Electric Charges And Fields Class 12th Physics Part I CBSE Solution

Class 12^th Physics Part I CBSE Solution

Exercises

1. What is the force between two small charged spheres having charges of 2 × 10-7 C and 3 ×…
2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of…
3. Check that the ratio is dimensionless. Look up a Table of Physical Constants and determine…
4. Explain the meaning of the statement ‘electric charge of a body is quantised’.…
5. Why can one ignore quantisation of electric charge when dealing with macroscopic i.e.…
6. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon…
7. Four-point charges qA = 2 μC, qB = -5 μC, qC = 2 μC, and qD = -5 μC are located at the…
8. An electrostatic field line is a continuous curve. That is, a field line cannot have…
9. Explain why two field lines never cross each other at any point?
10. Two point charges qA = 3 μC and qB = -3 μC are located 20 cm apart in vacuum. (a) What is…
11. A system has two charges qA = 2.5 × 10-7C and qB = -2.5 × 10-7 C located at points A: (0,…
12. An electric dipole with dipole moment 4 × 10-9 C m is aligned at 30° with the direction of…
13. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10-7 C. (a)…
14. Two insulated charged copper spheres A and B have their centres separated by a distance of…
15. What is the force of repulsion if each sphere is charged double the above amount, and the…
16. Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the…
17. Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give…
18. Consider a uniform electric field E = 3 × 10^3 î N/C. (a) What is the flux of this field…
19. What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side…
20. Careful measurement of the electric field at the surface of a black box indicates that the…
21. A point charge + 10 μC is a distance 5 cm directly above the centre of a square of side 10…
22. A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What…
23. A point charge causes an electric flux of -1.0 × 10^3 Nm^2 /C to pass through a spherical…
24. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm…
25. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of…
26. An infinite line charge produces a field of 9 × 10^4 N/C at a distance of 2 cm. Calculate…
27. Two large, thin metal plates are parallel and close to each other. On their inner faces,…

Additional Exercises

1. An oil drop of 12 excess electrons is held stationary under a constant electric field of…
2. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field…
3. In a certain region of space, electric field is along the z-direction throughout. The…
4. A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that…
5. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric…
6. Obtain the formula for the electric field due to a long thin wire of uniform linear charge…
7. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter)…
8. Consider an arbitrary electrostatic field configuration. A small test charge is placed…
9. A particle of mass m and charge (-q) enters the region between the two charged plates…
10. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx=…

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Exercises

Question 1.

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

Answer:

The coulomb attraction formula used to find the force, F is given as,

 …(1)

Where, q₁ and q₂ are the charges.

r is the distance between the charges.

 is a constant and its value is 9x10⁹ N m² C^-2

ϵ₀ is the permittivity of free space. Its value is 8.85 x 10^-12 (F/m)

Now, the diagram is:

Since, both the charges are positive, thus, the nature of force will be repulsive. F₁₂ is the force on charge q2 caused by charge q1 andF₂₁ is the force on charge 1 due to charge 2 _.

Now, Given:

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(in m) =30/100=0.3 m

Putting the values in equation (1), we get,

F = 6 × 10^-3 N

Hence, the force between the given charged particles will be 6 × 10^-3 N. Since the nature of the charges is the same i.e. they are both positive. Hence, the force will be repulsive.

Question 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) The electrostatic force, F = 0.2 N

Charge on this present sphere, q₁ = 0.4 μC = 0.4 × 10^-6 C

Charge on the other sphere, q₂ = –0.8 μC = -0.8 × 10^-6 C

The electrostatic force between the spheres can be given by the equation,

 …(1)

Where, q₁ and q₂ are the charges.

r is the distance between the charges.

 is a constant and its value is 9x10⁹ N m² C^-2

ϵ₀ is the permittivity of free space. Its value is 8.85 x 10^-12 (F/m)

Therefore, from equation (1),

⇒ r² = 144 x 10^-4 m²

Taking square root both the sides,

⇒ r = 

r = 0.12 m

Hence, the distance between the two spheres is 0.12 m.

(b) Since the nature of the charges on the spheres is opposite, the force between them will be attractive in nature. The spheres will attract each other with the same amount of force.

Therefore, the force on the second sphere due to the first sphere will be 0.2 N.

Question 3.

Check that the ratio  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 

It is the ratio of electric force i.e.to gravitational force i.e.

 between a proton and an electron with the distance between them as constant.

