Class 12 Physics Chapter 1: Electric Charges and Field - NCERT Solutions

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

Charge on the first sphere, q₁ = 2 × 10 ^{- 7} C

Charge on the second sphere, q₂ = 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.

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

Charge on this sphere, q₁ = 0.4 μC = 0.4 × 10⁻⁶ C

Charge on the second sphere, q₂ = − 0.8 μC = − 0.8 × 10⁻⁶ C

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.

(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.

The given ratio is

Where,

G = Gravitational constant

Its unit is N m² 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 m² C⁻².

Hence, the given ratio is dimensionless.

e = 1.6 × 10⁻¹⁹ C

G = 6.67 × 10⁻¹¹ N m² kg^-2

m_e= 9.1 × 10⁻³¹ kg

m_p = 1.66 × 10⁻²⁷ kg

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

(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.

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.

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.

 (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.

(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,

E₁ = along OB

Where,

= Permittivity of free space

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

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.

 

(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 × 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 × 10⁻³ N along OA.

According to the figure given below both the charges can be located in a coordinate frame of reference.

At A, amount of charge, q_A = 2.5 × 10⁻⁷C (Given)

At B, amount of charge, q_B = −2.5 × 10⁻⁷ C (Given)

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.

Electric dipole moment, p = 4 × 10⁻⁹ C m (Given)

Angle made by p with a uniform electric field, θ = 30° (Given)

Electric field, E = 5 × 10⁴ N C⁻¹ (Given)

Torque acting on the dipole is given by the relation,

τ = pE sinθ

= 4 × 10^-9 × 5 × 10⁴ × sin 30

= 20 × 10^-5 ×

 

= 10^-4 Nm

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

 

 

 

(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 × 10¹²

Therefore, the number of electrons transferred from wool to polythene is 1.87 × 10¹².

(b) Yes. There is a transfer of mass occur. This is because an electron has mass,

as we know

m_e = 9.1 × 10⁻³ kg

Total mass transferred from wool to polythene,

m = m_e × 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.

(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

=9x10⁹N m² C²

∴

=1.52X10^-2N Therefore, the force between the two spheres is 1.52 × 10⁻² N.

(b) As given in the question

After doubling the charge of sphere,

charge on sphere A would be,

q_A = 2 × 6.5 × 10⁻⁷ C

and charge on sphere B would be,

q_B = 1.3 × 10⁻⁶ C

The distance between the spheres is halved as given:

∴

Now,force of repulsion between the two spheres,

= 16 × 1.52 × 10⁻²

= 0.243 N

Therefore, the force of repulsion between the two spheres is 0.243 N.

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, q/2 amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is q/2.

When sphere C with charge q/2 is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as,

1/2 (q + q/2) = 3q/4

Hence, charge on each of the spheres, C and B, is 3q/4.

Force of repulsion between sphere A having charge q/2 and sphere B having 3q/4 charge is

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

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.

(a) Electric field intensity, = 3 × 10³ î N/C

Magnitude of electric field intensity, = 3 × 10³ N/C

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

Area of the square, A = s² = 0.01 m²

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 × 10³ × 0.01 × cos0°

= 30 N m²/C

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

Flux, Φ =

= 3 × 10³ × 0.01 × cos60°

= 15 N m²/C

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.

(a) Net outward flux through the surface of the box, Φ = 8.0 × 10³ N m²/C

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

∈0 = Permittivity of free space

= 8.854 × 10⁻¹² N⁻¹C² m⁻²

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.

(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.

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⁻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 × 10⁵ N m² C⁻¹.

Net electric flux (Φ_Net) 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 × 10⁻⁶ C

∴

= 2.26 × 10⁵ N m² C⁻¹

The net electric flux through the surface is 2.26 ×10⁵ N m²C⁻¹.

(a) Electric flux, Φ = −1.0 × 10³ N m²/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., −10³ N m²/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⁻¹C² m⁻²

∴ 

= −1.0 × 10³ × 8.854 × 10⁻¹²

= −8.854 × 10⁻⁹ C

= −8.854 nC

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

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 × 10³ N/C

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

∈0 = Permittivity of free space

And,

= 9 × 10⁹ N m² C⁻²

∴

= 6.67 × 10⁹ C

= 6.67 nC

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

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

Radius of the sphere, r = 1.2 m

Surface charge density, = 80.0 μC/m2 = 80 × 10−6 C/m2

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

∈₀  = 8.854 × 10⁻¹² N⁻¹C² m⁻²

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 × 10⁸ N C⁻¹ m².

