Physics Part I Solutions for Class 12 Science Physics Chapter 1 Electric Charges And Fields are provided here with simple step-by-step explanations. These solutions for Electric Charges And Fields are extremely popular among Class 12 Science students for Physics Electric Charges And Fields Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Part I Book of Class 12 Science Physics Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Physics Part I Solutions. All Physics Part I Solutions for class Class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 46:

#### Question 1.1:

What is the force
between two small charged spheres having charges of 2 × 10^{−7}
C and 3 × 10^{−7} C placed 30 cm apart in air?

#### Answer:

Repulsive force of
magnitude 6 × 10^{−3} N

Charge on the first
sphere, *q*_{1} = 2 × 10^{−7} C

Charge on the second
sphere, *q*_{2} = 3 × 10^{−7} C

Distance between the
spheres, *r *= 30 cm = 0.3 m

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^{−3} N.
The charges are of same nature. Hence, force between them will be
repulsive.

#### Page No 46:

#### 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 between the two spheres? (b) What is the force on the second sphere due to the first?

#### Answer:

**(a) **Electrostatic force on the first sphere, *F* = 0.2 N

Charge on this sphere, *q*_{1} = 0.4 μC = 0.4 × 10^{−6 }C

Charge on the second sphere, *q*_{2}_{ }= − 0.8 μC = − 0.8 × 10^{−6}^{ }C

Electrostatic force between the spheres is given by the relation,

Where, ∈_{0} = Permittivity of free space

${r}^{2}=\frac{{q}_{1}{q}_{2}}{4\pi {\in}_{0}F}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times 0.4\times {10}^{-6}\times 0.8\times {10}^{-6}}{0.2}\phantom{\rule{0ex}{0ex}}=144\times {10}^{-4}\phantom{\rule{0ex}{0ex}}r=\sqrt{144\times {10}^{-4}}=0.12m$

The distance between the two spheres is 0.12 m.

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

#### Page No 46:

#### Question 1.3:

Check that the ratio *ke*^{2}/*G m*_{e}*m*_{p}_{ }is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

#### Answer:

The given ratio is** **.

Where,

G = Gravitational constant

Its unit is N m^{2 }kg^{−2}.

*m*_{e} and *m*_{p} = 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 m^{2 }C^{−2}.

Hence, the given ratio is dimensionless.

*e *= 1.6 × 10^{−19 }C

G = 6.67 × 10^{−11} N m^{2 }kg^{-2}

*m*_{e}= 9.1 × 10^{−31 }kg

*m*_{p} = 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.

#### Page No 46:

#### Question 1.4:

**(a)** Explain the
meaning of the statement ‘electric charge of a body is
quantised’.

**(b)** Why can one ignore quantisation of electric charge when
dealing with macroscopic i.e., large scale charges?

#### Answer:

**(a)** Electric charge of a body is quantized. This means that
only integral (1, 2, …., *n*) number of electrons can be
transferred from one body to the other. Charges are not transferred
in fraction. Hence, a body possesses total charge only in integral
multiples of electric charge.

**(b) **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 is ignored and it is considered that electric charge is
continuous.

#### Page No 46:

#### 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 is 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.

#### Page No 46:

#### Question 1.6:

Four
point charges *q*_{A
}= 2 μC, *q*_{B
}= −5 μC, *q*_{C}
= 2 μC, and *q*_{D}
= −5
μC are located at the 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 = cm

AO = OC = DO = OB = cm

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 of attraction 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 force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.

#### Page No 46:

#### Question 1.7:

**(a)** An
electrostatic field line is a continuous curve. That is, a field line
cannot have sudden breaks. Why not?

**(b)**
Explain why two field lines never cross each other at any point?

#### Answer:

**(a)** 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 moves continuously and does not jump from one point to the
other.

**(b)** 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.

#### Page No 46:

#### Question 1.8:

Two point charges *q*_{A}
= 3 μC and *q*_{B}
= −3 μC are located 20 cm apart in vacuum.