This is because

Here, G = Gravitational constant. (Unit = Nm²kg^-2 )

m_e and m_p are the mass of electron and proton respectively. (Unit = Kg)

e = Electric charge (Unit = Coulomb)

K is a constant = 

 is the permittivity of space. Its units are Nm² C^-2.

Let us substitute the units into the ratio in order to deduce the dimensions,

All the units get cancelled, and we see that the given ratio is dimensionless i.e. M⁰ L⁰ T⁰

Let us now calculate the value of the given ratio,

G = 6.67 × 10¹¹ Nm² Kg^-2

e = 1.6 × 10¹⁹ C

m_e = 9.1 × 10^-31 kg

m_p = 1.66 × 10^-27 kg

On substituting these values into the ratio, we get

Note:

This implies that the electric force between a proton and an electron is 10³⁹ times greater than the gravitational force.

Question 4.

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

Answer:

Electric charge of a body is quantised.

This statement means that charge on a body can take only integral values i.e. (1, 2, 3, 4, …. N) number of electrons can be transferred from one body to another. The charges cannot be transferred in fractions. It can only be an integral multiple of the charge of one electron.

Therefore, as a result, a body will possess a charge that is an integral multiple of the electric charge of an electron.

Question 5.

Why can one ignore quantisation of electric charge when dealing with macroscopic i.e. large-scale charges?

Answer:

When considered on a macroscopic level or a large scale, we tend to ignore the quantisation of electric charges. This is because on a large scale, the charges that we deal with are extremely huge when compared to the magnitude of an individual charge. Hence, we can say that the quantisation of electric charge is not useful on a macroscopic scale. It is therefore ignored and considered to be of continuous nature.

Question 6.

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:

i. When two bodies are rubbed against each other, this process results in the production of charges on the bodies involved.

ii. The charges on each body will be of equal magnitude but of opposite nature. This happens because charges are always created in pairs.

This phenomenon of charging bodies by rubbing them against each other is termed as charging by friction. However, the net charge of the system – of both the bodies – is zero.

iii. This is because equal amounts of charge of opposite nature cancel each other out.

iv. When a glass rod is rubbed with a silk cloth, then opposite charges appear on both the bodies. This is in accordance with the law of conservation of energy. (Net charge of isolated system remains constant) This phenomenon is also observed when other pairs of bodies are also rubbed with each other.

Question 7.

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:

Consider a square of side 10 cm. Four charges are placed in its corners. Let O be the center of the square.

The diagram is:

AB = BC = CD = DA = 10 cm

Diagonal of a square = √2 × Side of a square

Therefore,

AC = BD = 10√2 cm

AO = OC = DO = OB = (10√2)/2 cm = 5√2cm

i. A charge of magnitude 1μC is placed at O, the center of the square.
ii. The force experienced at center due to side charge at B is same in magnitude but in a direction opposite to the direction to the force at center due to charge at D.
iii. The force experienced at center due to side charge at A is same in magnitude but in a direction opposite to the direction to the force at center due to charge at C .



Force on the 1 u C in the center is = F_a + F_b + F_c + F_{d =} F_a + (- F_a) + F_c + ( - F_d) = 0

Therefore, we can conclude that the resultant force by the four charges kept in the corners on the charge that is kept in the center will be zero.

Question 8.

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

Answer:

An electric field line always points in the direction that a positive test charge would accelerate in if placed upon that line. A charge will experience a continuous force when placed in an electrostatic field. Therefore, an electrostatic field line will be a continuous curve. The field lines do not have any sudden breaks as the test charge or any charge will move continuously and does not jump from any point to another.

Question 9.

Explain why two field lines never cross each other at any point?

Answer:

An electric field line represents the direction in which a positive test charge would accelerate in if placed on the line. If two field lines intersect, then this will imply that the electric field intensity points to two directions at the same point. This is not possible.

Hence, two electric field lines will never cross each other.

Question 10.

Two point charges q_A = 3 μ_C and q_B = –3 μ_C 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 diagram is shown as below:

AB = 20 cm

∵ O is the midpoint of the line AB.

∴ AO = OB = 10 cm = 0.1m

Let the Net electric field at the point O = E

The electric field at a point caused by charge q, is given as,

 …(1)

Where, q is the charge,

r is the distance between the charge and the point at which the field is being calculated

 is a constant and its value is 9x10⁹ N m² C^-2

ϵ₀ is the permittivity of free space. Its value is 8.85 x 10^-12 (F/m)

∴ The electric field at the point O caused by q_A = 3 μC

The direction of E_A will be along OB.

