HC Verma Solutions for Class 12 Physics Chapter 29 - Electric Field and Potential
HC Verma Solutions for Class 12 Physics Chapter 29 - Electric Field and Potential

HC Verma Physics books are the most preferred books among students of CBSE schools. Students can be found referring to the chapters as well as practice questions at the end of each of these chapters, in the books. Students follow these textbooks religiously since quite a few questions in these also appear in exams.

Contents

HC Verma Solutions for Class 12 Physics Chapter 29 – Electric Field and Potential

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Page No 119:

Question 1:

The charge on a proton is +1.6 × 10−19 C and that on an electron is −1.6 × 10−19 C. Does it mean that the electron has 3.2 × 10−19 C less charge than the proton?

Answer:

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.

Question 2:

Is there any lower limit to the electric force between two particles placed at a separation of 1 cm?

Answer:

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.

Question 3:

Consider two particles A and B of equal charges placed at some distance. Particle A is slightly displaced towards B. Does the force on B increase as soon as particle A is displaced? Does the force on particle A increase as soon as it is displaced?

Answer:

Electrostatic force follows the inverse square law,

F=kr2. 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.

Question 4:

Can a gravitational field be added vectorially to an electric field to get a total field?

Answer:

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.

Question 5:

Why does a phonograph record attract dust particles just after it is cleaned?

Answer:

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.

Question 6:

Does the force on a charge due to another charge depend on the charges present nearby?

Answer:

Coulomb’s Law states that the force between two charged particle is given by

F=q1q24π∈0r2,
where
q1 and q2 are the charges on the charged particles
r = separation between the charged particles

∈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.

Question 7:

In some old texts it is mentioned that 4π lines of force originate from each unit positive charge. Comment on the statement in view of the fact that 4π is not an integer.

Answer:

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.

Question 8:

Can two equipotential surfaces cut each other?

Answer:

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.

Question 9:

If a charge is placed at rest in an electric field, will its path be along a line of force? Discuss the situation when the lines of force are straight and when they are curved.

Answer:

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.

Question 10:

Consider the situation shown in the figure. What are the signs of q1 and q2? If the lines are drawn in proportion to the charges, what is the ratio q1/q2?
Figure

Answer:

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

q1q2=618⇒q1q2=136 lines are entering q​​​1 and 18 are coming out of q​​​2.

Question 11:

A point charge is taken from a point A to a point B in an electric field. Does the work done by the electric field depend on the path of the charge?

Answer:

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.

Question 12:

It is said that the separation between the two charges forming an electric dipole should be small. In comparison to what should this separation be small?

Answer:

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.

Question 13:

The number of electrons in an insulator is of the same order as the number of electrons in a conductor. What is then the basic difference between a conductor and an insulator?

Answer:

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.

Question 14:

When a charged comb is brought near a small piece of paper, it attracts the piece. Does the paper become charged when the comb is brought near it?

Answer:

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.

Question 1:

Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that
Figure
(a) EA > EB > EC
(b) EA = EB = EC
(c) EA = EC > EB
(d) EA = EC < EB

Answer:

(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

Question 2:

When the separation between two charges is increased, the electric potential energy of the charges
(a) increases
(b) decreases
(c) remains the same
(d) may increase or decrease

Answer:

(d) may increase or decrease

The electric potential energy, E, between the two charges, q1and q2, separated by the distance, r, is given as

E=kq1q2r,where k=constantAs 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.

Question 3:

If a positive charge is shifted from a low-potential region to a high-potential region, the electric potential energy
(a) increases
(b) decreases
(c) remains the same
(d) may increase of decrease

Answer:

(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:

Question 4:

Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is situated while moving from A to B. The potential
(a) continuously increases
(b) continuously decreases
(c) increases then decreases
(d) decreases then increases

Answer:

(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=q4π∈0x+q4π∈0(r-x)⇒V=qr4π∈0x(r-x)Now, differentiating V with respect to x, we get

dVdx=-qr(r-2x)4π∈0x2(r-x)2Therefore, x = r/2.
It can be observed that

dVdx<0for x < r/2.  Thus, the potential is decreasing first. At
x = r/2, the potential is minimum.
As

dVdx>0for x > r/2, the potential is increasing after x = r/2.

Question 5:

The electric field at the origin is along the positive x-axis. A small circle is drawn with the centre at the origin, cutting the axes at points A, B, C and D with coordinates (a, 0), (0, a), (−a, 0), (0, −a), respectively. Out of the points on the periphery of the circle, the potential is minimum at
(a) A
(b) B
(c) C
(d) D

Answer:

(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).

Question 6:

If a body is charged by rubbing it, its weight
(a) remains precisely constant
(b) increases slightly
(c) decreases slightly
(d) may increase slightly or may decrease slightly

Answer:

(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.

Question 7:

An electric dipole is placed in a uniform electric field. The net electric force on the dipole
(a) is always zero
(b) depends on the orientation of the dipole
(c) can never be zero
(d) depends on the strength of the dipole

Answer:

(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.

Question 8:

Consider the situation in the figure. The work done in taking a point charge from P to A is WA, from P to B is WB and from P to C is WC.
(a) WA < WB < WC
(b) WA > WB > WC
(c) WA = WB = WC
(d) None of these
Figure

Answer:

(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 P to C, WC = WA + WAB + WBC = WA​

Hence, WA = WB = WC

Question 9:

A point charge q is rotated along a circle in an electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is
(a) zero
(b) positive
(c) negative
(d) zero if the charge Q is at the centre, otherwise non-zero

Answer:

(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.

Question 1:

Mark out the correct options.
(a) The total charge of the universe is constant.
(b) The total positive charge of the universe is constant.
(c) The total negative charge of the universe is constant.
(d) The total number of charged particles in the universe is constant.

Answer:

(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.