**(a)** What is
the electric field at the midpoint O of the line AB joining the two
charges?

**(b)** If a
negative test charge of magnitude 1.5 × 10^{−9}
C is placed at this point, what is the force experienced by the test
charge?

#### Answer:

**(a) **The
situation is represented in the given figure. O is the mid-point of
line AB.

Distance between the two charges, AB = 20 cm

∴AO = OB = 10 cm

Net
electric field at point O = *E*

Electric field at point O caused by +3μC charge,

*E*_{1}
=
along OB

Where,

= Permittivity of free space

Magnitude of electric field at point O caused by −3μC charge,

*E*_{2}
=
=
along OB

=
5.4 × 10^{6} N/C along OB

Therefore,
the electric field at mid-point O is 5.4 × 10^{6} N C^{−1}
along OB.

**(b) **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 × 10^{6}

= 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^{−3}
N along OA.

#### Page No 46:

#### Question 1.9:

A system has two charges *q*_{A} = 2.5 × 10^{−7} C and *q*_{B} = −2.5 × 10^{−7} C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

#### Answer:

Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge, *q*_{A} = 2.5 × 10^{−}^{7}C

At B, amount of charge, *q*_{B} = −2.5 × 10^{−7} C

Total charge of the system,

*q* = *q*_{A} + *q*_{B}

= 2.5 × 10^{−7} 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,

*p* = *q*_{A} × *d =* *q*_{B} × *d*

= 2.5 × 10^{−7} × 0.3

= 7.5 × 10^{−8} C m along positive *z*-axis

Therefore, the electric dipole moment of the system is 7.5 × 10^{−8} C m along positive *z*−axis.

#### Page No 46:

#### Question 1.10:

An
electric dipole with dipole moment 4 × 10^{−9}
C m is aligned at 30° with the direction of a uniform electric
field of magnitude 5 × 10^{4}
N C^{−1}.
Calculate the magnitude of the torque acting on the dipole.

#### Answer:

Electric dipole moment,
*p* = 4 × 10^{−9} C m

Angle made by *p*
with a uniform electric field, *θ* = 30°

Electric field, *E*
= 5 × 10^{4} N C^{−1}

Torque acting on the dipole is given by the relation,

τ = *pE*
sin*θ*

Therefore, the
magnitude of the torque acting on the dipole is 10^{−4}
N m.

#### Page No 46:

#### Question 1.11:

A polythene piece rubbed with wool is found to
have a negative charge of 3 × 10^{−7}
C.

**(a)** Estimate
the number of electrons transferred (from which to which?)

**(b)** Is
there a transfer of mass from wool to polythene?

#### Answer:

**(a) **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 × 10^{12}

Therefore,
the number of electrons transferred from wool to polythene is 1.87 ×
10^{12}.

**(b) **Yes.

There is a transfer of mass taking place. This is because an electron has mass,

*m*_{e}
= 9.1 × 10^{−3} kg

Total mass transferred to polythene from wool,

*m*
= *m*_{e}* × n*

= 9.1 × 10^{−31} × 1.85 × 10^{12}

= 1.706 × 10^{−18} kg

Hence, a negligible amount of mass is transferred from wool to polythene.

#### Page No 46:

#### Question 1.12:

**(a)** 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.

**(b)** What is
the force of repulsion if each sphere is charged double the above
amount, and the distance between them is halved?

#### Answer:

**(a) **Charge on
sphere A, *q*_{A} = Charge on sphere B, *q*_{B}
= 6.5 × 10^{−7} C

Distance
between the spheres, *r* = 50 cm = 0.5 m

Force of repulsion between the two spheres,

Where,

∈_{0}
= Free space permittivity

=
9 × 10^{9} N m^{2} C^{−2}

∴

= 1.52 × 10^{−2} N

Therefore,
the force between the two spheres is 1.52 × 10^{−2}
N.

**(b) **After doubling the charge, charge on sphere A, *q*_{A}
= Charge on sphere B, *q*_{B} = 2 × 6.5 ×
10^{−7} C = 1.3 × 10^{−6} C

The distance between the spheres is halved.