Also, the electric field at O caused by q_B = –3 μC

The direction of E_B will be along OB.

Note: Both the field are acting in the same direction along OB, this is because the charges are of opposite nature, ∴ the force will be attractive.

Hence, we can sum them up to find the resultant electric field.

E_net = E_A + E_B

⇒ E_net = 2E_A

∴ E_net = 5.4 × 10² N/C(along OB)

(b) The diagram is:

q = 1.5 × 10^–9 C

The force that is experienced by q when placed at O, F.

F_O = q × E_Net

Where E is the Electric field at the point O.

∴ F = 1.5 × 10^–9 C x 5.4 × 10² N/C

⇒ F = 8.1 x 10^-7 N

The negative test charge will be repelled by the force placed at B and attracted by the force placed at A. Therefore, this force will be in the direction of OA.

Question 11.

A system has two charges q_A = 2.5 × 10^–7C 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: 

The charge at point A, q_A = 2.5 × 10^–7C

The charge at point B, q_B = –2.5 × 10^–7 C

∴ the total charge of the system, Q_net = q_{A +} q_B

Q_net = 2.5 × 10^–7C + (–2.5 × 10^–7 C)

Q_net = 0

The distance between these two charges = Distance between point A and B, d = 15 + 15

d = 30 cm = 0.3 m

The electric dipole moment can be defined as the measure of the separation of positive as well as negative charges within a specific system. It tells us about the given system’s overall polarity.

The electric dipole moment of the system, p = q_A x d = q_b x d

p = 2.5 × 10^–7C x 0.3

∴ p = 7.5 x 10^-8 C-m along the positive z-axis

Hence, the electric dipole moment of this system is 7.5 x 10^-8 C-m along the positive z axis.

Question 12.

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 × 10⁴ NC^–1. Calculate the magnitude of the torque acting on the dipole.

Answer:

The electric dipole moment can be defined as the measure of the separation of positive as well as negative charges within a specific system. It tells us about the given system’s overall polarity.

Electric dipole moment, p = 4 × 10^–9 C-m

Angle between p and uniform electric field,  = 30

Electric field, E = 5 × 10⁴ NC^–1

The torque acting on a dipole is given by:

τ = p × E

τ = p×E×sinθ(θ=30)

= 4 × 10^–9 C m x 5 × 10⁴ NC^–1 x (0.5)

= 10^-4 N m

The magnitude of torque acting on the given dipole is 10^-4 N m.

Question 13.

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:

Given:

Charge on polythene piece = -3 × 10⁷ C

(a) Number of electrons transferred from wool to plastic.

Explanation: We know that all materials are electrically neutral. As per the question, after rubbing the polythene piece is found to be negatively charged which suggests that negatively charged particles i.e. electrons were transferred from wool to plastic.

Let number of electrons transferred from wool to plastic be ‘n’

From quantization of charges we know that

q = n × e …(1)

Where, q is the total charge

e = charge of individual particle

n = number of charges (either positive or negative)

e = -1.6 × 10^-19 C

From equation (1) we get,

q = n × e

⇒ n = q/e

⇒ 

⇒ n = 1.87 × 10¹²

The number of electrons transferred is 1.8 × 10¹²

(b) Yes, mass is transferred because each electron has a mass of 9.1 × 10^-31 Kg.

Mass transferred = n × m_e …(1)

Where, n = number of electrons

M_e = mass of electron (9.1 × 10^-31Kg)

Now, putting the values of M_e and n in equation (1).

⇒ M = 1.87 × 10¹² × 9.1 × 10^-31Kg

⇒ M = 1.706 × 10^-18 Kg

Hence, the mass transferred from wool to plastic is 1.7 × 10^-18 Kg

Question 14.

Two insulated charged copper spheres A and B have their centres 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.

Answer:

Given:

Distance between centres of spheres, r = 50 cm = .5m

Charge on each sphere, q = 6.5 × 10^-7 C

Mutual force of electrostatic repulsion

…(1)

Where, F= mutual force of attraction

q₁ = charge on sphere 1

q₂ = charge on sphere 2

r = distance between centres

Where, ε₀ is the permittivity of the free space.

Now, putting the values of q₁, q₂ and r in equation (1).

