 C
Separation between the charges, r = 0.1 m
Mass of particle B, m = 80 g = 0.08 kg
Let the magnitude of charge of particle B be q2.
By Coulomb's Law, force on B due to A,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$
Force of friction on B = $\mu mg$
Since B is at equilibrium,
Coulomb force = Frictional force

∴ The range of the charge = ∓8.71×10-8 C

#### Page No 122:

Given:
Magnitude of charge on particles A and B, q =  2.0 × 10−6 C
Mass of particles A and B, m = 100 g = 0.1 kg
Let the separation between the charges be r along the plane.
By Coulomb's Law, force (F) on B due to A,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}^{2}}{{r}^{2}}$
For equilibrium:

For equilibrium along the plane,

So, the charge should be placed at a distance of 27 cm from the bottom.

#### Page No 122:

Let the charge q be placed at a distance x on the  perpendicular bisector of AB.
As shown in the figure, the horizontal component of force is balanced.

$\mathrm{sin}\theta =\frac{x}{\sqrt{{\left(\frac{d}{2}\right)}^{2}+{x}^{2}}}$
Total vertical component of force, $F\text{'}=2F\mathrm{sin}\theta$
$F\text{'}=2×\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}×\frac{qQ}{{\left(\frac{d}{2}\right)}^{2}+{x}^{2}}×\frac{x}{\sqrt{{\left(\frac{d}{2}\right)}^{2}+{x}^{2}}}\phantom{\rule{0ex}{0ex}}⇒F\text{'}=\frac{1}{2\mathrm{\pi }{\epsilon }_{0}}×\frac{qQx}{{\left[{\left(\frac{d}{2}\right)}^{2}+{x}^{2}\right]}^{3/2}}$
For maximum force, $\frac{d\mathrm{F}\text{'}}{dx}=0$

#### Page No 122:

(a) The charge q is displaced by a distance x on the perpendicular bisector of AB.
As shown in the figure, the horizontal component of the force is balanced.

$\mathrm{sin}\theta =\frac{x}{\sqrt{{\left(\frac{d}{2}\right)}^{2}+{x}^{2}}}$
Total vertical component of the force, $F\text{'}=2F\mathrm{sin}\theta$
$F\text{'}=2×\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}×\frac{qQ}{{\left(\frac{d}{2}\right)}^{2}+{x}^{2}}×\frac{x}{\sqrt{{\left(\frac{d}{2}\right)}^{2}+{x}^{2}}}\phantom{\rule{0ex}{0ex}}⇒F\text{'}=\frac{1}{2\mathrm{\pi }{\epsilon }_{0}}×\frac{qQx}{{\left[{\left(\frac{d}{2}\right)}^{2}+{x}^{2}\right]}^{3/2}}$
This is the net electric force experienced by the charge q.

(b) When x < < d:

(c) For the particle to execute simple harmonic motion:
F' = mw2x

#### Page No 122:

Net force
$=\frac{1}{4\mathrm{\pi }{\in }_{0}}\left[\frac{Qq}{{\left(\frac{d}{2}-x\right)}^{2}}-\frac{Qq}{{\left(\frac{d}{2}+x\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{Qq}{4\mathrm{\pi }{\in }_{0}}\frac{\left[{\left(\frac{d}{2}\right)}^{2}+{x}^{2}+xd-{\left(\frac{d}{2}\right)}^{2}-{x}^{2}+xd\right]}{{\left[{\left(\frac{d}{2}\right)}^{2}-{x}^{2}\right]}^{2}}$
when,
x << d
So, net force =

#### Page No 122:

Given:
Magnitude of the charges, q =  1.0 × 10−6 C
Electric force, F = 1.5 × 10−3 N
We know that F = qE

#### Page No 122:

Given:
Magnitude of the charges,
and
Separation between the charges, r = 0.2 m
(a) Let the electric field be zero at a distance x from A, as shown in the figure.