∴

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.

#### Page No 47:

#### 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 with 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, *r* = 0.5 m

Initially, the charge
on each sphere, *q* = 6.5 × 10^{−7} C

When sphere A is touched with an uncharged sphere C, amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is.

When sphere C with
charge
is brought in contact with sphere B with charge *q*, total
charges on the system will divide into two equal halves given as,

Each sphere will share each half. Hence, charge on each of the spheres, C and B, is.

Force of repulsion between sphere A having charge and sphere B having charge =

Therefore, the force of
attraction between the two spheres is 5.703 × 10^{−3}
N.

#### Page No 47:

#### 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 negatively charged. It can 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.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

#### Page No 47:

#### Question 1.15:

Consider
a uniform electric field **E **=
3 × 10^{3}
îN/C. (a) What is the flux of this
field through a square of 10 cm on a side whose plane is parallel to
the *yz *plane?
(b) What is the flux through the same square if the normal to its
plane makes a 60° angle with the *x*-axis?

#### Answer:

**(a) **Electric
field intensity,
= 3 × 10^{3 }î N/C

Magnitude
of electric field intensity,
=
3 × 10^{3} N/C

Side
of the square, *s* = 10 cm = 0.1 m

Area
of the square, *A* = s^{2} = 0.01 m^{2}

The
plane of the square is parallel to the *y-z* plane. Hence, angle
between the unit vector normal to the plane and electric field, *θ*
= 0°

Flux
(*Φ*) through the plane
is given by the relation,

Φ =

= 3 × 10^{3} × 0.01 × cos0°

= 30 N m^{2}/C

**(b) **Plane makes
an angle of 60° with the *x*-axis. Hence, *θ* =
60°

Flux,
*Φ* =

= 3 × 10^{3} × 0.01 × cos60°

= 15 N m^{2}/C

#### Page No 47:

#### 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.

#### Page No 47:

#### 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^{3}
N m^{2}/C.
(a) What is the net charge inside the box? (b) If the net outward
flux through the surface of the box were zero, could you conclude
that there were no charges inside the box? Why or Why not?

#### Answer:

**(a) **Net outward
flux through the surface of the box, *Φ*
= 8.0 × 10^{3} N m^{2}/C

For
a body containing net charge *q*, flux is given by the relation,

∈_{0}
= Permittivity of free space

= 8.854 × 10^{−12} N^{−1}C^{2
}m^{−2}

*q*
= ∈_{0}*Φ*

= 8.854 × 10^{−12} × 8.0 × 10^{3}

= 7.08 × 10^{−8}

= 0.07 μC

Therefore, the net charge inside the box is 0.07 μC.

**(b) **No

Net flux piercing out through a body depends on the net charge contained in the body. If 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.

#### Page No 47:

#### 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.

Hence, electric flux through one face of the cube i.e., through the square,

Where,

∈_{0} =
Permittivity of free space

= 8.854 ×
10^{−12} N^{−1}C^{2 }m^{−2}

*q* = 10 μC
= 10 × 10^{−6} C

∴

= 1.88 × 10^{5} N m^{2} C^{−1}

Therefore, electric
flux through the square is 1.88 × 10^{5} N m^{2}
C^{−1}.

#### Page No 48:

#### 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
(*Φ*_{Net})
through the cubic surface is given by,

Where,

∈_{0} =
Permittivity of free space

= 8.854 ×
10^{−12} N^{−1}C^{2 }m^{−2}

*q* = Net charge
contained inside the cube = 2.0 μC
= 2 × 10^{−6} C

∴

= 2.26 ×
10^{5} N m^{2} C^{−1}

The net electric flux
through the surface is 2.26 ×10^{5} N m^{2}C^{−1}.

#### Page No 48:

#### Question 1.20:

A
point charge causes an electric flux of −1.0 × 10^{3}
Nm^{2}/C
to pass through a spherical 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:

**(a) **Electric
flux, *Φ* = −1.0
× 10^{3} N m^{2}/C

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., −10^{3}
N m^{2}/C.