⇒ F = 1.52 × 10^-2 N

Hence the mutual force between two spheres is 1.52 × 10^-2N. Since the sign is positive so the force is repulsive in nature.

Question 15.

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

Answer:

Force if charges on each sphere is doubled and the distance is halved.

Given:

Q₁ = 2 × q

Q₂ = 2 × q

R = 0.5 × r

Using Coulomb’s law, we get,

…(1)

Where, ε₀ is the permittivity of the free space.

⇒ 

⇒ 

⇒ F = 0.243 N

The mutual force is 0.243N and it is repulsive in nature.

The new force is found to be 16 times the force in case I.

Question 16.

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:

Note: If a conducting uncharged material is brought in contact with a charged surface then the charges are shared uniformly between the two bodies.

Given: Charge on sphere 1, q₁ = 6.5 × 10^-7 C

Charge on sphere 2, q₂ = 6.5 × 10^-7 C

Charge on sphere 3, q₃ = 0

Step 1: The uncharged sphere is brought in contact with sphere 1. Since sphere 1 has charge ‘q’, it gets distributed among sphere 1 and sphere 3.

Now, charge on sphere 1 = q/2

Charge on sphere 2 = q/2

At this point the sphere 3 which was initially uncharged has a charge “q/2”.

Step 2: Now sphere 3 is brought in contact with sphere 2 due to which 1/4 × q will flow from sphere 2 to sphere 3. Now sphere 2 and sphere 3 have “3/4 × q” charge.

Now, q₁ = 1/2 × 6.5 × 10^-7 C

⇒ 3.25 × 10^-7 C

q₂ = 3/4 × 6.5 × 10^-7 C

⇒ 4.87 × 10^-7 C

q₃ = 3/4 × 6.5 × 10^-7 C

⇒ 4.87 × 10^-7 C

We know that,

…(1)

Where, F= mutual force of attraction

q₁ = charge on sphere 1

q₂ = charge on sphere 2

r = distance between centres

Where, ε₀ is the permittivity of the free space.

Plugging the values of q₁, q₂ and r in equation (1), we get

⇒ 

⇒ F = 5.703 × 10^-3 N

The repulsive force between sphere 1 and sphere 2 is 5.703 × 10^-3 N.

Question 17.

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:

1. Particles 1 and 2 are deflected towards the positively charged plate, hence it can be concluded that particles 1 and 2 are negatively charged.

2. Particle 3 is deflected towards the negatively charged plate hence it is positively charged.

The deflection of a particle is proportional to the charge to mass ratio (q: m) of the particle. We observe that particle 3 has maximum deflection, so it has the highest charge to mass ratio.

Explanation: Force on a particle in electric field

F = q × E …(1)

Where, q = charge on particle

E = magnitude of electric field

m × a = q × E …(2)

Where m = mass of particle

a = acceleration of particle

From equation (2)

⇒ 

The electric field in the given case is constant, hence we can write

⇒  …(3)

From the equation (3) we can conclude that acceleration of a charge particle with charge q and mass m in uniform electric field is proportional to the charge to mass ratio of the particle.

Question 18.

Consider a uniform electric field E = 3 × 10³î N/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:

Given:

Electric field E = 3 × 10³ N/C

Side of square, s = 10 cm

a) Flux of field through square whose plane is parallel to yz plane.

We understand that the normal to the plane is parallel to the direction of field.

So, θ = 0°

Φ = E . A

Φ = E × A × cos(θ) …(1)

Where, E = Electric field

A = Area through which we have to calculate flux

θ = Angle between normal to surface and the Electric field

A = s²

A = .01 m²

Plugging values, of E, A and θ in equation (1)

Φ = 3 × 10³ NC^-1 × 0.01m² × cos0°

Φ = 30 Nm²C^-1

b) If normal to its (square’s) plane makes 60° with the X axis.

θ = 60°

Φ = E × A × cos(θ)

Φ = 3 × 10³ NC^-1 × 0.01m² × cos60°

Φ = 15 Nm²C^-1

Question 19.

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:

Flux, Φ = q/ε₀

Where, q = net charged enclosed

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

Since no charge is enclosed by the cube, i.e. q = 0, the net flux through the Cube is zero.

Note: The number of field lines entering the cube is equal to the number of field lines leaving the cube. Hence the net flux becomes zero.

Question 20.