Question 2:

A point charge is brought inside an electric field. The electric field at a nearby point
(a) will increase if the charge is positive
(b) will decrease if the charge is negative
(c) may increase if the charge is positive
(d) may decrease if the charge is negative

Answer:

(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.

Question 3:

The electric field and the electric potential at a point are E and V, respectively.
(a) If E = 0, V must be zero.
(b) If V = 0, E must be zero.
(c) If E ≠ 0, V cannot be zero.
(d) If V ≠0, E cannot be zero.

Answer:

None of the above.

Electric field,

E=-dVdr, 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.

Question 4:

Electric potential decreases uniformly from 120 V to 80 V, as one moves on the x-axis from x = −1 cm to x = +1 cm. The electric field at the origin
(a) must be equal to 20 Vcm−1
(b) may be equal to 20 Vcm−1
(c) may be greater than 20 Vcm−1
(d) may be less than 20 Vcm−1

Answer:

(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=-dVdr

⇒E=-40 V-2⇒E=20 Vcm-1This 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.

Question 5:

Which of the following quantities does not depend on the choice of zero potential or zero potential energy?
(a) Potential at a point
(b) Potential difference between two points
(c) Potential energy of a two-charge system
(d) Change in potential energy of a two-charge system

Answer:

(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).

Question 6:

An electric dipole is placed in an electric field generated by a point charge.
(a) The net electric force on the dipole must be zero.
(b) The net electric force on the dipole may be zero.
(c) The torque on the dipole due to the field must be zero.
(d) The torque on the dipole due to the field may be zero.

Answer:

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

τ = r×F=rFsinθ,where

θis the angle between the force F and the arm of the couple.
So, for

θ=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.

Question 7:

A proton and an electron are placed in a uniform electric field.
(a) The electric forces acting on them will be equal.
(b) The magnitudes of the forces will be equal.
(c) Their accelerations will be equal.
(d) The magnitudes of their accelerations will be equal.

Answer:

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

We know:

F→=qE→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,

F→=qE→=ma→⇒a→=qE→mAs 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.

Question 8:

The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,
(a) it is uniform in the region
(b) it is proportional to r
(c) it is proportional to r2
(d) it increases as one moves away from the origin

Answer:

(c) it is proportional to r2

Given:
 E

∝r  and V = 0 at r =0

⇒E = kr
Also, E =

-dVdr

∴ V=-∫Edr⇒V=-∫krdr⇒V=-kr22+CUsing the condition, V = 0 at r = 0, we get  C =0.
Therefore,

V=-kr22⇒V ∝ r2

Page No 121:

Question 1:

Find the dimensional formula of ε0.

Answer:

By Coulomb’s Law,

F=14πε0q1q2r2

⇒ε0=14πFq1q2r2
Using [F] = [MLT−2]
[r] = [M0L1T0]
[q] = [M0L0T1A1], we get
[ε0] = [M−1L−3T4A2]

Question 2:

A charge of 1.0 C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times your weight is this force?

Answer:

Given:
q1 = q2 = q = 1.0 C
Distance between the charges, r = 2 km = 2 × 103 m
By Coulomb’s Law, electrostatic force,

F=14πε0q1q2r2

F=9×109×1×12×1032  =2.25×103 NLet my mass, m, be 50 kg.
Weight of my body, W = mg
⇒ W = 50 × 10 N = 500 N
Now,

Weight of my bodyForce between the charges=5002.25×103=14.5So, the force between the charges is 4.5 times the weight of my body.

Question 3:

At what separation should two equal charges, 1.0 C each, be placed, so that the force between them equals the weight of a 50 kg person?

Answer:

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=14πε0q1q2r2⇒490=9×109×1×1r2⇒r2=9×109490⇒r=949×108=37×104 m = 4.3×103 m

Question 4:

Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges, so that the force between them equals the weight of a 50 kg person?

Answer:

Let the magnitude of each charge be q.
Separation between them, r  = 1 m
Force between them, F = 50 × 9.8 = 490 N
By Coulomb’s Law force,

F=14πε0q1q2r2

⇒490=9×109×q212⇒q2=54.4×10-9⇒q=54.4×10-9= 23.323×10-5 COr q= 2.3×10-4 C

Question 5:

Find the electric force between two protons separated by a distance of 1 fermi (1 fermi = 10−15 m). The protons in a nucleus remain at a separation of this order.

Answer:

We know:
Charge on a proton, q = 1.6 × 10−19 C
Given, separation between the charges, r = 10−15 m
By Coulomb’s Law, electrostatic force,

F=14πε0q1q2r2

⇒F=9×109×1.6×10-19210-152⇒F=230 N

Question 6:

Two charges 2.0 × 10−6 C and 1.0 × 10−6 C are placed at a separation of 10 cm. Where should a third charge be placed, such that it experiences no net force due to these charges?

Answer:

Given:

q1=2.0×10-6 Cq2=1.0×10-6 CLet 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=14πε0Q1Q2r2Force on charge q due to q1,

F=9×109×2.0×10-6×qx2Force on charge q due to q2,

F’=9×109×10-6×q10-x2According to the question,

F-F’=0

⇒F=F’⇒9×109×2×10-6×qx2=9×109×10-6×q10-x2⇒x2=210-x2⇒x2-40x+200=0⇒x=20±102⇒x=5.9 cm  (∵ x≠20+102)So, the third charge should be placed at a distance of 5.9 cm from q1.

Question 7:

Suppose the second charge in the previous problem is −1.0 × 10−6 C. Locate the position where a third charge will not experience a net force.