**(b) **Electric
flux is given by the relation,

Where,

*q*
= Net charge enclosed by the spherical surface

∈_{0}
= Permittivity of free space = 8.854 × 10^{−12}
N^{−1}C^{2 }m^{−2}

∴

= −1.0 × 10^{3} × 8.854 × 10^{−12}

= −8.854 × 10^{−9} C

= −8.854 nC

Therefore, the value of the point charge is −8.854 nC.

#### Page No 48:

#### 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^{3 }N/C and points radially inward, what is the net charge on the sphere?

#### Answer:

Electric field intensity (*E*) at a distance (*d*) from the centre of a sphere containing net charge *q* is given by the relation,

Where,

*q* = Net charge = 1.5 × 10^{3} N/C

*d *= Distance from the centre = 20 cm = 0.2 m

∈_{0} = Permittivity of free space

And, = 9 × 10^{9} N m^{2} C^{−2}

∴

= 6.67 × 10^{−}^{9} C

= 6.67 nC

Therefore, the net charge on the sphere is 6.67 nC.

#### Page No 48:

#### Question 1.22:

A
uniformly charged conducting sphere of 2.4 m diameter has a surface
charge density of 80.0 μC/m^{2}.
(a) Find the charge on the sphere. (b) What is the total electric
flux leaving the surface of the sphere?

#### Answer:

**(a) **Diameter of
the sphere, *d* = 2.4 m

Radius
of the sphere, *r* = 1.2 m

Surface
charge density,
=
80.0 μC/m^{2} = 80 ×
10^{−6} C/m^{2}

Total charge on the surface of the sphere,

*Q*
= Charge density × Surface area

=

= 80 × 10^{−6} × 4 × 3.14 ×
(1.2)^{2}

= 1.447 × 10^{−3} C

Therefore,
the charge on the sphere is 1.447 × 10^{−3} C.

**(b) **Total electric flux ()
leaving out the surface of a sphere containing net charge *Q *is
given by the relation,

Where,

∈_{0}
= Permittivity of free space

= 8.854 × 10^{−12} N^{−1}C^{2
}m^{−2}

*Q
*= 1.447 × 10^{−3} C

= 1.63 × 10^{8} N C^{−1} m^{2}

Therefore,
the total electric flux leaving the surface of the sphere is 1.63 ×
10^{8} N C^{−1} m^{2}.

#### Page No 48:

#### Question 1.23:

An
infinite line charge produces a field of 9 × 10^{4}
N/C at a distance of 2 cm. Calculate the linear charge density.

#### Answer:

Electric field produced
by the infinite line charges at a distance *d* having linear
charge density *λ* is
given by the relation,

Where,

*d* = 2 cm = 0.02
m

*E* = 9 ×
10^{4} N/C

∈_{0} =
Permittivity of free space

= 9 × 10^{9} N m^{2} C^{−2}

= 10 μC/m

Therefore, the linear charge density is 10 μC/m.

#### Page No 48:

#### 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^{−22
}C/m^{2}.
What is **E**:
(a) in the outer region of the first plate, (b) in the outer region
of the second plate, and (c) between the plates?

#### Answer:

The situation is represented in the following figure.

A and B are two
parallel plates close to each other. Outer region of plate A is
labelled as **I**, outer region of plate B is labelled as **III**,
and the region between the plates, A and B, is labelled as **II.**

Charge density of plate
A, *σ* = 17.0 ×
10^{−22} C/m^{2}

Charge density of plate
B, *σ* = −17.0 ×
10^{−22} C/m^{2}

In the regions, **I**
and **III**, electric field *E* is zero. This is because
charge is not enclosed by the respective plates.

Electric field *E*
in region **II** is given by the relation,

Where,

∈_{0} =
Permittivity of free space = 8.854 × 10^{−12}
N^{−1}C^{2 }m^{−2}

∴

= 1.92 ×
10^{−10} N/C

Therefore, electric
field between the plates is 1.92 × 10^{−10} N/C.