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 × 10³ Nm²/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:

Given:

(a) Φ = 8.0 × 10³ Nm²C^-1

Let net charge inside the box = q

We know that,

Flux, Φ = q/ε₀ ..(1)

Where, q = net charged enclosed

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

Plugging values of Φ and ε₀ in equation (1) we get,

q = Φ × ε₀

⇒ q = 8.0 × 10³ Nm²C^-1 × 8.85 × 10^-12N^-1C²m^-2

⇒ q = 7.08 × 10^-8 C

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

b) No, we cannot conclude that the body doesn’t have any charge. The flux is due to the Net charge of the body. There may still be equal amount of positive and negative charges. So, it is not necessary that if flux is zero then there will be no charges.

Question 21.

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:

Given:

q = + 10 

s = 10 cm

Assume the charge to be enclosed by a cube, where the square is one of its sides.

Now, let us find the total flux through the imaginary cube.

We know that,

Flux, Φ = q/ε …(1)

Where, q = net charged enclosed

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

Now plugging the values of q and ε₀ in equation (1)

⇒ Φ = 11.28 × 10⁵ Nm^-2C^-1

We understand that flux through all the faces of cube will be equal;

Let flux through the square = Φ_a

Hence,

Φ_a = Φ/6

Explanation: The net flux will be distributed equally among all 6 faces of the cube. Hence, the square will have one sixth of the total flux.

Φ_a = 1.88 Nm^-2C^-1

The flux through the square is.

Question 22.

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:

Given:

Total charge inside the cube, q = 2.0 μC

Edge length of Cube, a = 9.0 cm

We know that,

Flux, Φ = q/ε₀

Where, q = net charged enclosed

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

⇒ Φ = 2.26 × 10⁵ Nm²C^-1

The net electric flux through the cubic Gaussian surface is Φ = 2.26 × 10⁵ Nm²C^-1

NOTE: The flux through a Gaussian surface depends on the net charge enclosed by it. Geometry has no effect on the total flux.

Question 23.

A point charge causes an electric flux of –1.0 × 10³ Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred 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:

Given:

Φ = -1.0 × 10³ Nm²C^-1

r₁ = 10.0 cm.

a) Flux if the radius of the Gaussian surface is doubled.

If the radius is doubled then the flux would remain same i.e. -1.0 × 10³ Nm²C^-1_.

The geometry of the Gaussian surface doesn’t affect the total flux through it. The net charge enclosed by Gaussian surface determines the net flux.

b) Let value of point charge enclosed by Gaussian surface = q

We know that,

Flux, Φ = q/ε₀ …(1)

Where, q = net charged enclosed

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

Now plugging, the values Φ and ε₀ in equation (1).

q = Φ × ε₀

⇒ q = -1.0 × 10³Nm²C^-1 × 8.85 × 10^-12N^-1m^-2C²

⇒ q = -8.85 × 10^-9 C

The charge enclosed by the surface is -8.85 × 10^-9 C.

Question 24.

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 × 10³ N/C and points radially inward, what is the net charge on the sphere?

Answer:

Given:

Radius of charged sphere, r = 10 cm

Electric field, 20 cm away from centre of sphere, E = 1.5 × 10³ NC^-1

We know that, electric field intensity at a point P, located at a distance R, due to net charge q is given by,

 …(1)

Now plugging the values of q and R in equation (1)

q = 4 × π × ε₀ × R² × E

⇒ q = 4 × 3.14 × 8.85 × 10^-12N^-1m^-2C^-2 × (0.2m)² × 1.5 × 10³C

⇒ q = 6.67 × 10^-9C

The net charge on the sphere is -6.67 × 10^-9C. Since the electric field points radially inwards, we can infer that charges on sphere are negative.

Question 25.

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:

Given: radius of sphere, r = 1.2 m

Surface charge density, σ = 80.0 μC/m²

a) Let charge on sphere = q

We understand that,

Total charge, Q = Surface charge density × surface area.

Surface area of sphere, S = 4πr²

S = 18.08 m²

Q = 80 × 10^-6Cm^-2 × 18.08m^-2

⇒ Q = 1.447 × 10^-3C

b) Let total electric flux leaving the surface of sphere = 

We know that,

Flux, Φ = q/ε₀ …(1)

Where, Q = net charged.

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

By, plugging the values of q and ε₀ in equation (1), we get,

⇒ 

⇒ Φ = 1.63 × 10⁸ Nm²C^-1

The total flux through the sphere is 1.63 × 10⁸ Nm²C^-1 .

Question 26.

An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.