Answer:

Given:

q1=2×10-6 Cq2=-1×10-6 CSince 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=14πε0Q1Q2r2So, force on charge q due to q1,

F=9×109×2.0×10-6×qx2Force on charge q due to q2,

F’=9×109×10-6×q10-x2According to the question,F-F’=0

⇒F=F’⇒9×109×2×10-6×qx2=9×109×10-6×qx-102⇒x2=2x-102⇒x2-40x+200=0⇒x=20±102 m⇒x=34.14 cm  (∵x≠20-102)

Question 8:

Two charged particles are placed 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?

Answer:

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=14πε0q1q2r2

⇒F=9×109×1.6×10-19210-22⇒F=2.3×10-24 N

Question 9:

Estimate the number of electrons in 100 g of water. How much is the total negative charge on these electrons?

Answer:

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 =

6.023×102418×100                                                       = 3.34 × 1025
∴ Total charge = 3.34 × 1025 × 1.6 × 10−19
= 5.34 × 106 C

Question 10:

Suppose all the electrons of 100 g water are lumped together to form a negatively-charged particle and all the nuclei are lumped together to form a positively-charged particle. If these two particles are placed 10.0 cm away from each other, find the force of attraction between them. Compare it with your weight.

Answer:

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 =

6.023×102418×100                                                           = 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=14πε0q1q2r2

=9×109×5.34×106×5.34×10610-2=2.56×1025 NThis 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).

Question 11:

Consider a gold nucleus to be a sphere of radius 6.9 fermi in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at largest separation. Why do these protons not fly apart under this repulsion?

Answer:

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 =

1.6×10-19 CBy Coulomb’s Law, force of repulsion,

F=14πε0q1q2r2

⇒F=9×109×1.6×10-1922R2=1.2 NInside 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.

Question 12:

Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other during rubbing?

Answer:

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=14πε0q1q2r2Let the no. of electrons transferred from one sphere to the other be n. Then,
F=14πε0nq2r2

⇒0.1=9×10-9×n2×1.6×10-19210-4⇒n2=0.1×10-49×109×1.6×1.6×10-38⇒n=2×1011

Question 13:

An NaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75 × 10−8 cm, find the force of attraction between them. State the assumptions (if any) that you have made.

Answer:

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 =

-1.6×10-19C
Net charge on the sodium ion, q2 =

1.6×10-19C
By Coulomb’s Law, force,

F=14πε0q1q2r2

=9×109×1.6×10-1922.75×10-102=3.05×10-9 N

Question 14:

Find the ratio of the electrical and gravitational forces between two protons.

Answer:

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,

G=6.67×10-11 N-m2/kg2Electrostatic force,

Fe=14πε0q2r2Gravitational force,

Fg=Gm2r2⇒FeFg=q24πε0Gm2            =9×109×1.6×10-1926.67×10-11×1.67×10-272            =1.23×1036

Question 15:

Suppose an attractive nuclear force acts between two protons which may be written as F=Ce−kr/r2. (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi−1 and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.

Answer:

(a) Given, nuclear force of attraction,

F=Ce-Krr2Here e−Kr is just a pure number, i.e. a dimensionless quantity. So,

C=F×r2C=MLT-2×L2

C=ML3T-2

And K=1r=L-1(b)By Coulomb’s Law, electric force,

F=14πε0e2r2Taking

r=5×10-15 m, we get

F=9×109×1.6×10-1925×10-152Andnuclear force, F = Ce−kr/r2
Taking r = 5 × 10-15 m and k = 1 fermi−1, we get

F=C×10-55×10-152Comparing both the forces, we get

C=3.4×10-26 N-m2

Question 16:

Three equal charges, 2.0 × 10−6 C each, are held at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the other two.

Answer:

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,

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

F’=FBAsinθ+FCAsinθF’=2FsinθGiven: r = 5 cm =0.05 m
By Coulomb’s Law, force,

F=14πε0q1q2r2

F’=2×9×109×2×10-62×sin60°0.052F’=2×9×109×2×10-62×320.052F‘ = 24.9 N

Question 17:

Four equal charges of 2.0 × 10−6 C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb’s force experienced by one of the charges due to the other three.

Answer:

Given,
Magnitude of the charges,

q=2×10-6 CSide of the square,

a=5 cm=0.05 mBy Coulomb’s Law, force,

F=14πε0q1q2r2
So, force on the charge at A due to the charge at B,

F→BA=9×109×2×10-620.052         =9×109×4×10-1225×10-4      =14.4 NForce on the charge at A due to the charge at C,

F→CA=9×109×2×10-622×0.052      =9×109×4×10-1225×2×10-4      =7.2 NForce on the charge at A due to the charge at D,

F→DA=F→BAThe resultant force at A, F‘=

F→BA+F→CA+F→DAThe resultant force of

F→DA and F→BAwill be

2FBAin the direction of

F→CA. Hence, the resultant force,

F’=14.42+7.2   =27.56 N

Question 18:

A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom (1 angstrom = 10−10 m and is abbreviated as Å) with the proton at the centre. The hydrogen atom is said to be in the ground state in this case. Find the magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state.

Answer:

Given:
Separation between the two charges, r = 0.53 Å = 0.53 × 10−10 m

q1=q2=eBy Coulomb’s Law, force,

F=14πε0q1q2r2

F=9×109×1.6×10-1920.53×10-102  =8.2×10-8 N

Question 19:

Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.

Answer:

Given:
Separation between the two charges, r = 0.53 Å = 0.53 × 10−10 m
By Coulomb’s Law, force,

F=14πε0q1q2r2Here,

q1=q2=e

⇒F=9×109×1.6×10-1920.53×10-102  =8.2×10-8 NNow, mass of an electron, Me = 9.12 × 10−31 kg
The necessary centripetal force is provided by the Coulombian force.

