Question 1.1:
What is the force between two small charged spheres having charges of 2 × 10
× 10−7 C placed 30 cm apart in air?
Answer
Repulsive force of magnitude 6 × 10
Charge on the first sphere, q1
Charge on the second sphere,
Distance between the spheres,
Electrostatic force between the spheres is given by the relation,
Where, ∈0 = Permittivity of free space
Hence, force between the two small charged spheres is 6 × 10
same nature. Hence, force between them will be repulsive.
Question 1.2:
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of
charge − 0.8 μC in air is 0.2 N. (a) What is the distance betwe
What is the force on the second sphere due to the first?
Answer
10−3 N
1 = 2 × 10−7 C
econd q2 = 3 × 10−7 C
r = 30 cm = 0.3 m
10−3 N. The charges are of
between the two spheres? (b)
10−7 C and 3
en
Electrostatic force on the first sphere,
Charge on this sphere, q1 = 0.4 μC = 0.4
Charge on the second sphere,
Electrostatic force between the spheres is given by the relation,
Where, ∈0 = Permittivity of free space
The distance between the two spheres is 0.12 m.
Both the spheres attract each other
second sphere due to the first is 0.2 N.
Question 1.3:
Check that the ratio ke2/G me
and determine the value of this ratio. What does the
Answer
F = 0.2 N
× 10−6 C
q2 = − 0.8 μC = − 0.8 × 10−6 C
with the same force. Therefore, the force on the
emp is dimensionless. Look up a Table of Physical Constants
ratio signify?
The given ratio is .
Where,
G = Gravitational constant
Its unit is N m2 kg−2.
me and mp = Masses of electron and proton.
Their unit is kg.
e = Electric charge.
Its unit is C.
∈0 = Permittivity of free space
Its unit is N m2 C−2.
Hence, the given ratio is dimensionless.
e = 1.6 × 10−19 C
G = 6.67 × 10−11 N m2 kg-2
me= 9.1 × 10−31 kg
mp = 1.66 × 10−27 kg
Hence, the numerical value of the given ratio is
This is the ratio of electric force to the gravitational force between a proton and an
electron, keeping distance between them constant.
Question 1.4:
Explain the meaning of the statement ‘electric charge of a body is quantised’.
Why can one ignore quantisation of electric charge when dealing with macroscopic i.e.,
large scale charges?
Answer
Electric charge of a body is quantized. This means that only integral (1, 2, ….,
of electrons can be transferred from one body to the other. Charges are not transferred in
fraction. Hence, a body possesses total
charge.
In macroscopic or large scale charges, the charges used are huge as compared to the
magnitude of electric charge. Hence, quantization of electric charge is of no use on
macroscopic scale. Therefore, it
continuous.
Question 1.5:
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar
phenomenon is observed with many other pairs of bodies. Explain how this observation
consistent with the law of conservation of charge.
Answer
Rubbing produces charges of equal magnitude but of opposite nature on the two bodies
because charges are created in pairs. This phenomenon of charging is called charging by
friction. The net charge on the system of two rubbed bodies is zero. This is because equal
amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk
cloth, opposite natured charges appear on both the bodies. This phenomenon is in
consistence with the law of conservation of energy. A similar phenomenon is observed
with many other pairs of bodies.
charge only in integral multiples of electric
is ignored and it is considered that electric charge is
et e n) number
is
Question 1.6:
Four point charges qA = 2 μC,
corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at
the centre of the square?
Answer
The given figure shows a square of side 10 cm with four charges placed at its corners. O
is the centre of the square.
Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD =
AO = OC = DO = OB =
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude
but opposite in direction relative to the force of repulsion between the charges placed at
corner C and centre O. Hence, they will cancel each other. Similarly, force
between charges placed at corner B and centre O is equal in magnitude but opposite in
direction relative to the force of attraction between the charges placed at corner D and
centre O. Hence, they will also cancel each other. Therefore, net f
charges placed at the corner of the square on 1 μC charge at centre O is zero.
qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the
cm
cm
of attraction
force caused by the four
orce
Question 1.7:
An electrostatic field line is a continuous curve. That is, a field line cannot have sudden
breaks. Why not?
Explain why two field lines never cross each other at any point?
Answer
An electrostatic field line is a continuous curve because a charge experiences a
continuous force when traced in an electrostatic field. The field line cannot have sudden
breaks because the charge mo
other.
If two field lines cross each other at a point, then electric field intensity will show two
directions at that point. This is not possible. Hence, two field lines never cross each other.
Question 1.8:
Two point charges qA = 3 μC and
What is the electric field at the midpoint O of the line AB joining the two charges?
If a negative test charge of magnitude 1.5 × 10
force experienced by the test charge?
Answer
The situation is represented in the given figure. O is the mid
Distance between the two charges, AB = 20 cm
moves continuously and does not jump from one point to the
qB = −3 μC are located 20 cm apart in vacuum.
10−9 C is placed at this point, what is the
mid-point of line AB.
ves
∴AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by +3μC charge,
E1 = along OB
Where,
= Permittivity of free space
Magnitude of electric field at point O caused by −3μC charge,
E2 = = along OB
= 5.4 × 106 N/C along OB
Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.
A test charge of amount 1.5 × 10−9 C is placed at mid-point O.
q = 1.5 × 10−9 C
Force experienced by the test charge = F
∴F = qE
= 1.5 × 10−9 × 5.4 × 106
= 8.1 × 10−3 N
The force is directed along line OA. This is because the negative test charge is repelled by
the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 × 10
Question 1.9:
A system has two charges qA
(0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric
dipole moment of the system?
Answer
Both the charges can be located in a coordinate frame of r
figure.
At A, amount of charge, qA = 2.5 × 10
At B, amount of charge, qB =
Total charge of the system,
q = qA + qB
= 2.5 × 107 C − 2.5 × 10−7 C
= 0
Distance between two charges at points A and B,
d = 15 + 15 = 30 cm = 0.3 m
Electric dipole moment of the system is given by,
10−3 N along OA.
A = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A:
reference as shown in the given
10−7C
−2.5 × 10−7 C
eference
p = qA × d = qB × d
= 2.5 × 10−7 × 0.3
= 7.5 × 10−8 C m along positive
Therefore, the electric dipole moment of the system is 7.5 × 10
z−axis.
Question 1.10:
An electric dipole with dipole moment 4 × 10
a uniform electric field of magnitude 5 × 10
acting on the dipole.