Then,
$\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}}{{x}^{2}}+\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{2}}{{\left(0.2-\mathrm{x}\right)}^{2}}=0\phantom{\rule{0ex}{0ex}}⇒\frac{2}{{x}^{2}}+\frac{-4}{{\left(0.2-x\right)}^{2}}=0$
or x = 48.3 cm from A along BA.

(b) Let the potential be zero at a distance x from A.

x = from A along AB
and x = 20 cm from A along BA.

#### Page No 122:

Given:
Magnitude of the electric field, E = 5.0 NC−1 at a distance, r = 40 cm = 0.4 m
Let the magnitude of the charge be q.

#### Page No 122:

Mass of the particle,
Charge on the particle,
Let the magnitude of the electric field be E.
The particle stays suspended. Therefore,
Downward gravitational force = Upward electric force
That is, mg  = qE

The direction of the electric field will be upward to balance the downward gravitational force.

#### Page No 122:

Given:
Magnitude of charges, q =  1.0 × 10−8 C
Side of the triangle,

Let be the electric fields at the centre due to the charges at A, B and C, respectively.
The distance of the centre is the same from all the charges. So,

Resolving into vertical and horizontal components (with $\theta =30°$)
The horizontal components cancel each other, as shown in the figure.
So, the net electric field at the centre,

Now,
${h}^{2}={l}^{2}-{\left(\frac{l}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}{h}^{2}={\left(0.2\right)}^{2}-{\left(0.1\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒h=\frac{\sqrt{3}}{10}$
Let the distance of the centre from each charge be r.
For an equilateral triangle,

Potential at the centre,

$V=3×\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{q}{r}\phantom{\rule{0ex}{0ex}}V=\frac{3×9×{10}^{9}×{10}^{-8}}{0.115}$

#### Page No 122:

Consider an element of angular width $d\theta$ at a distance r from the charge q on the circular ring, as shown in the figure.

So, charge on the element,
$dq=\frac{Q}{2\pi R}Rd\mathrm{\theta }=\frac{Q}{2\mathrm{\pi }}d\mathrm{\theta }$
Electric field due to this charged element,

By the symmetry, the Esinθ component of all such elements on the ring will vanish.
So, net electric field,
$d{E}_{net}=dE\mathrm{cos}\theta =\frac{Qd\theta }{8{\pi }^{2}{\epsilon }_{0}{\left({R}^{2}+{x}^{2}\right)}^{3/2}}$
Total force on the charged particle,

According to the question,

Comparing this with the condition of simple harmonic motion, we get

#### Page No 122:

Consider an element of angular width dθ, as shown in the figure.

$dq=\frac{Q}{L}Rd\mathrm{\theta }$
The net electric field has a vertical component only.

#### Page No 122:

Net electric field = ∫dE cos θ
$\int \frac{1}{4\mathrm{\pi }{\in }_{0}}\frac{\frac{q}{l}.dx}{\left(75-{x}^{2}\right)}.\frac{x}{\sqrt{75-{x}^{2}}}\phantom{\rule{0ex}{0ex}}=\underset{-5}{\overset{+5}{\int }}\frac{1}{4\mathrm{\pi }{\in }_{0}}.\frac{q}{l}\frac{xdx}{\sqrt{75-{x}^{2}.}\left(75-{x}^{2}\right)}\phantom{\rule{0ex}{0ex}}=5.2×{10}^{7}\mathrm{N}/\mathrm{C}$

#### Page No 122:

Let the total charge of the ring be Q.
Radius of the ring = R
The electric field at distance x from the centre of ring,

For maximum value of electric field,
$\frac{dE}{dx}=0$
From equation (1),
$\frac{dE}{dx}=\frac{Q}{4\pi {\epsilon }_{0}}\left[1\left({R}^{2}+{x}^{2}{\right)}^{-3/2}-x\frac{3}{2}\left({R}^{2}+{x}^{2}{\right)}^{-5/2}2x\right]=0$
$⇒{R}^{2}+{x}^{2}-3{x}^{2}=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}={R}^{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{R}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}$

#### Page No 122:

As the wire is bent to form a regular hexagon, it forms an equipotential surface, as shown in the figure.
Hence, the charge at each point is equal and the net electric field at the centre is 0.