⇒ Fe=Mev2r⇒v2=0.4775×1013          =4.775×1012 ⇒v=2.18×106 m/s

Question 20:

Ten positively-charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm, 30 cm, …., 100 cm. the first particle has a charge 1.0 × 10−8 C, the second 8 × 10−8 C, the third 27 × 10−8 C and so on. The tenth particle has a charge 1000 × 10−8 C.  Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Answer:

By Coulomb’s Law, force (F) on charge q due to one charge,

F=14πε0q1qr2So, net force due to ten charges,

F=F1+F2+F3+…..+F10   =9×109×1.00.102+8.202+27.302+……10001.0210-8   =9×109×10-810-21+2+3+……10    =9×103×55    =4.95×105 N

Question 21:

Two charged particles with charge 2.0 × 10−8 C each are joined by an insulating string of length 1 m and the system is kept on a smooth horizontal table. Find the tension in the string.

Answer:

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=FBy Coulomb’s Law,        

F=14πε0q2r2

⇒T=9×109×2.0×10-8212         =3.6×10-6 N

Question 22:

Two identical balls, each with a charge of 2.00 × 10−7 C and a mass of 100 g, are suspended from a common point by two insulating strings, each 50 cm long. The balls are held 5.0 cm apart and then released. Find (a) the electric force on one of the charged balls (b) the components of the resultant force on it along and perpendicular to the string (c) the tension in the string (d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.

Answer:

Given:
Magnitude of the charges, q =  2.0 × 10−7 C
Separation between the charges,

r=5×10-2 mLength of the string,

l=50 cm=50×10-2 mMass of the balls, m = 100 g = 0.1 kg


(a) By Coulomb’s Law, the electric force,

F=14πε0q2r2

=14.4 N×10-2 N=0.144 N(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

T2=F2+mg2    =0.1442+0.1×9.82⇒T=0.986 N

Question 23:

Two identical pith balls are charged by rubbing one against the other. They are suspended from a horizontal rod through two strings of length 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0 × 10−8 C.

Answer:


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,

sinθ=120cosθ=1-1202From equation (1) and (2),

tanθ=Fmg∴ m=Fgtanθ⇒m=14πε0q2r2×1gtanθ⇒m=9×109×2×10-829×10-4×0.98×tanθ ⇒m=0.0082 kg=8.2 g

T=Fsinθ=9×109×2×10-829×10-4×120⇒T=8.2×102 N

Question 24:

Two small spheres, each with a mass of 20 g, are suspended from a common point by two insulating strings of length 40 cm each. The spheres are identically charged and the separation between the balls at equilibrium is found to be 4 cm. Find the charge on each sphere.

Answer:


Tcosθ = mg   …(i)
Tsinθ = Fe   …(ii)
Here,

Fe=14πε0q2r2

tanθ=239.9Dividing equation (ii) by (i), we get

tanθ=Femg⇒tanθ=14πε0q2r2.1mg⇒q2=tanθ.r2.mg1/4πε0⇒q2=17×10-16⇒q=4.123×10-8 C

Page No 122:

Question 25:

Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length l. Find the mass of each ball if the angle between the strings is 2θ in equilibrium.

Answer:


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)

⇒tanθ=Fmg⇒m=Fgtanθ    …3Here, separation between the two charges,

r=2lsinθ

F=14πε0q2r2=14πε0×q22lsinθ2Substituting the value of F in equation (3), we get

m=q2cotθ16πε0gl2sin2θ

Question 26:

A particle with a charge of 2.0 × 10−4 C is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. The mass of the bob is 100 g. What charge should the bob be given so that the string becomes loose?

Answer:

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’ .

⇒Fe=14πε0qq’r2=T

⇒9×109×q’×2×10-410-2=0.98⇒q’=5.4×10-9 C

Question 27:

Two particles A and B with charges q and 2q, respectively, are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table under electrical forces. What should be the charge on C and where should it be clamped?

Answer:

For equilibrium,

F→AC+F→CB=0
Let the charge at point c be θ.

qθ4π ∈0x2+2qθ4π ∈0d-x2=0

Again, F→AC=F→CBSo, 1×2=2d-x2

Or 2×2=d-x2Or 2x=d-xOr x=(2-1)dFor a charge at rest,

F→AC=F→CB14π ∈0qθ[(2-1)d]2+14π ∈0q×2qd2=0Or θ=(6-42)q

Question 28:

Two identically-charged particles are fastened to the two ends of a spring of spring constant 100 N m−1 and natural length 10 cm. The system rests on a smooth horizontal table. If the charge on each particle is 2.0 × 10−8 C, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.

Answer:

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=14πε0q2r2The spring force due to extension x,

F=KxFor equilibrium,
Electrostatic force = Spring force

14π ε0×q2x+0.12=K0.1+x⇒0.1+x3=9×109×2.0×10-8210-2                     =36×109×10-1610-2                     = 36×10-5⇒x=3.6×10-6 mYes, 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.

Question 29:

A particle A with a charge of 2.0 × 10−6 C is held fixed on a horizontal table. A second charged particle of mass 80 g stays in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is μ = 0.2. Find the range within which the charge of this second particle may lie.

Answer:

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=14πε0q1q2r2Force of friction on B =

μmgSince B is at equilibrium,
Coulomb force = Frictional force

⇒F=μmg⇒ 9×109×2×10-5×q20.12=0.2×0.08×9.8⇒ q2=8.71×10-8 C∴ The range of the charge = ∓8.71×10-8 C

Question 30:

A particle A with a charge of 2.0 × 10−6 C and a mass of 100 g is placed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, with the same charge and mass, be placed on the incline so that it may remain in equilibrium?

Answer:

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=14πε0q2r2For equilibrium:

For equilibrium along the plane,

F=mgsinθ⇒ 9×109×2×10-62r2=0.1×9.8×12⇒r2=7.34×10-2 m⇒ r=0.271  mSo, the charge should be placed at a distance of 27 cm from the bottom.