Answer
Electric dipole moment, p = 4 × 10
Angle made by p with a uniform electric field,
Electric field, E = 5 × 104 N C
Torque acting on the dipole is given by the relation,
τ = pE sinθ
Therefore, the magnitude of the torque acting
Question 1.11:
A polythene piece rubbed with wool is found to have a negative charge of 3 × 10
Estimate the number of electrons transferred (from which to which?)
z-axis
10−8 C m along positive
10−9 C m is aligned at 30° with the direction of
104 N C−1. Calculate the magnitude of the torque
10−9 C m
θ = 30°
C−1
on the dipole is 10−4 N m.
. 10−7 C.
Is there a transfer of mass from wool to polythene?
Answer
When polythene is rubbed against wool, a number of electrons get transferred from wool
to polythene. Hence, wool becomes positively charged and polythene becomes negatively
charged.
Amount of charge on the polythene piece, q = −3 × 10−7 C
Amount of charge on an electron, e = −1.6 × 10−19 C
Number of electrons transferred from wool to polythene = n
n can be calculated using the relation,
q = ne
= 1.87 × 1012
Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.
Yes.
There is a transfer of mass taking place. This is because an electron has mass,
me = 9.1 × 10−3 kg
Total mass transferred to polythene from wool,
m = me × n
= 9.1 × 10−31 × 1.85 × 1012
= 1.706 × 10−18 kg
Hence, a negligible amount of mass is transferred from wool to polythene.
Question 1.12:
Two insulated charged copper spheres A and B have their centers separated by a distance
of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 ×
10−7 C? The radii of A and B are negligible compared to the distance of separation.
What is the force of repulsion if each sphere is charged double the above amount, and the
distance between them is halved?
Answer
Charge on sphere A, qA = Charge on sphere B,
Distance between the spheres,
Force of repulsion between the two spheres,
Where,
∈0 = Free space permittivity
= 9 × 109 N m2 C−2
∴
= 1.52 × 10−2 N
Therefore, the force between the two spheres
After doubling the charge, charge on sphere A,
10−7 C = 1.3 × 10−6 C
The distance between the spheres is halved.
qB = 6.5 × 10−7 C
r = 50 cm = 0.5 m
is 1.52 × 10−2 N.
qA = Charge on sphere B, qB
= 2 × 6.5 ×
∴
Force of repulsion between the two spheres,
= 16 × 1.52 × 10−2
= 0.243 N
Therefore, the force between the two spheres is 0.243 N.
Question 1.13:
Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the
same size but uncharged is brought in contact with the first, then brought in contact w
the second, and finally removed from both. What is the new force of repulsion between A
and B?
Answer
Distance between the spheres, A and B,
Initially, the charge on each sphere,
When sphere A is touched with an uncharged sphere C,
transfer to sphere C. Hence, charge on each of the spheres, A and C, is
When sphere C with charge
charges on the system will divide into two equal halves given as,
r = 0.5 m
q = 6.5 × 10−7 C
amount of charge from A will
.
is brought in contact with sphere B with charge
with
q, total
Each sphere will each half. Hence, charge on each of the spheres, C and B, is
Force of repulsion between sphere A having charge
Therefore, the force of attraction bet
Question 1.14:
Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give
the signs of the three charges. Which particle has the highest charge to mass ratio?
Answer
Opposite charges attract each other and same charges repel each other. It can be observed
that particles 1 and 2 both move towards the positively charged plate and repel away from
the negatively charged plate. Hence, these two particles are neg
also be observed that particle 3 moves towards the negatively charged plate and repels
away from the positively charged plate. Hence, particle 3 is positively charged.
and sphere B having charge
between the two spheres is 5.703 × 10−3 N.
negatively charged. It can
.
=
atively
The charge to mass ratio (emf) is directly proportional to the d
deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has
the highest charge to mass ratio.
Question 1.15:
Consider a uniform electric field
through a square of 10 cm on a side whose plane is parallel to the
the flux through the same square if the normal to its plane makes a 60° angle with the
axis?
Answer
Electric field intensity, = 3 × 10
Magnitude of electric field intensity,
Side of the square, s = 10 cm = 0.1 m
Area of the square, A = s2 = 0.01 m
The plane of the square is parallel to the
normal to the plane and electric field,
Flux (Φ) through the plane is given by the relation,
Φ =
= 3 × 103 × 0.01 × cos0°
= 30 N m2/C
Plane makes an angle of 60° with the
Flux, Φ =
= 3 × 103 × 0.01 × cos60°
displacement or amount of
E = 3 × 103 îN/C. (a) What is the flux of this field
yz plane? (b) What is
103 î N/C
= 3 × 103 N/C
m2
y-z plane. Hence, angle between the unit vector
θ = 0°
) x-axis. Hence, θ = 60°
isplacement /xplane.
= 15 N m2/C
Question 1.16:
What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side
20 cm oriented so that its faces are parallel to the coordinate planes?
Answer
All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field
lines entering the cube is equal to the number of field lines piercing out of the cube. As a
result, net flux through the cube is zero.
Question 1.17:
Careful measurement of the electric field at the surface of a black box indicates that the
net outward flux through the surface of the box is 8.0 × 10
charge inside the box? (b) If the net outward flux through the surfac
zero, could you conclude that there were no charges inside the box? Why or Why not?
Answer
Net outward flux through the surface of the box,
For a body containing net charge
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
103 N m2/C. (a) What is the net
surface of the box were
Φ = 8.0 × 103 N m2/C
q, flux is given by the relation,
/e
q = ∈0Φ
= 8.854 × 10−12 × 8.0 × 103
= 7.08 × 10−8
= 0.07 μC
Therefore, the net charge inside the box is 0.07 μC.
No
Net flux piercing out through a body depends on the
net flux is zero, then it can be inferred that net charge inside the body is zero. The body
may have equal amount of positive and negative charges.
Question 1.18:
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10
cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square?
(Hint: Think of the square as one face of a cube with edge 10 cm.)
Answer
The square can be considered as one face of a cube of edge 10 cm with a centre where
charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through
all its six faces.
net charge contained in the body. If
Hence, electric flux through one face of the c
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = 10 μC = 10 × 10−6 C
∴
= 1.88 × 105 N m2 C−1
Therefore, electric flux through the square is 1.88 × 10
Question 1.19:
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge.