#### Page No 48:

#### Question 1.25:

An
oil drop of 12 excess electrons is held stationary under a constant
electric field of 2.55 × 10^{4}
N C^{−1}
in Millikan’s oil drop experiment. The density of the oil is
1.26 g cm^{−3}.
Estimate the radius of the drop. (*g *=
9.81 m s^{−2};
*e *=
1.60 × 10^{−19}
C).

#### Answer:

Excess electrons on an
oil drop, *n* = 12

Electric field
intensity, *E* = 2.55 × 10^{4} N C^{−1}

Density of oil, *ρ*
= 1.26 gm/cm^{3} = 1.26 × 10^{3} kg/m^{3}

Acceleration due to
gravity, g = 9.81 m s^{−2}

Charge on an electron,
*e* = 1.6 × 10^{−19} C

Radius of the oil drop
= *r*

Force (*F*) due to
electric field *E* is equal to the weight of the oil drop (*W*)

*F = W*

*Eq *= *m*g

*Ene*

Where,

*q* = Net charge
on the oil drop = *ne*

*m *= Mass of the
oil drop

= Volume of the oil drop × Density of oil

= 9.82 ×
10^{−4} mm

Therefore, the radius
of the oil drop is 9.82 × 10^{−4} mm.

#### Page No 48:

#### Question 1.26:

Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

**(a)**

** **

**(b)**

**(c)**

**(d)**

**(e)**

#### Answer:

**(a) **The field lines showed in (a) do not represent
electrostatic field lines because field lines must be normal to the
surface of the conductor.

**(b) **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.

**(c) **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.

**(d) **The field lines showed in (d) do not represent
electrostatic field lines because the field lines should not
intersect each other.

**(e) **The field lines showed in (e) do not represent
electrostatic field lines because closed loops are not formed in the
area between the field lines.

#### Page No 49:

#### Question 1.27:

In
a certain region of space, electric field is along the z-direction
throughout. The magnitude of electric field is, however, not constant
but increases uniformly along the positive *z*-direction,
at the rate of 10^{5}
NC^{−1}
per metre. What are the force and torque experienced by a system
having a total dipole moment equal to 10^{−7}
Cm in the negative *z*-direction?

#### Answer:

Dipole moment of the
system, *p* = *q × dl *= −10^{−7}
C m

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-direction i.e., opposite to the direction of
electric field. Hence, the angle between electric field and dipole
moment is 180°.

Torque (*τ*)
is given by the relation,

τ = *pE*
sin180°

= 0

Therefore, the torque experienced by the system is zero.

#### Page No 49:

#### Question 1.28:

(a)
A conductor A with a cavity as shown in Fig. 1.36(a) is given a
charge *Q*.
Show that the entire charge must appear on the outer surface of the
conductor. (b) Another conductor B with charge *q
*is inserted into the cavity keeping
B insulated from A. Show that the total charge on the outside surface
of A is *Q *+
*q *[Fig.
1.36(b)]. (c) A sensitive instrument is to be shielded from the
strong electrostatic fields in its environment. Suggest a possible
way.

#### Answer:

**(a) **Let us consider a Gaussian surface that is lying wholly
within a conductor and enclosing the cavity. The electric field
intensity *E* inside the charged conductor is zero.

Let
*q* is the charge inside the conductor and
is
the permittivity of free space.

According to Gauss’s law,

Flux,

Here,
*E *= 0

Therefore, charge inside the conductor is zero.

The
entire charge *Q* appears on the outer surface of the conductor.

**(b) **The outer surface of conductor A has a charge of amount *Q*.
Another conductor B having 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 +*q*
is induced on the outer surface of conductor A. Therefore, total
charge on the outer surface of conductor A is* Q *+ *q*.

**(c) **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.

#### Page No 49:

#### Question 1.29:

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is , where is the unit vector in the outward normal direction, and is the surface charge density near the hole.

#### Answer:

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.