Question 31:

Two particles A and B, each with a charge Q, are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB, so that it experiences maximum force? What is the magnitude of this maximum force?

Answer:

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.

sinθ=xd22+x2Total vertical component of force,

F’=2Fsinθ

F’=2×14πε0×qQd22+x2×xd22+x2⇒F’=12πε0×qQxd22+x23/2For maximum force,

dF’dx=0qQ2πε0×d22+x2-3/2-x32d22+x2-5/22x=0⇒ d22+x2-3×2=0⇒2×2=d22⇒ x=d22∴ F’max=12πε0qQd22d22+d2223/2                   =3.08Qq4πε0d2

Question 32:

Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C of mass m and charge q is kept at the middle point of the line AB. (a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it? (b) Assuming x<<d, show that this force is proportional to x. (c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement? Find the time period of the oscillations if these conditions are satisfied.

Answer:

(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.

sinθ=xd22+x2Total vertical component of the force,

F’=2Fsinθ

F’=2×14πε0×qQd22+x2×xd22+x2⇒F’=12πε0×qQxd22+x23/2This is the net electric force experienced by the charge q.

(b) When x < < d:

F’=12πε0qQxd23   ∵x2<<d22⇒F’=4πε0qQxd3(c) For the particle to execute simple harmonic motion:
F’ = mw2x

⇒4πε0qQxd3=m2πT2x⇒T2=mπ3ε0d3Qq⇒ T =mπ3ε0d3Qq1/2

Question 33:

Repeat the previous problem if the particle C is displaced through a distance x along the line AB.

Answer:


Net force

=14π∈0Qqd2-x2-Qqd2+x2=Qq4π∈0d22+x2+xd-d22-x2+xdd22-x22when,
x << d
So, net force =

qQ4π∈02xdd4=qQ4π∈02xd3Or m2πT2x=2xqQ4π∈0d3T=π3∈0md32Qq1/2

Question 34:

The electric force experienced by a charge of 1.0 × 10−6 C is 1.5 × 10−3 N. Find the magnitude of the electric field at the position of the charge.

Answer:

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

⇒E=Fq=1.5×10-31.0×10-6          =1.5×103 N/C

Question 35:

Two particles A and B possessing charges of +2.00 × 10−6 C and of −4.00 × 10−6 C, respectively, are held fixed at a separation of 20.0 cm. Locate the points (s) on the line AB, where (a) the electric field is zero (b) the electric potential is zero.

Answer:

Given:
Magnitude of the charges,

q1=2.0×10-6 Cand

q2=-4.0×10-6 CSeparation 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,

14πε0q1x2+14πε0q20.2-x2=0⇒2×2+-40.2-x2=0or x = 48.3 cm from A along BA.

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

14πε0q1x+14πε0q20.2±x=0⇒ 2x-40.2±x=0Or 1x=20.2±x⇒0.2±x=2x⇒ x =

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

Question 36:

A point charge produces an  electric field of magnitude 5.0 NC−1 at a distance of 40 cm from it. What is the magnitude of the charge?

Answer:

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.

E=14πε0qr2⇒5.0=9×109×q0.42⇒q=8.9×10-11 C

Question 37:

A water particle of mass 10.0 mg and with a charge of 1.50 × 10−6 C stays suspended in a room. What is the magnitude of electric field in the room? What is its direction?

Answer:


Mass of the particle,

m=10 mg=10×10-5 kgCharge on the particle,

q=1.5×10-6 CLet the magnitude of the electric field be E.
The particle stays suspended. Therefore,
Downward gravitational force = Upward electric force
That is, mg  = qE

⇒ E=mgq=10×10-5×101.5×10-6        = 100015=66.7 N/CThe direction of the electric field will be upward to balance the downward gravitational force.

Question 38:

Three identical charges, each with a value of 1.0 × 10−8 C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the centre of the triangle.

Answer:

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

l=20 cm = 0.2 m
Let

EA,EB and ECbe 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,

EA=EB=EC=E sayResolving

EB and ECinto vertical and horizontal components (with

θ=30°)
The horizontal components cancel each other, as shown in the figure.
So, the net electric field at the centre,

Enet=EA-EBsinθ+EAsinθ      =E1-sin30°-sin30°⇒Enet=0Now,

h2=l2-l22h2=0.22-0.12⇒h=310Let the distance of the centre from each charge be r.
For an equilateral triangle,

r=23h⇒r=23×1.73210=1.15×10-1 mPotential at the centre,

V=VA+VB+VC∵ VA=VB=VC, V=3VA

V=3×14πε0qrV=3×9×109×10-80.115

V=23×102=2.3×103 V

Question 39:

A positive charge Q is distributed uniformly over a circular ring of radius R. A particle of mass m, and a negative charge q, is placed on its axis at a distance x from the centre. Find the force on the particle. Assuming x << R, find the time period of oscillation of the particle if it is released from there.

Answer:

Consider an element of angular width

dθat a distance r from the charge q on the circular ring, as shown in the figure.

So, charge on the element,

dq=Q2πRRdθ=Q2πdθElectric field due to this charged element,

dE=dq4πε01r2     =Q2πdθ4πε01R2+x2By the symmetry, the Esinθ component of all such elements on the ring will vanish.
So, net electric field,

dEnet=dEcosθ=Qdθ8π2ε0R2+x23/2Total force on the charged particle,

F=∫qdEnet   =qQ8π2ε0xR2+x23/2∫02πdθ    =xQq4πε0R2+x23/2According to the question,

x<<RF=Qq x4πε0R3Comparing this with the condition of simple harmonic motion, we get

F=mω2x⇒mω2=Qq4πε0R3⇒m2πT2=Qq4πε0R3⇒T=16π3ε0mR3Qq1/2

Question 40:

A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.

Answer:

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

dq=QLRdθThe net electric field has a vertical component only.