What is the net electric flux through the surface?
Answer
Net electric flux (ΦNet) through the cubic surface is given by,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = Net charge contained inside the cube = 2.0 μC = 2
cube i.e., through the square,
105 N m2 C−1.
) × 10−6 C
∴
= 2.26 × 105 N m2 C−1
The net electric flux through the surface is 2.26 ×10
Question 1.20:
A point charge causes an electric flux of
Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the
Gaussian surface were doubled, how much flux would pass through the surface? (b) What
is the value of the point charge?
Answer
Electric flux, Φ = −1.0 × 103
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the net charge enclosed inside a
body. It does not depend on the size of the body. If the radius of the Gaussian surface is
doubled, then the flux passing through the surface remains the same i.e.,
Electric flux is given by the relation,
Where,
q = Net charge enclosed by the spherical surface
∈0 = Permittivity of free space = 8.854 × 10
∴
= −1.0 × 103 × 8.854 × 10−12
×105 N m2C−1.
−1.0 × 103 Nm2/C to pass through a spherical
N m2/C
−10
10−12 N−1C2 m−2
/103 N m2/C.
= −8.854 × 10−9 C
= −8.854 nC
Therefore, the value of the point charge is
Question 1.21:
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm
from the centre of the sphere is 1.5 × 10
charge on the sphere?
Answer
Electric field intensity (E) at a distance (
charge q is given by the relation,
Where,
q = Net charge = 1.5 × 103 N/C
d = Distance from the centre = 20 cm = 0.2 m
∈0 = Permittivity of free space
And, = 9 × 109 N m2 C
∴
= 6.67 × 109 C
= 6.67 nC
−8.854 nC.
103 N/C and points radially inward, what is the net
) d) from the centre of a sphere containing net
C−2
)
Therefore, the net charge on the sphere is 6.67 nC.
Question 1.22:
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of
80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving
the surface of the sphere?
Answer
Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surface charge density, = 80.0 μC/m
Total charge on the surface of the sphere,
Q = Charge density × Surface area
=
= 80 × 10−6 × 4 × 3.14 × (1.2)
= 1.447 × 10−3 C
Therefore, the charge on the sphere is 1
Total electric flux ( ) leaving out the surface of a sphere containing net charge
given by the relation,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
. m2 = 80 × 10−6 C/m2
2
1.447 × 10−3 C.
Q is
Q = 1.447 × 10−3 C
= 1.63 × 108 N C−1 m2
Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10
Question 1.23:
An infinite line charge produces a field of 9 × 10
linear charge density.
Answer
Electric field produced by the infinite line charges at a distance
density λ is given by the relation,
Where,
d = 2 cm = 0.02 m
E = 9 × 104 N/C
∈0 = Permittivity of free space
= 9 × 109 N m2 C−2
= 10 μC/m
104 N/C at a distance of 2 cm. Calculate the
d having linear charge
108 N C−1 m2.
Therefore, the linear charge density is 10 μC/m.
Question 1.24:
Two large, thin metal plates are parallel and close to each other. On their inner faces, the
plates have surface charge densities of opposite signs and of magnitude 17.0 × 10
C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the
second plate, and (c) between the plates?
Answer
The situation is represented in the following figure.
A and B are two parallel plates close to each other. O
I, outer region of plate B is labelled as
labelled as II.
Charge density of plate A, σ = 17.0 × 10
Charge density of plate B, σ =
In the regions, I and III, electric field
the respective plates.
Electric field E in region II is given by the relation,
Where,
∈0 = Permittivity of free space = 8.854 × 10
: Outer region of plate A is labelled as
, III, and the region between the plates, A and B, is
10−22 C/m2
−17.0 × 10−22 C/m2
, E is zero. This is because charge is not enclosed by
10−12 N−1C2 m−2
10−22
uter ,
∴
= 1.92 × 10−10 N/C
Therefore, electric field between the plates is 1.92 × 10
Question 1.25:
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55
× 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm
Estimate the radius of the drop. (
Answer
Excess electrons on an oil drop,
Electric field intensity, E = 2.55 × 10
Density of oil, ρ = 1.26 gm/cm
Acceleration due to gravity, g = 9.81 m s
Charge on an electron, e = 1.6 × 10
Radius of the oil drop = r
Force (F) due to electric field
F = W
Eq = mg
Ene
Where,
q = Net charge on the oil drop =
m = Mass of the oil drop
10−10 N/C.
g = 9.81 m s−2; e = 1.60 × 10−19 C).
n = 12
104 N C−1
cm3 = 1.26 × 103 kg/m3
s−2
10−19 C
) E is equal to the weight of the oil drop (W)
ne
cm−3.
= Volume of the oil drop × Density of oil
= 9.82 × 10−4 mm
Therefore, the radius of the oil drop is 9.82 × 10
Question 1.26:
Which among the curves shown in Fig. 1.35
lines?
(a)
(b)
10−4 mm.
ccaannnnoott ppoossssiibbllyy rreepprreesseenntt eelleeccttrroossttaattiicc ffiieelldd
(c)
(d)
(e)
Answer
The field lines showed in (a) do not represent electrostatic field lines because field lines
must be normal to the surface of the conductor.
The field lines showed in (b) do not represent electrostatic field lines because the field
lines cannot emerge from a negative charge and cannot terminate at a positive charge.
The field lines showed in (c) represent electrostatic field lines. This is because the field
lines emerge from the positive charges and repel each other.
The field lines showed in (d) do not represent electrostatic field lines because the field
lines should not intersect each other.
The field lines showed in (e) do not represent electrostatic field
are not formed in the area between the field lines.
Question 1.27:
In a certain region of space, electric field is along the z
magnitude of electric field is, however, not constant but increases
positive z-direction, at the rate of 10
experienced by a system having a total dipole moment equal to 10
direction?
Answer
Dipole moment of the system,
Rate of increase of electric field per unit length,
Force (F) experienced by the system is given by the relation,
F = qE
= −10−7 × 10−5
= −10−2 N
The force is −10−2 N in the negative z
field. Hence, the angle between electric field and dipole moment is 180°.