#### Page No 122:

We know that the electric field is zero at the centre of a uniformly charged circular wire.
That is, the electric field due to a small element dl of wire + electric field due to the remaining wire = 0
Let the charge on the small element dl be dq. So,
$dq=\frac{Q}{2\pi a}dL$
Electric field due to a small element at the centre,
$E=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{dq}{{a}^{2}}\phantom{\rule{0ex}{0ex}}⇒E=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}{\mathrm{a}}^{2}}\frac{Q}{2\pi a}dL=\frac{QdL}{8{\pi }^{2}{\epsilon }_{0}{a}^{3}}$
So, the electric field at the centre due to the remaining wire = $\frac{QdL}{8{\mathrm{\pi }}^{2}{\epsilon }_{0}{a}^{3}}$ (Opposite the direction of the electric field due to the small element)

#### Page No 122:

Electric field at a point due to a given
charge, E = $\frac{q}{4\mathrm{\pi }{\in }_{0}{d}^{2}}$,
where q is the charge and R the distance between the point and the charge.

#### Page No 122:

Given:
Magnitude of the charge, q =  2.0 × 10−8 C
Mass of the bob, m = 80 mg = 80 × 10−6 kg
Electric field, E = 20 kVm−1 = 20 × 103 Vm−1
Let the direction of the electric field be from the left to right. Tension in the string is T.
From the figure,

Tcosθ = mg    ...(1)
Tsinθ = qE    ...(2)

$⇒\mathrm{tan}\theta =\frac{5}{9.8}\phantom{\rule{0ex}{0ex}}⇒\theta =27°$
From equation (1),

#### Page No 122:

Given:
Charge of the particle = q
Velocity of projection = u
Electric field intensity = E
Mass of the particle = m
We know that the force experience by a charged particle in an electric field is qE.
Acceleration produced, a = $\frac{qE}{m}$ (Negative because the particle is thrown against the electric field)
Let the distance covered by the particle be s.
Then v2 = u2 + 2as
[Here, a = deceleration, v = final velocity]
Here, 0 = u2 − 2as

#### Page No 122:

Given:
Charge of the particle, q = 2.5 × 10−4 C
Initial velocity, u = 0
Electric field intensity, E = 1.2 × 104 N/C
Mass of the particle, m = 1 g = 10−3 kg
Distance travelled, s = 40 cm = 4 × 10−1 m

(a) Electric force,

Force of gravity,

(b) Acceleration of the particle,

Let t be the time taken by the particle to cover the distance s = 40 cm. Then,

(c) Using the third equation of motion, we get

(d) Work done by the electric force,

#### Page No 123:

Given:
Charge of the ball, q = 4.9 × 10−5 C
Electrical field intensity, E = 2 × 104 N/C
Mass of the ball, m = 100 gm
Force of gravity, Fg = mg
Electrical force, Fe = Eq
The particle moves due to the resultant force of Fg and Fe.

Fg = Fe
⇒ tanθ = 1
θ = 45°
θ is the angle made by the horizontal with the resultant.
Hence, the path of the ball is straight and is along the resultant force at an angle of 45° with the horizontal
Vertical displacement in t = 2 s,

Both the forces are same.
So, vertical displacement in 2 s = Horizontal displacement in 2 s
Net displacement

#### Page No 123:

Given:
Charge of the particle, q = 4.0 × 10−6 C
Electrical field intensity, E = 2.5 × 104 NC−1
Mass of the particle, m = 40 g = 0.04 kg
When no electrical field is applied,
time period, ${T}_{1}=2\mathrm{\pi }\sqrt{\frac{l}{g}}$
When upward electrical field is applied,
time period, ${T}_{2}=2\mathrm{\pi }\sqrt{\frac{l}{g\mathit{-}a}}$,
where a is the upward acceleration due to electrical field, which is given by