So, Enet=∫dE sinθ            =14πε0QL∫0πRdθR2sinθ

=-14πε0QLRcosθ0π=2Q4πε0LR=Q2ε0L2  ∵πR=L

Question 41:

A 10-cm long rod carries a charge of +50 μC distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both ends of the rod.

Answer:


Net electric field = ∫dE cos θ

∫14π∈0ql.dx75-x2.x75-x2=∫-5+514π∈0.qlxdx75-x2.75-x2=5.2×107N/C

Question 42:

Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.

Answer:

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,

E=Qx4πε0R2+x23/2    …(1)For maximum value of electric field,

dEdx=0From equation (1),

dEdx=Q4πε01(R2+x2)-3/2-x32(R2+x2)-5/22x=0

⇒R2+x2-3×2=0⇒3×2=R2⇒x=R2

Question 43:

A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the centre? You may answer this part without making any numerical calculations.

Answer:

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.

Question 44:

A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre due to the remaining wire.

Answer:

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=Q2πadLElectric field due to a small element at the centre,

E=14πε0dqa2⇒E=14πε0a2Q2πadL=QdL8π2ε0a3So, the electric field at the centre due to the remaining wire =

QdL8π2ε0a3(Opposite the direction of the electric field due to the small element)

Question 45:

A positive charge q is placed in front of a conducting solid cube at a distance d from its centre. Find the electric field at the centre of the cube to the charges appearing on its surface.

Answer:

Electric field at a point due to a given
charge, E =

q4π∈0d2,
where q is the charge and R the distance between the point and the charge.

Question 46:

A pendulum bob of mass 80 mg and carrying a charge of 2 × 10−8 C is at rest in a uniform, horizontal electric field of 20 kVm−1. Find the tension in the thread.

Answer:

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)

⇒tanθ=qEmg            =20×103×2×10-880×10-6×9.8

⇒tanθ=59.8⇒θ=27°From equation (1),

T=mgcosθ=80×10-6×9.8cos27°⇒T=8.8×10-4 N

Question 47:

A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?

Answer:

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 =

qEm(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

⇒ s=u22a⇒s=mu22qE

Question 48:

A particle of mass 1 g and charge 2.5 × 10−4 C is released from rest in an electric field of 1.2 × 10 4 N C−1. (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis? (b) How long will it take for the particle to travel a distance of 40 cm? (c) What will be the speed of the particle after travelling this distance? (d) How much is the work done by the electric force on the particle during this period?

Answer:

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,

Fe = qE⇒Fe=2.5 ×10-4×1.2×104=3 NForce of gravity,

Fg=mg⇒Fg=9.8×10-3 N(b) Acceleration of the particle,

a=Fem=310-3=3×103 m/s2Let t be the time taken by the particle to cover the distance s = 40 cm. Then,

s=12at2⇒ t=2sa=1.63×10-2 s(c) Using the third equation of motion, we get

v2=u2+2as⇒v2=0+2×3×103×4×10-1⇒v=4.9×10=49 m/s(d) Work done by the electric force,

W=Fes=3×4×10-1    =12 ×10-1 J =1.20 J

Page No 123:

Question 49:

A ball of mass 100 g and with a charge of 4.9 × 10−5 C is released from rest in a region where a horizontal electric field of 2.0 × 104 N C−1 exists. (a) Find the resultant force acting on the ball. (b) What will be the path of the ball? (c) Where will the ball be at the end of 2 s?

Answer:

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.

R2=Fg2+Fe2   =(0.1×9.8)2+(4.9×10-5×2×104)2   =0.9604+96.04×10-2   =1.9208 N⇒R=1.3859 N 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,

y=12gt2⇒y=12×9.8×2×2=19.6 mBoth the forces are same.
So, vertical displacement in 2 s = Horizontal displacement in 2 s
Net displacement

=19.62+19.62=768.932=27.7 m

Question 50:

The bob of a simple pendulum has a mass of 40 g and a positive charge of 4.0 × 10−6 C. It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude 2.5 × 104 NC−1 is switched on. How much time will it now take to complete 20 oscillations?

Answer:

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,

T1=2πlgWhen upward electrical field is applied,
time period,

T2=2πlg-a,
where a is the upward acceleration due to electrical field, which is given by

a=qEm⇒a=4×10-6×2.5×10440×10-3       =2.5 m/s2

∴T2T1=gg-a⇒T2T1=9.89.8-2.5⇒T2=45×9.87.3=52 s

Question 51:

A block of mass m with a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k, as shown in the figure. A horizontal electric field E, parallel to the spring, is switched on. Find the amplitude of the resulting SHM of the block.
Figure

Answer:

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=-qEk=qEk

Question 52:

A block of mass containing a net positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in the figure. The distance of the block from the wall is d. A horizontal electric field E towards the right is switched on. Assuming elastic collisions (if any), find the time period of the resulting oscillatory motion. Is it a simple harmonic motion?
Figur

Answer:

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

d=ut+12at2⇒t=2mdqE   (Using u=0, and a=qEm)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=22dmqE=8dmqE

Question 53:

A uniform electric field of 10 N C−1 exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm.

Answer:

Electric field intensity, E = 10 N/C,
Change in height, ds = 50 cm =

12m
Change in electric potential, dV = E.ds = 10 ×

12= 5 V

Question 54:

12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. How much is the potential difference  VB − VA?

Answer:

Given:
Charge, q = 0.01 C,
Work done, W = 12 J
Now, work done = potential difference × charge
W = (VB − VA) × q
VB − VA =

120.01= 1200 V

Question 55:

Two equal charges, 2.0 × 10−7 C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process?