Torque (τ) is given by the relation,
lines because closed loops
z-direction throughout. The
uniformly along the
105 NC−1 per metre. What are the force and torque
10−7 Cm in the negative
p = q × dl = −10−7 C m
) z-direction i.e., opposite to the direction of electric
) zdirection
τ = pE sin180°
= 0
Therefore, the torque experienced by the system is
Question 1.28:
A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge
entire charge must appear on the outer surface of the conductor. (b) Another conductor B
with charge q is inserted into the cavity keeping
charge on the outside surface of A is
be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Answer
Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing
the cavity. The electric field intensity
Let q is the charge inside the conductor and
According to Gauss’s law,
Flux,
Here, E = 0
zero.
Q. Show that the
B insulated from A. Show that the total
Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to
E inside the charged conductor is zero.
is the permittivity of free space.
.
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
The outer surface of conductor A has a charge of amount
charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of
amount −q will be induced in the inner surface of conductor A and +
outer surface of conductor A. Therefore, total charge on the outer surface of conductor A
is Q + q.
A sensitive instrument can be shielded from the strong electrostatic field in its
environment by enclosing it fully inside a metallic surface. A closed metallic body acts as
an electrostatic shield.
Question 1.29:
A hollow charged conductor
field in the hole is
and is the surface charge density near the hole.
Answer
Let us consider a conductor with a cavity or a
Let E is the electric field just outside the conductor,
charge density, and is the permittivity of free space.
Charge
According to Gauss’s law,
Q. Another conductor B having
q is induced on the
has a tiny hole cut into its surface. Show that the electric
, where is the unit vector in the outward normal direction,
hole. Electric field inside the cavity is zero.
q is the electric charge,
. is the
Therefore, the electric field just outside the conductor is
superposition of field due to the cavity
conductor . These fields are equal and opposite inside the conductor, and equal in
magnitude and direction outside
Therefore, the field due to the rest of the conductor is
Hence, proved.
Question 1.30:
Obtain the formula for the electric field due to a long thin wire of uniform linear charge
density λ without using Gauss’s law. [
necessary integral.]
Answer
Take a long thin wire XY (as shown in the figure) of uniform linear charge density
Consider a point A at a perpendicular distance
shown in the following figure.
. This field is a
and the field due to the rest of the charged
the conductor.
.
Hint: Use Coulomb’s law directly and evaluate the
l from the mid-point O of the wire, as
.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
Electric field due to the piece,
The electric field is resolved into two rectangular components. is the
perpendicular component and is the parallel component.
When the whole wire is considered, the component is cancelled.
Only the perpendicular component affects point A.
Hence, effective electric field at point A due to the element dx is dE1.
On differentiating equation (2), we obtain
From equation (2),
Putting equations (3) and (4) in equation (1), we obtain
The wire is so long that tends from
By integrating equation (5), we obtain the value of field
Therefore, the electric field due to long wire is
Question 1.31:
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter)
are themselves built out of more elementary units called quarks. A proton and a neutron
consist of three quarks each. Two types of quarks, the so called ‘up’ quark (d
of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (
electrons build up ordinary matter. (Quarks of other types have also been found which
give rise to different unusual varieties of matter.) Suggest a poss
a proton and neutron.
Answer
to .
E1 as,
.
, (−1/3) e, together with
possible quark composition of
denoted by u)
, ible
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of
.
Charge due to n up quarks
Number of down quarks in a proton = 3 − n
Each down quark has a charge of .
Charge due to (3 − n) down quarks
Total charge on a proton = + e
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3 − n = 3 − 2 = 1
Therefore, a proton can be represented as ‘uud’.
A neutron also has three quarks. Let there be n up quarks in a neutron, each having a
charge of .
Charge on a neutron due to n up quarks
Number of down quarks is 3 − n,each having a charge of .
Charge on a neutron due to
Total charge on a neutron = 0
Number of up quarks in a neutron,
Number of down quarks in a neutron = 3
Therefore, a neutron can be represented as ‘udd’.
Question 1.32:
Consider an arbitrary electrostatic field configuration. A small test charge is placed at a
null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test
charge is necessarily unstable.
Verify this result for the simple configu
sign placed a certain distance apart.
Answer
Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and
displaced from its position in any direction, then it experiences a
null point, where the electric field is zero. All the field lines near the null point are
directed inwards towards the null point. There is a net inward flux of electric field
through a closed surface around the null point. Accor
electric field through a surface, which is not enclosing any charge, is zero. Hence, the
equilibrium of the test charge can be stable.
Two charges of same magnitude and same sign are placed at a certain distance. The mid
point of the joining line of the charges is the null point. When a test charged is displaced
along the line, it experiences a restoring force. If it is displaced normal t
down quarks =
n n = 1
− n = 2
configuration of two charges of the same magnitude and
restoring force towards a
According to Gauss’s law, the flux of
to the joining line,
ration ding midpoint
o
then the net force takes it away from the null point. Hence, the charge is unstable because
stability of equilibrium requires restoring force in all directions.
Question 1.33:
A particle of mass m and charge (
initially moving along x-axis with speed
plate is L and an uniform electric field
vertical deflection of the particle at the far edge of the plate is
Compare this motion with motion of a projectile in gravitational field discussed in
Section 4.10 of Class XI Textbook of Physics.
Answer
Charge on a particle of mass
Velocity of the particle = vx
Length of the plates = L
Magnitude of the uniform electric field between the plates =
Mechanical force, F = Mass (
Therefore, acceleration,
Time taken by the particle to cross the field of length
t
In the vertical direction, initial velocity,
According to the third equation of motion, vertical deflection
obtained as,
(−q) enters the region between the two charged plates
vx (like particle 1 in Fig. 1.33). The length of
E is maintained between the plates. Show that the
he qEL2/ (2m ).
m = − q
E
m) × Acceleration (a)
L is given by,
u = 0
s of the particle can be
)
Hence, vertical deflection of the particle at the far edge of the plate is
. This is similar to the motion of horizontal projectiles under gravity.
Question 1.34:
Suppose that the particle in Exercise in 1.33 is an electron projected with velocity
× 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 10
electron strike the upper plate? (|
Answer
Velocity of the particle, vx = 2.0 × 10
Separation of the two plates,
Electric field between the two plates,
Charge on an electron, q = 1.6 × 10
Mass of an electron, me = 9.1 × 10
Let the electron strike the upper plate at the end of plate
Therefore,
102 N/C, where will the
e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)
106 m/s
d = 0.5 cm = 0.005 m
E = 9.1 × 102 N/C
10−19 C
10−31 kg
L, when deflection is
vx= 2.0
, s.
Therefore, the electron will strike the other plate after travelling 11.66 cm.