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 × 10⁴ N/C

∈₀ = Permittivity of free space

= 9 × 10⁹ N m² C⁻²

= 10 μC/m

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

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/m²

Charge density of plate B, σ = −17.0 × 10⁻²² C/m²

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⁻¹C² m⁻²

∴

= 1.92 × 10⁻¹⁰ N/C

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

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 10⁴ N C ^{- 1}

Density of oil, ÃÂ = 1.26 gm/cm³ = 1.26 × 10³ kg/m³

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 = mg

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.

(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.

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 (Ä) is given by the relation,

Ä = pE sin180°

= 0

Therefore, the torque experienced by the system is zero.

(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 \(\epsilon_{0}\) is the permittivity of free space.

 

According to Gauss’s law,

 

Flux, \(\phi = \overrightarrow{E}.\overrightarrow{ds} = \frac{q}{\epsilon_{0}}\)

 

Here, E = 0

 

\(\frac{q}{\epsilon_{0}} = 0\)

 

\(\because\epsilon_{0} \neq 0\)

 

\(\therefore q=0\)

 

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.

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, σ is the charge density and ε 0 is the permittivity of free space.

Charge q = σ × ds

According to Gauss’s law, flux, ∅ = E. ds = q/ε₀

⇒ E. ds = σ × ds / ε₀ 

∴ E= σ/ 2ε₀ n̂

Therefore, the electric field just outside the conductor is σ/ 2ε₀ n̂. This field is a superposition of field due to the cavity E` and the field due to the rest of the charged conductor E`. These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor.

∴  E`+ E` = E

⇒ E` = E/2  = σ/2ε₀ n̂

Hence, the field due to the rest of the conductor is σ/2ε₀ n̂.

Take a long thin wire XY (as shown in the following figure) of uniform linear charge density λ.

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.

∴ q = λdx

Electric field due to the piece,

However,

The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component. When the whole wire is considered, the component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point A.

Hence, effective electric field at point A due to the element dx is dE₁ .

∴ d E₁ = 1/4πε₀ x λdx. cos θ/(l² + x²)                  ...(1)

In ∆AZO, tanθ = x/l ⇒ x = l.tanθ                          ...(2)

On differentiating equation (2), we obtain

dx/dθ = l x sec ² θ ⇒ dx = l x sec ² θdθ              ...(3)

From equation (2), we have

x² + l² = l²tan²θ + l² = l² (tan² θ + 1) = l² sec² θ          ...(4)

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

The wire is so long that θ tends from − π/2 to π/2.

By integrating equation (5), we obtain the value of field E₁ as,

 

Therefore, the electric field due to long wire is λ/2πε₀ l .

A proton has three quarks. Let there be n up quarks in a proton, each having a charge of (+2/3) e.

Charge due to n up quarks = (2/3 e) n

Number of down quarks in a proton = 3 − n 

Each down quark has a charge of - 1/3e 

Charge due to (3 − n) down quarks = (-1/3 e) (3-n)

Total charge on a proton = + e

Therefore, e = (2/3 e) n + (- 1/3 e) (3-n)

⇒ e = 2ne/3 - e + ne/3

⇒ 2e = ne ⇒ n = 2

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.

Charge on a neutron due to n up quarks = (+ 2/3 e) n

Number of down quarks is 3 − n, and each having a charge of - 1/3 e.

Charge on a neutron due to (3 – n) down quarks = (- 1/3 e) (3-n)

Total charge on a neutron = 0

Therefore, 0 = (2/3 e) n + (-1/3 e) (3-n)

⇒ 0 = 2ne/3 - e + ne/3

⇒ e = ne ⇒ n = 1

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’.

(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.

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 10⁴ N C⁻¹

Density of oil, ρ = 1.26 gm/cm³ = 1.26 × 10³ kg/m³

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 = mg

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.

(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.

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.