Answer:

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,

VA=2×14πε0qrPotential at B,

VB=2×14πε0qr’Work done,

W=qVB-VA

⇒W=2×10-72×2×10-74πε010.1-10.2⇒W=3.6×10-3 J

Question 56:

An electric field of 20 NC−1 exists along the x-axis in space. Calculate the potential difference VB − VA where the points A and B are
(a) A = (0, 0); B = (4 m, 2m)
(b) A = (4 m, 2 m); B = (6 m, 5 m)
(c) A = (0, 0); B = (6 m, 5 m)
Do you find any relation between the answers of parts (a), (b) and (c)?

Answer:

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 VB − VA = 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, VB − VA = E.ds = 20 × (0 − 4 m) = − 80 V

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

(c) A = (0, 0), B = (6 m, 5 m)
⇒ VB − VA = 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)

Question 57:

Consider the situation of the previous problem. A charge of −2.0 × 10−4 C is moved from point A to point B. Find the change in electrical potential energy UB − UA for the cases (a), (b) and (c).

Answer:

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 (UB − UA) between the points = dV × q
⇒ UB − UA = (−80) × (−2.0 × 10−4)
⇒ UB − UA = 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 (UB − UA) between the points = dV × q
⇒ UB − UA = (−40) × (−2 × 10−4)
⇒ UB − UA = 80 × 10−4 = 0.008 J

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

Question 58:

An electric field

E→=(i→20+j→30) NC-1exists in space. If the potential at the origin is taken to be zero, find the potential at (2 m, 2 m).

Answer:

Given:

E→=(i→20+j→30) N/C

r→=(2i→+2j→)So,

V=-E→.r→

⇒V=-(i→20+30j→). (2i→+2j→)⇒V=-(2×20+2×30)=-100 V

Question 59:

An electric field

E→=i→Ax exists in space, where A = 10 V m−2. Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.

Answer:

Given:
Electric field intensity,

E→=i^Ax=10xi^
Potential,

dV=-E→. dx→=-10xdxOn integrating, we get

V=10×x22=-5x2100V=5×100=500 VSo, at the origin, the potential is 500 V.

Question 60:

The electric potential existing in space is

V(x, y, z)=A(xy+yz+zx).(a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 SI units, find the magnitude of the electric field at (1 m, 1 m, 1 m).

Answer:

Given:
Electric potential,

V(x,y,z)=A(xy+yz+zx)

A=voltm2⇒A=ML2I-1T-3L2⇒A=[MT-3I-1](b) Let E be the electric field.

dV=-E→.dr→⇒A(y+z)dx+A(z+x)dy+A(x+y)dz=-E(dxi^+dyj^+dzk^)⇒[A(y+z)i^+A(z+x)j^+A(x+y)k^] [dxi^+dyj^+dzk^]=-Edxi^+dyj^+dzk^Equating now, we get

E→=-A(y+z)i^-A(z+x)j^-A(x+y)k^(c) Given: A = 10 V/m2

r=(1 m, 1 m, 1 m)E→=-10 (2)i^-10 (2)j^-10 (2)k^   =-20i^-20j^-20k^Magnitude of electric field,

E=202+202+202=1200=34.64=35 N/C

Question 61:

Two charged particles, with equal charges of 2.0 × 10−5 C, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

Answer:

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=14πε0q1q2r

∆U=9×109×2×10-5×2×10-510×10-2=36 J

Question 62:

Some equipotential surface is shown in the figure. What can you say about the magnitude and the direction of the electric field?
Figure

Answer:

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

So, the angle between

E → and dx→=

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

E →.dx→=-dV

⇒Edxcos(90°+30°)=-dV⇒E(10×10-2)cos120°=-10⇒E=200 V/mThe 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

E →and

dr→= 0

°Potential at point A,

VA=14πε0qr=60 V

⇒q4πε0=60×r⇒q4πε0=0.6So, electric field,

E=q4πε0×1r2=0.6r2The electric field is radially outward, decreasing with increasing distance.

Question 63:

Consider a circular ring of radius r, uniformly charged with linear charge density λ. Find the electric potential at a point on the axis at a distance x from the centre of the ring. Using this expression for the potential, find the electric field at this point.

Answer:

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’=r2+x2
Electricity potential,

V=14πε0qr’

⇒V=14πε02πrλr2+x2

⇒V=12ε0rλ(r2+x2)1/2
Due to symmetry at point P, vertical component of electric field vanishes.
So, net electric field = Ecosθ

⇒E=rλ2ε0(r2+x2)1/2x(r2+x2)⇒E=rλx2ε0(r2+x2)3/2

Question 64:

An electric field of magnitude 1000 NC−1 is produced between two parallel plates with a separation of 2.0 cm, as shown in the figure. (a) What is the potential difference between the plates? (b) With what minimum speed should an electron be projected from the lower place in the direction of the field, so that it may reach the upper plate? (c) Suppose the electron is projected from the lower place with the speed calculated in part (b). The direction of projection makes an angle of 60° with the field. Find the maximum height reached by the electron.
Figure

Answer:

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,

V=E.l=1000×0.02=20 V(b) The acceleration of an electron,

a=eEm⇒a=1.6×10-19×10009.1×10-31=1.75×1014 m/s2Final velocity, v = 0

v2=u2-2al⇒0=u2-2×1.75×1014×0.02⇒ u2=0.04×1.75×1014⇒u=2.64×106 m/s(c) Now, u = ucos60

°Final velocity, v = 0
Let the maximum height reached be s.

∴v2=u2-2 as⇒s=0.497 ×10-2≈0.005 m=0.50 cm

Page No 124:

Question 65:

A uniform field of 2.0 NC−1 exists in space in the x-direction. (a) Taking the potential at the origin to be zero, write an expression for the potential at a general point (x, y, z). (b) At which point, the potential is 25 V? (c) If the potential at the origin is taken to be 100 V, what will be the expression for the potential at a general point? (d) What will be the potential at the origin if the potential at infinity is taken to be zero? Is it practical to choose the potential at infinity to be zero?