What is the force between two small charged spheres having charges of 2 × 10
× 10−7 C placed 30 cm apart in air?
Answer
Repulsive force of magnitude 6 × 10
Charge on the first sphere, q1
Charge on the second sphere,
Distance between the spheres,
Electrostatic force between the spheres is given by the relation,
Where, ∈0 = Permittivity of free space
Hence, force between the two small charged spheres is 6 × 10
same nature. Hence, force between them will be repulsive.
Question 1.2:
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of
charge − 0.8 μC in air is 0.2 N. (a) What is the distance betwe
What is the force on the second sphere due to the first?
Answer
10−3 N
1 = 2 × 10−7 C
econd q2 = 3 × 10−7 C
r = 30 cm = 0.3 m
10−3 N. The charges are of
between the two spheres? (b)
10−7 C and 3
en
Electrostatic force on the first sphere,
Charge on this sphere, q1 = 0.4 μC = 0.4
Charge on the second sphere,
Electrostatic force between the spheres is given by the relation,
Where, ∈0 = Permittivity of free space
The distance between the two spheres is 0.12 m.
Both the spheres attract each other
second sphere due to the first is 0.2 N.
Question 1.3:
Check that the ratio ke2/G me
and determine the value of this ratio. What does the
Answer
F = 0.2 N
× 10−6 C
q2 = − 0.8 μC = − 0.8 × 10−6 C
with the same force. Therefore, the force on the
emp is dimensionless. Look up a Table of Physical Constants
ratio signify?
The given ratio is .
Where,
G = Gravitational constant
Its unit is N m2 kg−2.
me and mp = Masses of electron and proton.
Their unit is kg.
e = Electric charge.
Its unit is C.
∈0 = Permittivity of free space
Its unit is N m2 C−2.
Hence, the given ratio is dimensionless.
e = 1.6 × 10−19 C
G = 6.67 × 10−11 N m2 kg-2
me= 9.1 × 10−31 kg
mp = 1.66 × 10−27 kg
Hence, the numerical value of the given ratio is
This is the ratio of electric force to the gravitational force between a proton and an
electron, keeping distance between them constant.
Question 1.4:
Explain the meaning of the statement ‘electric charge of a body is quantised’.
Why can one ignore quantisation of electric charge when dealing with macroscopic i.e.,
large scale charges?
Answer
Electric charge of a body is quantized. This means that only integral (1, 2, ….,
of electrons can be transferred from one body to the other. Charges are not transferred in
fraction. Hence, a body possesses total
charge.
In macroscopic or large scale charges, the charges used are huge as compared to the
magnitude of electric charge. Hence, quantization of electric charge is of no use on
macroscopic scale. Therefore, it
continuous.
Question 1.5:
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar
phenomenon is observed with many other pairs of bodies. Explain how this observation
consistent with the law of conservation of charge.
Answer
Rubbing produces charges of equal magnitude but of opposite nature on the two bodies
because charges are created in pairs. This phenomenon of charging is called charging by
friction. The net charge on the system of two rubbed bodies is zero. This is because equal
amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk
cloth, opposite natured charges appear on both the bodies. This phenomenon is in
consistence with the law of conservation of energy. A similar phenomenon is observed
with many other pairs of bodies.
charge only in integral multiples of electric
is ignored and it is considered that electric charge is
et e n) number
is
Question 1.6:
Four point charges qA = 2 μC,
corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at
the centre of the square?
Answer
The given figure shows a square of side 10 cm with four charges placed at its corners. O
is the centre of the square.
Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD =
AO = OC = DO = OB =
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude
but opposite in direction relative to the force of repulsion between the charges placed at
corner C and centre O. Hence, they will cancel each other. Similarly, force
between charges placed at corner B and centre O is equal in magnitude but opposite in
direction relative to the force of attraction between the charges placed at corner D and
centre O. Hence, they will also cancel each other. Therefore, net f
charges placed at the corner of the square on 1 μC charge at centre O is zero.
qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the
cm
cm
of attraction
force caused by the four
orce
Question 1.7:
An electrostatic field line is a continuous curve. That is, a field line cannot have sudden
breaks. Why not?
Explain why two field lines never cross each other at any point?
Answer
An electrostatic field line is a continuous curve because a charge experiences a
continuous force when traced in an electrostatic field. The field line cannot have sudden
breaks because the charge mo
other.
If two field lines cross each other at a point, then electric field intensity will show two
directions at that point. This is not possible. Hence, two field lines never cross each other.
Question 1.8:
Two point charges qA = 3 μC and
What is the electric field at the midpoint O of the line AB joining the two charges?
If a negative test charge of magnitude 1.5 × 10
force experienced by the test charge?
Answer
The situation is represented in the given figure. O is the mid
Distance between the two charges, AB = 20 cm
moves continuously and does not jump from one point to the
qB = −3 μC are located 20 cm apart in vacuum.
10−9 C is placed at this point, what is the
mid-point of line AB.
ves
∴AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by +3μC charge,
E1 = along OB
Where,
= Permittivity of free space
Magnitude of electric field at point O caused by −3μC charge,
E2 = = along OB
= 5.4 × 106 N/C along OB
Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.
A test charge of amount 1.5 × 10−9 C is placed at mid-point O.
q = 1.5 × 10−9 C
Force experienced by the test charge = F
∴F = qE
= 1.5 × 10−9 × 5.4 × 106
= 8.1 × 10−3 N
The force is directed along line OA. This is because the negative test charge is repelled by
the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 × 10
Question 1.9:
A system has two charges qA
(0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric
dipole moment of the system?
Answer
Both the charges can be located in a coordinate frame of r
figure.
At A, amount of charge, qA = 2.5 × 10
At B, amount of charge, qB =
Total charge of the system,
q = qA + qB
= 2.5 × 107 C − 2.5 × 10−7 C
= 0
Distance between two charges at points A and B,
d = 15 + 15 = 30 cm = 0.3 m
Electric dipole moment of the system is given by,
10−3 N along OA.
A = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A:
reference as shown in the given
10−7C
−2.5 × 10−7 C
eference
p = qA × d = qB × d
= 2.5 × 10−7 × 0.3
= 7.5 × 10−8 C m along positive
Therefore, the electric dipole moment of the system is 7.5 × 10
z−axis.
Question 1.10:
An electric dipole with dipole moment 4 × 10
a uniform electric field of magnitude 5 × 10
acting on the dipole.