Answer:

Given:

E = 9 × 10⁴ NC^-1

d = 2 cm

Let the linear charge density = λ Coulomb/metre

We know that, Electric field produced by a line charge with a linear charge density σ, at a distance d is given by,

 …(1)

Where, λ = linear charge density.

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

d = distance

From equation (1) we have,

⇒ λ = E × 2 × π × ε₀ × d

⇒ λ = 9 × 10⁴ NC^-1 × 2 × π × 8.85 × 10^-12N^-1 m^-2C² × 0.02m

⇒ λ = 10 ×10^-8 Cm^-1

The linear charge density is 10^-7 Cm^-1.

Question 27.

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/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:

Given:

Surface charge density on plate A, σ_A = -17.0 × 10^-22 Cm^-2

Surface charge density on plate B, σ_B = 17.0 × 10^-22 Cm^-2

The arrangement of plates are as shonw:

Let electric field in region 1 = E₁

Region 2 = E₂

Region 3 = E₃

The electric field in region I and region III is zero because no charge is present in these regions.

E₁ = 0

E₃ = 0

We know that,

E₂ = σ/ε₀ ...(1)

Where, σ Surface charge density

ε₀ = permittivity of free space

ε₀ = 8.85 × 10^-12N^-1 m^-2C²

Now, plugging the Values in equation (1).

⇒ 

⇒ E₂ = 1.92 × 10^-10 NC^-1

The electric field in the region enclosed by the plates is found to be 1.92 × 10^-10 NC^-1.

The electric field in region II is 1.92 × 10^-10 NC^-1.

---

Additional Exercises

Question 1.

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC^–1 in Millikan’s oil drop experiment. The density of the oil is 1.26 gcm^-3. Estimate the radius of the drop.

(g = 9.81 m s^–2; e = 1.60 × 10^–19 C).

Answer:

If the oil drop is stationary, the net force on it must be Zero or the resultant of all the forces on oil drop is zero.

There are two forces acting on the oil drop, its weight or force due to Earth’s gravity which is pulling it vertically downwards, and Electrostatic force which is acting in vertically upward direction.

Both forces should be equal in magnitude and opposite direction so that they cancel each other.

The arrangement is shown in the figure

Note: Electric Field is in vertically downward direction, because force on an negatively charged body in opposite direction of the field, so force on the drop is in vertically upward direction, which balances weight of the drop acting in vertically downward direction.

There are nine excess electrons which make the drop negatively charged because the electron is negatively charged, the net magnitude charge on a body is given by

q = n × E

n is the excess of electrons or protons on the body,

the charge of an electron is denoted by e

e = 1.60 × 10^–19 C

here since there are twelve excess electrons so

n = 12

i.e. q = 12 × 1.60 × 10^–19 C

= 1.92 × 10^-18 C

Now the magnitude electrostatic force on a charged particle held in an electric field is given by

F = q × E

where, F force is acting on a particle having charge q held in an electric field of magnitude E

here, charge on the oil drop is

q = 1.92 × 10^-18 C

magnitude of Electric field is

E = 2.55 × 10⁴NC^–1

So Electrostatic force on the oil drop is

F = 1.92 × 10^-18 C × 2.55 × 10⁴NC^–1

= 4.896 × 10^-14 N

This force is acting in vertically upward direction, so Weight should have same magnitude of Force and is acting in vertically downward direction.

Let us assume oil drop to be Spherical in shape, so the volume of drop will be

 (Volume of Sphere)

Where r is the radius of the Spherical drop

Now we know mass of an object whose volume and density are known is given by

m = V × d

where, m is the mass

V is the volume and d is the density

Here density of the drop is

d = 1.26 gcm^-3

= 1.26 × 10³ Kgm^-3

so mass of the drop is

m =  × 1.26 × 10³

Now the downward pull on a body due to earth’s gravitational force is the weight of body given by

W = m × g

Where W is the weight of a body having mass m

The acceleration due to gravity is denoted by g

g = 9.81 ms^-2

so the weight or the downward gravitational force on oil drop is

W =  × 1.26 × 10³ Kgm^-3× 9.81ms^-2

Both the forces should be equal in magnitude so equating them

i.e. putting F = W

4.896 × 10^-14 N = (4/3 πr³) × 1.26 × 10³ Kgm^-3 × 9.81ms^-2

we get ,

(1N = 1kgms^-2)

So, r 

⇒r = 0.98110^-6 m

= 9.81 × 10^-7 m

= 981 × 10^-4 mm

i.e. radius of the oil drop is 98110^-4 mm

Question 2.