#### Page No 123:

Given:
Charge of the block = q
Electrical field intensity = E
Mass of the block = m
Spring constant = k
Electric force, F = qE
Spring force, F = − kx (where x is the amplitude)
Therefore, qE = − kx
$⇒x=\left|-\frac{qE}{k}\right|=\frac{qE}{k}$

#### Page No 123:

For motion to be simple harmonic,acceleration should be proportional to the displacement and should be directed in a direction opposite to the displacement.
When the block is moving towards the wall, the acceleration is along displacement.
So, the block does not undergo SHM.
Time taken to reach the wall is given by

Since it is an elastic collision, the time taken by the block to move towards the wall is the time taken to move away from it till the velocity is zero.
Total time, T = 2t
$⇒T=2\sqrt{\frac{2dm}{qE}}=\sqrt{\frac{8dm}{qE}}$

#### Page No 123:

Electric field intensity, E = 10 N/C,
Change in height, ds = 50 cm = $\frac{1}{2}$ m
Change in electric potential, dV = E.ds = 10 × $\frac{1}{2}$ = 5 V

#### Page No 123:

Given:
Charge, q = 0.01 C,
Work done, W = 12 J
Now, work done = potential difference × charge
W = (VB − VA) × q
VB − VA$\frac{12}{0.01}$ = 1200 V

#### Page No 123:

Let the third charge be moved from point A to point B.

Magnitude of the charges, q = 2.0 × 10−7 C
Distance of both the charges from point A, r = 10 cm =0.1 m
Distance of both the charges from point B, r' = 20 cm =0.2 m
Potential at A, ${V}_{A}=2×\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{q}{r}$
Potential at B, ${V}_{B}=2×\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{q}{r\text{'}}$
Work done,  $W=q\left({V}_{B}-{V}_{A}\right)$

#### Page No 123:

Given:
Electric field intensity, E = 20 N/C
The electric field is along the x-axis. So, while calculating the potential difference between points B and A using the formula VBVA = E.ds, we will use the difference of the x-coordinates of these point as ds.
(a) A = (0, 0) B = (4 m, 2 m).
So, VBVA = E.ds = 20 × (0 − 4 m) = − 80 V

(b) A = (4 m, 2 m), B = (6 m, 5 m)
VBVA = E.ds = 20 × (4 − 6) = − 40 V.

(c) A = (0, 0), B = (6 m, 5 m)
VBVA = E.ds = 20 × (0 − 6) = − 120 V.

Potential difference between points (0, 0) and (6 m, 5 m) = Potential difference between points (0, 0) and (4 m, 2 m) + Potential difference between points (4 m, 2 m) and (6 m, 5 m)

#### Page No 123:

Given:
Magnitude of charge, q =  −2.0 × 10−4 C

(a) The electric field is along the x direction.
Thus, potential difference between (0, 0) and (4, 2),
dV = −E.dx = −20 × 4 = −80 V
Potential energy (UBUA) between the points = dV × q
UBUA = (−80) × (−2.0 × 10−4)
UBUA = 160 × 10−4 = 0.016 J

(b) A = (4 m, 2m), B = (6 m, 5 m)
dV = −E.dx = − 20 × 2 = −40 V
Potential energy (UBUA) between the points = dV × q
UBUA = (−40) × (−2 × 10−4)
UBUA = 80 × 10−4 = 0.008 J

(c) A = (0, 0) B = (6m, 5m)
dV = −E.dx = −20 × 6 = −120 V
Potential energy (UBUA) between the points A and B = dV × q
UBUA = (−120) × (−2 × 10−4)
UBUA = 240 × 10−4 = 0.024 J

#### Page No 123:

Given:

$\stackrel{\to }{r}=\left(2\stackrel{\to }{i}+2\stackrel{\to }{j}\right)$
So, $V=-\stackrel{\to }{E}.\stackrel{\to }{r}$

#### Page No 123:

Given:
Electric field intensity, $\stackrel{\to }{E}=\stackrel{^}{i}\mathrm{Ax}=10x\stackrel{^}{i}$
Potential,
On integrating, we get

So, at the origin, the potential is 500 V.