Answer
Electric dipole moment, p = 4 × 10
Angle made by p with a uniform electric field,
Electric field, E = 5 × 104 N C
Torque acting on the dipole is given by the relation,
τ = pE sinθ
Therefore, the magnitude of the torque acting
Question 1.11:
A polythene piece rubbed with wool is found to have a negative charge of 3 × 10
Estimate the number of electrons transferred (from which to which?)
z-axis
10−8 C m along positive
10−9 C m is aligned at 30° with the direction of
104 N C−1. Calculate the magnitude of the torque
10−9 C m
θ = 30°
C−1
on the dipole is 10−4 N m.
. 10−7 C.
Is there a transfer of mass from wool to polythene?
Answer
When polythene is rubbed against wool, a number of electrons get transferred from wool
to polythene. Hence, wool becomes positively charged and polythene becomes negatively
charged.
Amount of charge on the polythene piece, q = −3 × 10−7 C
Amount of charge on an electron, e = −1.6 × 10−19 C
Number of electrons transferred from wool to polythene = n
n can be calculated using the relation,
q = ne
= 1.87 × 1012
Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.
Yes.
There is a transfer of mass taking place. This is because an electron has mass,
me = 9.1 × 10−3 kg
Total mass transferred to polythene from wool,
m = me × n
= 9.1 × 10−31 × 1.85 × 1012
= 1.706 × 10−18 kg
Hence, a negligible amount of mass is transferred from wool to polythene.
Question 1.12:
Two insulated charged copper spheres A and B have their centers separated by a distance
of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 ×
10−7 C? The radii of A and B are negligible compared to the distance of separation.
What is the force of repulsion if each sphere is charged double the above amount, and the
distance between them is halved?
Answer
Charge on sphere A, qA = Charge on sphere B,
Distance between the spheres,
Force of repulsion between the two spheres,
Where,
∈0 = Free space permittivity
= 9 × 109 N m2 C−2
∴
= 1.52 × 10−2 N
Therefore, the force between the two spheres
After doubling the charge, charge on sphere A,
10−7 C = 1.3 × 10−6 C
The distance between the spheres is halved.
qB = 6.5 × 10−7 C
r = 50 cm = 0.5 m
is 1.52 × 10−2 N.
qA = Charge on sphere B, qB
= 2 × 6.5 ×
∴
Force of repulsion between the two spheres,
= 16 × 1.52 × 10−2
= 0.243 N
Therefore, the force between the two spheres is 0.243 N.
Question 1.13:
Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the
same size but uncharged is brought in contact with the first, then brought in contact w
the second, and finally removed from both. What is the new force of repulsion between A
and B?
Answer
Distance between the spheres, A and B,
Initially, the charge on each sphere,
When sphere A is touched with an uncharged sphere C,
transfer to sphere C. Hence, charge on each of the spheres, A and C, is
When sphere C with charge
charges on the system will divide into two equal halves given as,
r = 0.5 m
q = 6.5 × 10−7 C
amount of charge from A will
.
is brought in contact with sphere B with charge
with
q, total
Each sphere will each half. Hence, charge on each of the spheres, C and B, is
Force of repulsion between sphere A having charge
Therefore, the force of attraction bet
Question 1.14:
Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give
the signs of the three charges. Which particle has the highest charge to mass ratio?
Answer
Opposite charges attract each other and same charges repel each other. It can be observed
that particles 1 and 2 both move towards the positively charged plate and repel away from
the negatively charged plate. Hence, these two particles are neg
also be observed that particle 3 moves towards the negatively charged plate and repels
away from the positively charged plate. Hence, particle 3 is positively charged.
and sphere B having charge
between the two spheres is 5.703 × 10−3 N.
negatively charged. It can
.
=
atively
The charge to mass ratio (emf) is directly proportional to the d
deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has
the highest charge to mass ratio.
Question 1.15:
Consider a uniform electric field
through a square of 10 cm on a side whose plane is parallel to the
the flux through the same square if the normal to its plane makes a 60° angle with the
axis?
Answer
Electric field intensity, = 3 × 10
Magnitude of electric field intensity,
Side of the square, s = 10 cm = 0.1 m
Area of the square, A = s2 = 0.01 m
The plane of the square is parallel to the
normal to the plane and electric field,
Flux (Φ) through the plane is given by the relation,
Φ =
= 3 × 103 × 0.01 × cos0°
= 30 N m2/C
Plane makes an angle of 60° with the
Flux, Φ =
= 3 × 103 × 0.01 × cos60°
displacement or amount of
E = 3 × 103 îN/C. (a) What is the flux of this field
yz plane? (b) What is
103 î N/C
= 3 × 103 N/C
m2
y-z plane. Hence, angle between the unit vector
θ = 0°
) x-axis. Hence, θ = 60°
isplacement /xplane.
= 15 N m2/C
Question 1.16:
What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side
20 cm oriented so that its faces are parallel to the coordinate planes?
Answer
All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field
lines entering the cube is equal to the number of field lines piercing out of the cube. As a
result, net flux through the cube is zero.
Question 1.17:
Careful measurement of the electric field at the surface of a black box indicates that the
net outward flux through the surface of the box is 8.0 × 10
charge inside the box? (b) If the net outward flux through the surfac
zero, could you conclude that there were no charges inside the box? Why or Why not?
Answer
Net outward flux through the surface of the box,
For a body containing net charge
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
103 N m2/C. (a) What is the net
surface of the box were
Φ = 8.0 × 103 N m2/C
q, flux is given by the relation,
/e
q = ∈0Φ
= 8.854 × 10−12 × 8.0 × 103
= 7.08 × 10−8
= 0.07 μC
Therefore, the net charge inside the box is 0.07 μC.
No
Net flux piercing out through a body depends on the
net flux is zero, then it can be inferred that net charge inside the body is zero. The body
may have equal amount of positive and negative charges.
Question 1.18:
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10
cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square?
(Hint: Think of the square as one face of a cube with edge 10 cm.)
Answer
The square can be considered as one face of a cube of edge 10 cm with a centre where
charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through
all its six faces.
net charge contained in the body. If
Hence, electric flux through one face of the c
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = 10 μC = 10 × 10−6 C
∴
= 1.88 × 105 N m2 C−1
Therefore, electric flux through the square is 1.88 × 10
Question 1.19:
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge.
What is the net electric flux through the surface?