Let *E* is the
electric field just outside the conductor, *q* is the electric
charge,
is
the charge density, and
is
the permittivity of free space.

Charge

According to Gauss’s law,

Therefore, the electric field just outside the conductor is. This field is a superposition of field due to the cavity and the field due to the rest of the charged conductor. These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.

Therefore, the field due to the rest of the conductor is.

Hence, proved.

#### Page No 49:

#### 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. [*Hint: *Use Coulomb’s law directly and evaluate the 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 *l* from the mid-point O of the wire, as shown in the following figure.

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 *dE*_{1}.

On differentiating equation (2), we obtain

$\frac{dx}{d\theta}=lse{c}^{2}\theta \phantom{\rule{0ex}{0ex}}dx=lse{c}^{2}\theta d\theta $

From equation (2),

Putting equations (3) and (4) in equation (1), we obtain

The wire is so long that tends from to .

By integrating equation (5), we obtain the value of field *E*_{1} as,

Therefore, the electric field due to long wire is.

#### Page No 49:

#### 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 (denoted by
u) of charge (+2/3) *e*,
and the ‘down’ quark (denoted by d) of charge (−1/3)
*e*,
together with electrons build up ordinary matter. (Quarks of other
types have also been found which give rise to different unusual
varieties of matter.) Suggest a possible quark composition of a
proton and neutron.

#### Answer:

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 down quarks =

Total charge on a neutron = 0

Number of up quarks in
a neutron, *n* = 1

Number of down quarks
in a neutron = 3 − *n* = 2

Therefore, a neutron can be represented as ‘udd’.

#### Page No 50:

#### Question 1.32:

**(a)**
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.

**(b)** Verify
this result for the simple configuration of two charges of the same
magnitude and sign placed a certain distance apart.

#### Answer:

**(a) **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 restoring force towards 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. According to Gauss’s law, the flux of electric
field through a surface, which is not enclosing any charge, is zero.
Hence, the equilibrium of the test charge can be stable.

**(b) **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 to the
joining line, 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.

#### Page No 50:

#### Question 1.33:

A
particle of mass *m *and
charge (−*q*)
enters the region between the two charged plates initially moving
along *x*-axis
with speed *vx *(like
particle 1 in Fig. 1.33). The length of plate is *L
*and an uniform electric field *E
*is maintained between the plates.
Show that the vertical deflection of the particle at the far edge of
the plate is *qEL*^{2}/
(2*m*).

*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 *m* = − *q*

Velocity of the
particle = *v*_{x}

Length of the plates =*
L*

Magnitude of the
uniform electric field between the plates = *E*

Mechanical force, *F*
= Mass (*m*) ×
Acceleration (*a*)

Therefore, acceleration,

Time taken by the
particle to cross the field of length *L *is given by,

*t*

In the vertical
direction, initial velocity, *u* = 0

According to the third
equation of motion, vertical deflection *s *of the particle can
be obtained as,

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.

#### Page No 50:

#### Question 1.34:

Suppose
that the particle in Exercise in 1.33 is an electron projected with
velocity *v*_{x}= 2.0 × 10^{6}
m s^{−1}.
If *E *between
the plates separated by 0.5 cm is 9.1 × 10^{2}
N/C, where will the electron strike the upper plate? (| *e
*| =1.6 × 10^{−19}
C, *m*_{e
}= 9.1 × 10^{−31}
kg.)

#### Answer:

Velocity of the
particle, *v*_{x} = 2.0 ×
10^{6} m/s

Separation of the two
plates, *d* = 0.5 cm = 0.005 m

Electric field between
the two plates, *E* = 9.1 ×
10^{2} N/C

Charge on an electron,
*q* = 1.6 × 10^{−19}
C

Mass of an electron, *m*_{e
}= 9.1 × 10^{−31}
kg

Let the electron strike
the upper plate at the end of plate *L*, when deflection is *s*.

Therefore,

Therefore, the electron will strike the upper plate after travelling 1.6 cm.

View NCERT Solutions for all chapters of Class 12