Answer:

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

V=-E.r    =-Exi^+Eyj^+Ezk^.xi^+yj^+zk^⇒V=-Ex.x =-2x(b) From the above expression for V, we have

(25-0)=-2x⇒ x=25-2=-12.5 m(c) If potential at the origin is 100 V, then potential at a general point is given by

V-100=-2x⇒V=100-2x(d) Potential at infinity is given by

V’-V=-2x,where V is the potential at the origin.

∵V’=0 , x=∞,V=V’+2x=∞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.

Question 66:

How much work has to be done in assembling three charged particles at the vertices of an equilateral triangle, as shown in the figure?
Figure

Answer:

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

⇒W=U12+U13+U23⇒W=14πε0q1q2l+q1q3l+q2q3l        =9×10310-1×10-10[4×2+4×3+3×2]        =9×(8+12+6)        =9×26=234 J

Question 67:

The kinetic energy of a charged particle decreases by 10 J as it moves from a point at potential 100 V to a point at potential 200 V. Find the charge on the particle.

Answer:

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

⇒10=(200-100)×q⇒100×q=10⇒q=10100=0.1 C

Question 68:

Two identical particles, each with a charge of 2.0 × 10−4 C and mass of 10 g, are kept at a separation of 10 cm and then released. What would be the speed of the particles when the separation becomes large?

Answer:

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=14πε0q2r2

F=1.8×104 NF=m×a⇒a=1.8×106 m/s2Now,

v2=u2+2asv2=0+2×1.8×106×10×10-2v=36×104v=600 m/s

Question 69:

Two particles of masses 5.0 g each and opposite charges of +4.0 × 10−5 C and −4.0 × 10−5 C are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.

Answer:

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.

14πε0q1q2r=2×12mv2+14πε0q1q2r/2⇒-14πε0q2r=mv2-24πε0q2r⇒mv2=14πε0q2r⇒v=14πε0mq2r⇒v=9×10-9×4×10-520.005×1⇒v=53.66 m/s

Question 70:

A sample of HCI gas is placed in an electric field of 2.5 × 104 NC−1. The dipole moment of each HCI molecule is 3.4 × 10−30 Cm. Find the maximum torque that can act on a molecule.

Answer:

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,

Γ=PEsinθFor

Γmax, putting sinθ = 1, we get

Γmax=PE       =3.4×10-302.5×104       =8.5×10-26 Nm

Question 71:

Two particles A and B, of opposite charges 2.0 × 10−6 C and −2.0 × 10−6 C, are placed at a separation of 1.0 cm. (a) Write down the electric dipole moment of this pair. (b) Calculate the electric field at a point on the axis of the dipole 1.0 cm away from the centre. (c) Calculate the electric field at a point on the perpendicular bisector of the dipole and 1.0 m away from the centre.

Answer:

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

(a) Dipole moment, P = q × l

P=2×10-6×10-2   =2×10-8 C-m(b) Electric field at the axial point of the dipole,

E=14πε02Pr3E=9×109×2×2×10-80.013E = 36 × 107 N/C

(c) Electric field at at a point on the perpendicular bisector of the dipole,

E=14πε0Pr’3E=9×109×2×10-813E = 180 N/C

Question 72:

Three charges are arranged on the vertices of an equilateral triangle, as shown in the figure. Find the dipole moment of the combination.
Figure

Answer:

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,

Pnet=P2+P2+2Pcos60°      =(qd)2+(qd)2+2qd2cos60°     =[d2q2+d2q2+2q2d2×12]1/2     =[3q2d2]1/2=3qdThe net dipole moment is along the bisector of the angle at +q, away from the triangle.

Question 73:

Find the magnitude of the electric field at the point P in the configuration shown in the figure for d >> a.
Figure

Answer:

For figure in (a)
Taking

P=2qa, E=14π∈0.qd2For figure in (b)

(b) E1sin θ=E2 sinθSo, E=E1cos θ+E2cosθ=E1 cosθ+E2 cosθ=2E1 cos θ=2.14π∈0.qa(d2+a2)3/2=14π∈0.pd3 [since a<<d]

Question 74:

Two particles, carrying charges −q and +q and and of mass m each, are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at an end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is slightly tilted and then released. Neglecting gravity, find the time period of small oscillations.

Answer:

Consider the rod to be a simple pendulum
Time period of a simple pendulum,

T=2πla'(where l = length and a‘ = acceleration)
Now,

a’=Fm=Eqmand length l = a
∴ The time period,

T=2πmaqE

Question 75:

Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire of mass 6.4 g (take the atomic weight of copper to be 64 g mol−1).

Answer:

Atomic weight of copper = 64 grams
No. of moles in 64 g of copper = 1
∴ No. of moles in 6.4 g of copper =  0.1
No. of atoms in 1 mole of copper = 6 × 10−23 = Avogadro’s Number
No. of atoms in 0.1 mole = (6 × 10−23 × 0.1) = 6 × 1022
1 atom contributes 1 free electron.
∴ 6 × 1022 atoms contribute 6 × 1022 free electrons.

Chapterwise HC Verma Solutions Class 12 Physics :

About the Author – HC Verma

HC Verma, the author of many popular and well-renowned Physics books, was born on 8 April 1952. Passing out from one of the most prestigious colleges of the country, IIT Kanpur, he worked as an experimental physicist in the Department of Nuclear Physics.

His most famous works which he is known for include the two-volume Concepts of Physics. He also worked for the social upliftment of the economically weaker children through his organization named Shiksha Sopan. He is also the recipient of the Padma Shri, which is considered India’s fourth-highest civilian award. He received the same because of his contribution and valuable work in the field of Physics.