Answer
Net electric flux (ΦNet) through the cubic surface is given by,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = Net charge contained inside the cube = 2.0 μC = 2
cube i.e., through the square,
105 N m2 C−1.
) × 10−6 C
∴
= 2.26 × 105 N m2 C−1
The net electric flux through the surface is 2.26 ×10
Question 1.20:
A point charge causes an electric flux of
Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the
Gaussian surface were doubled, how much flux would pass through the surface? (b) What
is the value of the point charge?
Answer
Electric flux, Φ = −1.0 × 103
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the net charge enclosed inside a
body. It does not depend on the size of the body. If the radius of the Gaussian surface is
doubled, then the flux passing through the surface remains the same i.e.,
Electric flux is given by the relation,
Where,
q = Net charge enclosed by the spherical surface
∈0 = Permittivity of free space = 8.854 × 10
∴
= −1.0 × 103 × 8.854 × 10−12
×105 N m2C−1.
−1.0 × 103 Nm2/C to pass through a spherical
N m2/C
−10
10−12 N−1C2 m−2
/103 N m2/C.
= −8.854 × 10−9 C
= −8.854 nC
Therefore, the value of the point charge is
Question 1.21:
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm
from the centre of the sphere is 1.5 × 10
charge on the sphere?
Answer
Electric field intensity (E) at a distance (
charge q is given by the relation,
Where,
q = Net charge = 1.5 × 103 N/C
d = Distance from the centre = 20 cm = 0.2 m
∈0 = Permittivity of free space
And, = 9 × 109 N m2 C
∴
= 6.67 × 109 C
= 6.67 nC
−8.854 nC.
103 N/C and points radially inward, what is the net
) d) from the centre of a sphere containing net
C−2
)
Therefore, the net charge on the sphere is 6.67 nC.
Question 1.22:
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of
80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving
the surface of the sphere?
Answer
Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surface charge density, = 80.0 μC/m
Total charge on the surface of the sphere,
Q = Charge density × Surface area
=
= 80 × 10−6 × 4 × 3.14 × (1.2)
= 1.447 × 10−3 C
Therefore, the charge on the sphere is 1
Total electric flux ( ) leaving out the surface of a sphere containing net charge
given by the relation,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
. m2 = 80 × 10−6 C/m2
2
1.447 × 10−3 C.
Q is
Q = 1.447 × 10−3 C
= 1.63 × 108 N C−1 m2
Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10
Question 1.23:
An infinite line charge produces a field of 9 × 10
linear charge density.
Answer
Electric field produced by the infinite line charges at a distance
density λ is given by the relation,
Where,
d = 2 cm = 0.02 m
E = 9 × 104 N/C
∈0 = Permittivity of free space
= 9 × 109 N m2 C−2
= 10 μC/m
104 N/C at a distance of 2 cm. Calculate the
d having linear charge
108 N C−1 m2.
Therefore, the linear charge density is 10 μC/m.
Question 1.24:
Two large, thin metal plates are parallel and close to each other. On their inner faces, the
plates have surface charge densities of opposite signs and of magnitude 17.0 × 10
C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the
second plate, and (c) between the plates?
Answer
The situation is represented in the following figure.
A and B are two parallel plates close to each other. O
I, outer region of plate B is labelled as
labelled as II.
Charge density of plate A, σ = 17.0 × 10
Charge density of plate B, σ =
In the regions, I and III, electric field
the respective plates.
Electric field E in region II is given by the relation,
Where,
∈0 = Permittivity of free space = 8.854 × 10
: Outer region of plate A is labelled as
, III, and the region between the plates, A and B, is
10−22 C/m2
−17.0 × 10−22 C/m2
, E is zero. This is because charge is not enclosed by
10−12 N−1C2 m−2
10−22
uter ,
∴
= 1.92 × 10−10 N/C
Therefore, electric field between the plates is 1.92 × 10
Question 1.25:
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55
× 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm
Estimate the radius of the drop. (
Answer
Excess electrons on an oil drop,
Electric field intensity, E = 2.55 × 10
Density of oil, ρ = 1.26 gm/cm
Acceleration due to gravity, g = 9.81 m s
Charge on an electron, e = 1.6 × 10
Radius of the oil drop = r
Force (F) due to electric field
F = W
Eq = mg
Ene
Where,
q = Net charge on the oil drop =
m = Mass of the oil drop
10−10 N/C.
g = 9.81 m s−2; e = 1.60 × 10−19 C).
n = 12
104 N C−1
cm3 = 1.26 × 103 kg/m3
s−2
10−19 C
) E is equal to the weight of the oil drop (W)
ne
cm−3.
= Volume of the oil drop × Density of oil
= 9.82 × 10−4 mm
Therefore, the radius of the oil drop is 9.82 × 10
Question 1.26:
Which among the curves shown in Fig. 1.35
lines?
(a)
(b)
10−4 mm.
ccaannnnoott ppoossssiibbllyy rreepprreesseenntt eelleeccttrroossttaattiicc ffiieelldd
(c)
(d)
(e)
Answer
The field lines showed in (a) do not represent electrostatic field lines because field lines
must be normal to the surface of the conductor.
The field lines showed in (b) do not represent electrostatic field lines because the field
lines cannot emerge from a negative charge and cannot terminate at a positive charge.
The field lines showed in (c) represent electrostatic field lines. This is because the field
lines emerge from the positive charges and repel each other.
The field lines showed in (d) do not represent electrostatic field lines because the field
lines should not intersect each other.
The field lines showed in (e) do not represent electrostatic field
are not formed in the area between the field lines.
Question 1.27:
In a certain region of space, electric field is along the z
magnitude of electric field is, however, not constant but increases
positive z-direction, at the rate of 10
experienced by a system having a total dipole moment equal to 10
direction?
Answer
Dipole moment of the system,
Rate of increase of electric field per unit length,
Force (F) experienced by the system is given by the relation,
F = qE
= −10−7 × 10−5
= −10−2 N
The force is −10−2 N in the negative z
field. Hence, the angle between electric field and dipole moment is 180°.