#### Page No 123:

Given:
Electric potential, $V\left(x,y,z\right)=A\left(xy+yz+zx\right)$
$A=\frac{\mathrm{volt}}{{\mathrm{m}}^{2}}\phantom{\rule{0ex}{0ex}}⇒\left[A\right]=\frac{\left[{\mathrm{ML}}^{2}{\mathrm{I}}^{-1}{\mathrm{T}}^{-3}\right]}{\left[{\mathrm{L}}^{2}\right]}\phantom{\rule{0ex}{0ex}}⇒A=\left[{\mathrm{MT}}^{-3}{\mathrm{I}}^{-1}\right]$

(b) Let E be the electric field.

Equating now, we get
$\stackrel{\to }{E}=-A\left(y+z\right)\stackrel{^}{i}-A\left(z+x\right)\stackrel{^}{j}-A\left(x+y\right)\stackrel{\mathit{^}}{k}$

(c) Given: A = 10 V/m2

Magnitude of electric field,

#### Page No 123:

Magnitude of charges, q1 = q2 = 2 × 10−5 C
Initial potential energy = 0
Each is brought from infinity to within a separation of 10 cm.
That is, r = 10 × 10−2 m
So, increase in potential energy = final potential energy − initial potential energy.
That is, increase in potential energy, $∆U=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{r}$

#### Page No 123:

(a) The electric field is always perpendicular to the equipotential surface. (As shown in the figure)

So, the angle between = $90°+30°$
Change in potential in the first and second equipotential surfaces, dV = 10 V
So,

The electric field is making an angle of 120$°$ with the x axis.

(b) The electric field is always perpendicular to the equipotential surface.

So, the angle between and  = 0$°$
Potential at point A,
$⇒\frac{q}{4\mathrm{\pi }{\epsilon }_{0}}=60×r\phantom{\rule{0ex}{0ex}}⇒\frac{q}{4\mathrm{\pi }{\epsilon }_{0}}=0.6$
So, electric field, $E=\frac{q}{4\mathrm{\pi }{\epsilon }_{0}}×\frac{1}{{r}^{2}}=\frac{0.6}{{r}^{2}}$
The electric field is radially outward, decreasing with increasing distance.

#### Page No 123:

Given:
Radius of the ring = r
So, circumference = 2πr
Charge density = λ,
Total charge, q = 2πr × λ
Distance of the point from the centre of the ring = x
Distance of the point from the surface of the ring, $r\text{'}=\sqrt{{r}^{2}+{x}^{2}}$
Electricity potential, $V=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{q}{r\text{'}}$
$⇒V=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{2\mathrm{\pi }r\lambda }{\sqrt{{r}^{2}+{x}^{2}}}$
$⇒V=\frac{1}{2{\epsilon }_{0}}\frac{r\lambda }{\left({r}^{2}+{x}^{2}{\right)}^{1/2}}$

Due to symmetry at point P, vertical component of electric field vanishes.
So, net electric field = Ecosθ
$⇒E=\frac{r\lambda }{2{\epsilon }_{0}\left({r}^{2}+{x}^{2}{\right)}^{1/2}}\frac{x}{\left({r}^{2}+{x}^{2}\right)}\phantom{\rule{0ex}{0ex}}⇒E=\frac{r\lambda x}{2{\epsilon }_{0}\left({r}^{2}+{x}^{2}{\right)}^{3/2}}$

#### Page No 123:

Given:
Electric field intensity, E = 1000 N/C
Separation between the plates, l = 2 cm = 0.02 m

(a) The potential difference between the plates,

(b) The acceleration of an electron,

Final velocity, v = 0

(c) Now, u = ucos60$°$
Final velocity, v = 0
Let the maximum height reached be s.