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

Answer:

Electrostatic field lines have some properties and characteristics, which should be taken into account while drawing Electrostatic field lines, the Characteristic of electrostatic field lines are :-

1. They start from a positively charged body and end at a negatively charged body.

2. Tangent to the electrostatic field line at any point gives the direction of the electric intensity at that point.

3. Electrostatic field lines cannot intersect each other.

4. The electrostatic field lines are always normal to the surface of a conductor. There is no component of the electric filed intensity parallel to the surface of the conductor.

5. Electrostatic field lines do not form any closed loops.

These properties and characteristics must be taken into account while drawing electrostatic field lines, if any of these are violated then the following representation will be incorrect.

Hence, only (c) is the correct representation of electrostatic field lines and (a),(b),(d),(e) are incorrect representations of electrostatic field lines .

Question 3.

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 10⁵ NC^–1 per meter. 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:

A dipole is a system consisting of two charges equal in magnitude and opposite in nature i.e. a positive charge + q and a negative charge –q separated by some distance d

Electric dipole moment of a dipole is given by,

P = q × d

Where, ‘q’ is magnitude of either of charge (in column)

And ‘d’ is separation between the pair of charges (in metres)

Dipole moment is a vector quantity and its direction is from negative charge to positive charge

We are given dipole moment of the system,

P = 10^-7Cm , along negative Z axis

Here Electric Field is varying at the rate of 10⁵ NC^–1 per metre in positive Z direction

i.e. dE/dl = 10⁵ NC^–1m^-1 along Z direction

so the electric field at any point on z axis at distance of a metres is 

Now the total dipole moment equal to 10^–7 Cm in the negative z-direction.

The system and the Forces on it are as shown in the figure

Force on a charged particle in an electric field is given by

F = q × E

Where q is the magnitude of charge and E is the magnitude of Electric Field and the force is same as the direction of electric field in case of a positively charged particle and opposite to the direction of field in case of negatively charged particle

so Force on the + q charge located at a distance l from origin will be

directed towards positive Z axis

(since the electric field on z axis at distance of l metres from origin )

Force on the -q charge located at a distance l + d from origin will be

directed towards negative Z axis

(since the electric field on z axis at distance of l + d metres from origin)

So net force on the system would be

F = F _{+ q} – F_-q

(since F _{+ q} is directed towards positive Z axis and F_-q is directed towards negative Z axis )

So net force will be

Solving we get

We know,

q × d = P which is the dipole moment of the system so we get

Putting the values of P = 10^–7 Cm and  = 10⁵ NC^–1m^-1 We get

F = - (10^-7C × 10⁵NC^-1m^-1)

i.e. F = -10^-2 N

Negative sign states that force is along negative Z axis , so the net force on the system is 10^-2 N in negative Z direction.

We know Torque on an Electric Dipole is Vector or cross product of dipole moment P and electric field E and is given as

= PE

which can be further evaluated as

 = PE.sin(θ)

 is the Torque of the dipole

Where P is the magnitude of dipole moment

E is the magnitude of electric field

 is the angle between Dipole moment and electric field

 is a unit vector perpendicular to plane containing P and E

Here dipole moment P is along negative z axis and electric field E along positive Z axis so angle between them 

So, Sin = Sin180 = 0

So whatever be the magnitude of P and E putting  in the equation  = PESin

We get the torque τ = 0 Nm

So the torque of the system is zero

Question 4.

(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:

In Electrostatic we deal with charges at rest so there is no current inside or on the surface of the conductor i.e. in the static situation, the electric field is zero everywhere inside the conductor. This is the basic property of a conductor. A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift and re distribute themselves in such a way so that net electric field inside a conductor is zero.

We will use the above stated property to solve the problems

(a) Here in this situation a conductor with a cavity having no charge inside is given a charge Q on outer surface, we have to show that all charge resides on the external surface only and there is no charge on the inner surface.