Torque (τ) is given by the relation,
lines because closed loops
z-direction throughout. The
uniformly along the
105 NC−1 per metre. What are the force and torque
10−7 Cm in the negative
p = q × dl = −10−7 C m
) z-direction i.e., opposite to the direction of electric
) zdirection
τ = pE sin180°
= 0
Therefore, the torque experienced by the system is
Question 1.28:
A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge
entire charge must appear on the outer surface of the conductor. (b) Another conductor B
with charge q is inserted into the cavity keeping
charge on the outside surface of A is
be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Answer
Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing
the cavity. The electric field intensity
Let q is the charge inside the conductor and
According to Gauss’s law,
Flux,
Here, E = 0
zero.
Q. Show that the
B insulated from A. Show that the total
Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to
E inside the charged conductor is zero.
is the permittivity of free space.
.
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
The outer surface of conductor A has a charge of amount
charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of
amount −q will be induced in the inner surface of conductor A and +
outer surface of conductor A. Therefore, total charge on the outer surface of conductor A
is Q + q.
A sensitive instrument can be shielded from the strong electrostatic field in its
environment by enclosing it fully inside a metallic surface. A closed metallic body acts as
an electrostatic shield.
Question 1.29:
A hollow charged conductor
field in the hole is
and is the surface charge density near the hole.
Answer
Let us consider a conductor with a cavity or a
Let E is the electric field just outside the conductor,
charge density, and is the permittivity of free space.
Charge
According to Gauss’s law,
Q. Another conductor B having
q is induced on the
has a tiny hole cut into its surface. Show that the electric
, where is the unit vector in the outward normal direction,
hole. Electric field inside the cavity is zero.
q is the electric charge,
. is the
Therefore, the electric field just outside the conductor is
superposition of field due to the cavity
conductor . These fields are equal and opposite inside the conductor, and equal in
magnitude and direction outside
Therefore, the field due to the rest of the conductor is
Hence, proved.
Question 1.30:
Obtain the formula for the electric field due to a long thin wire of uniform linear charge
density λ without using Gauss’s law. [
necessary integral.]
Answer
Take a long thin wire XY (as shown in the figure) of uniform linear charge density
Consider a point A at a perpendicular distance
shown in the following figure.
. This field is a
and the field due to the rest of the charged
the conductor.
.
Hint: Use Coulomb’s law directly and evaluate the
l from the mid-point O of the wire, as
.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
Electric field due to the piece,
The electric field is resolved into two rectangular components. is the
perpendicular component and is the parallel component.
When the whole wire is considered, the component is cancelled.
Only the perpendicular component affects point A.
Hence, effective electric field at point A due to the element dx is dE1.
On differentiating equation (2), we obtain
From equation (2),
Putting equations (3) and (4) in equation (1), we obtain
The wire is so long that tends from
By integrating equation (5), we obtain the value of field
Therefore, the electric field due to long wire is
Question 1.31:
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter)
are themselves built out of more elementary units called quarks. A proton and a neutron
consist of three quarks each. Two types of quarks, the so called ‘up’ quark (d
of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (
electrons build up ordinary matter. (Quarks of other types have also been found which
give rise to different unusual varieties of matter.) Suggest a poss
a proton and neutron.
Answer
to .
E1 as,
.
, (−1/3) e, together with
possible quark composition of
denoted by u)
, ible
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of
.
Charge due to n up quarks
Number of down quarks in a proton = 3 − n
Each down quark has a charge of .
Charge due to (3 − n) down quarks
Total charge on a proton = + e
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3 − n = 3 − 2 = 1
Therefore, a proton can be represented as ‘uud’.
A neutron also has three quarks. Let there be n up quarks in a neutron, each having a
charge of .
Charge on a neutron due to n up quarks
Number of down quarks is 3 − n,each having a charge of .
Charge on a neutron due to
Total charge on a neutron = 0
Number of up quarks in a neutron,
Number of down quarks in a neutron = 3
Therefore, a neutron can be represented as ‘udd’.
Question 1.32:
Consider an arbitrary electrostatic field configuration. A small test charge is placed at a
null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test
charge is necessarily unstable.
Verify this result for the simple configu
sign placed a certain distance apart.
Answer
Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and
displaced from its position in any direction, then it experiences a
null point, where the electric field is zero. All the field lines near the null point are
directed inwards towards the null point. There is a net inward flux of electric field
through a closed surface around the null point. Accor
electric field through a surface, which is not enclosing any charge, is zero. Hence, the
equilibrium of the test charge can be stable.
Two charges of same magnitude and same sign are placed at a certain distance. The mid
point of the joining line of the charges is the null point. When a test charged is displaced
along the line, it experiences a restoring force. If it is displaced normal t
down quarks =
n n = 1
− n = 2
configuration of two charges of the same magnitude and
restoring force towards a
According to Gauss’s law, the flux of
to the joining line,
ration ding midpoint
o
then the net force takes it away from the null point. Hence, the charge is unstable because
stability of equilibrium requires restoring force in all directions.
Question 1.33:
A particle of mass m and charge (
initially moving along x-axis with speed
plate is L and an uniform electric field
vertical deflection of the particle at the far edge of the plate is
Compare this motion with motion of a projectile in gravitational field discussed in
Section 4.10 of Class XI Textbook of Physics.
Answer
Charge on a particle of mass
Velocity of the particle = vx
Length of the plates = L
Magnitude of the uniform electric field between the plates =
Mechanical force, F = Mass (
Therefore, acceleration,
Time taken by the particle to cross the field of length
t
In the vertical direction, initial velocity,
According to the third equation of motion, vertical deflection
obtained as,
(−q) enters the region between the two charged plates
vx (like particle 1 in Fig. 1.33). The length of
E is maintained between the plates. Show that the
he qEL2/ (2m ).
m = − q
E
m) × Acceleration (a)
L is given by,
u = 0
s of the particle can be
)
Hence, vertical deflection of the particle at the far edge of the plate is
. This is similar to the motion of horizontal projectiles under gravity.
Question 1.34:
Suppose that the particle in Exercise in 1.33 is an electron projected with velocity
× 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 10
electron strike the upper plate? (|
Answer
Velocity of the particle, vx = 2.0 × 10
Separation of the two plates,
Electric field between the two plates,
Charge on an electron, q = 1.6 × 10
Mass of an electron, me = 9.1 × 10
Let the electron strike the upper plate at the end of plate
Therefore,
102 N/C, where will the
e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)
106 m/s
d = 0.5 cm = 0.005 m
E = 9.1 × 102 N/C
10−19 C
10−31 kg
L, when deflection is
vx= 2.0
, s.
Therefore, the electron will strike the other plate after travelling 11.66 cm.
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