#### Page No 124:

(a) Given:
Electric field intensity, E = 2 N/C in the x-direction
(a) Potential at the origin = 0

(b) From the above expression for V, we have

(c) If potential at the origin is 100 V, then potential at a general point is given by
$V-100=-2x\phantom{\rule{0ex}{0ex}}⇒V=100-2x$

(d) Potential at infinity is given by
$V\text{'}-V=-2x,\phantom{\rule{0ex}{0ex}}$
where V is the potential at the origin.

It is not practical to take the potential at infinity to be zero because in that case, we have to take the potential at origin to be infinity and the calculations will become practically impossible.

#### Page No 124:

Let
Side of the equilateral triangle = l
As
Work done in assembling the charges = Net potential energy of the system

#### Page No 124:

By work-energy theorem,
Change in K.E. = Amount of work done
Change in K.E. = 10 J
Let the charge on the particle be q.
Change in potential, $∆V$ = (200 − 100) V
Work done = $∆V×q$

#### Page No 124:

Given:
Magnitude of charges, q  = 2.0 × 10−4 C
Mass of particles, m = 10 g = 0.01 kg
Separation between the charges, r = 10 cm = 0.1 m
Force of repulsion,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{{q}^{2}}{{r}^{2}}$

Now,

#### Page No 124:

Given:
Magnitude of charges, q =  4.0 × 10−5 C
Initial separation between charges, r = 1 m
Initial speed = 0; so, initial K.E. = 0
Mass of the particles, m = 5.0 g =0.005 kg
Let the required velocity of each particle be v.
By the law of conservation of energy,
Initial P.E. + Initial K.E. = Final P.E. + Final K.E.

#### Page No 124:

Given:
Electric field intensity, E = 2.5 × 104 NC−1
Dipole moment of an HCl molecule, P = 3.4 × 10−30 Cm
Torque on the molecule,
$\Gamma =PE\mathrm{sin}\theta$
For ${\Gamma }_{max}$, putting sinθ = 1, we get

#### Page No 124:

Given:
Magnitude of charge, q = 2.0 × 10−6 C
Separation between the charges, l = 1.0 cm

(a) Dipole moment, P = q × l

(b) Electric field at the axial point of the dipole,
$E=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{2P}{{r}^{3}}\phantom{\rule{0ex}{0ex}}E=\frac{9×{10}^{9}×2×2×{10}^{-8}}{{\left(0.01\right)}^{3}}$
E = 36 × 107 N/C

(c) Electric field at at a point on the perpendicular bisector of the dipole,
$E=\frac{1}{4\mathrm{\pi }{\epsilon }_{0}}\frac{P}{r{\text{'}}^{3}}\phantom{\rule{0ex}{0ex}}E=\frac{9×{10}^{9}×2×{10}^{-8}}{{1}^{3}}$
E = 180 N/C

#### Page No 124:

The system can be considered as a combination of two dipoles making an angle of 60 with each other.

Length of each dipole = d
So, the dipole moment, P = q × d
So, the resultant dipole moment,

The net dipole moment is along the bisector of the angle at +q, away from the triangle.

#### Page No 124:

For figure in (a)
Taking

For figure in (b)

#### Page No 124:

Consider the rod to be a simple pendulum
Time period of a simple pendulum,
$T=2\mathrm{\pi }\sqrt{\frac{l}{a\text{'}}}$
(where l = length and a' = acceleration)
Now, $a\text{'}=\frac{F}{m}=\frac{Eq}{m}$
and length l = a
∴ The time period,
$T=2\mathrm{\pi }\sqrt{\frac{ma}{qE}}$