From Gauss Theorem we know, net electric flux through an Gaussian surface enclosing a charge is equal to net charge enclosed by surface divided by permittivity of free space

Mathematically 

Where E is the electric field so  denotes the net Electric flux through the surface q is the net charge enclosed by the Gaussian surface and  is permittivity of free space which is a constant

Now if electric field at all the points on a Gaussian surface is zero

i.e. flux is zero or 

this means q/ϵ₀ = 0 or q = 0

i.e if electric field at all the points on a Gaussian surface is zero then the net charge enclosed by the Gaussian surface is also zero, or there is no charge inside the Gaussian surface

Now if we consider a Gaussian surface enclosing the cavity inside the conductor as shown in the figure

we know that net electric field everywhere inside a conductor is zero

so the electric flux through this Gaussian surface would also be zero, this means no charge is enclosed by the Gaussian surface, since cavity does not contain any charge so no charge should be induced on the cavity’s metal surface as well so whole of the charge Q is distributed on the outer surface only.

(b) Again using Gauss Theorem we know,

If electric field at all the points on a Gaussian surface is zero then the net charge enclosed by the Gaussian surface is also zero, or there is no charge inside the Gaussian surface

Now if we consider a Gaussian surface enclosing the cavity inside the conductor as shown in the figure

we know that net electric field everywhere inside a conductor is zero

so the electric flux through this Gaussian surface would also be zero, this means no charge is enclosed by the Gaussian surface, but the cavity has a charge q , this means a charge –q must be induced on the inner metal surface because net charge enclosed by Gaussian surface must be zero (q + (-q) = 0) , but the metal surface has only been given charge Q so net charge of system must only be q + Q , i.e. a charge q is induced on outer surface of the conductor as which makes the total charge on the conductor equal to Q only Q + (-q) + q = Q

and total charge of the system as

Q + (-q) + q + q = Q + q

This arrangement is also in accordance with the law of conservation of charge

note: Only Q charge has been given on the conductor’s outer surface , the charge appearing (Q + q) on it is only due to induction due to charge inside the cavity.

(c) To shield the sensitive instrument from strong electrostatic fields in its environment. We need to enclose it inside a metal piece’s Cavity, this process is also known as electrostatic shielding.

we consider a metal Block with a cavity having no charge inside and Instrument inside it as shown in the figure

any cavity inside a metal surface the electric field is zero no matter in which environment it is placed.

So in order to save the Instrument from External Electrostatic fiels we need to keep it inside a metal Cavity.

Question 5.

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (/2), where  is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

Answer:

Now in order to find the electric field inside the tiny hole on the surface of a charged hollow conductor, we will assume the hole to be filled up with the same conductor with same charge density i.e. Hollow conductor without any hole and will find out electric field at the point of hole on the conductor due to the whole conductor and due to the hole element and thus finding the contribution in electric field due the hole element and subtracting it with the total electric field of the conductor without any hole we will get the electric field of the rest of the conductor with hole.

Now consider a hollow spherical conductor with surface charge density  C/m² with a tiny on its surface now considering two points one just outside the surface of conductor near the hole say A and other inside the hollow cavity say B

As shown in the figure

Now finding the electric field at point A due to the whole conductor assuming hole to be filled up now the total charge on the conductor with total surface area S and surface charge density  will be

q =  × S

From Gauss Theorem we know, net electric flux through an Gaussian surface enclosing a charge is equal to net charge enclosed by surface divided by permittivity of free space

Mathematically: 

Where E is the electric field so  denotes the net Electric flux through the surface q is the net charge enclosed by the Gaussian surface and  is permittivity of free space which is a constant

Now assuming an Gaussian surface just enclosing the conductor

As shown in the figure

So we will have

Since electric field is always normal to a conductor surface so E and ds are in same direction i.e.  = 0

So, cos(θ) = 1 and due to spherical symmetry magnitude of electric field will same everywhere so we have

 = E × S = q/

i.e. E × S = ( × S) /

or E = /ϵ₀

now since electric field is always normal to the surface of a charged conductor we have

E = (/ϵ₀)

Where ń is the outward normal to the surface of conductor

Now let us assume contribution of the tiny hole to this electric field is E₁ and Electric field due to rest of the conductor is E₂

So, we have E = E₁ + E₂ (Just outside the conductor)

Now inside the hollow cavity inside the conductor at point B we know electric field is zero because of electrostatic shielding

Again assuming the electric field at that point to be contribution of the tiny hole to this electric field is E₁ and Electric field due to rest of the conductor is E₂ now since the hole will act as a point sized charge so field would be directed outwards i.e. in vertically downward direction, and rest of the conductor is below the point so field would be in vertically upward direction, i.e. field due to both hole and rest of conductor are opposite in direction and net field is zero inside a cavity in a conductor

So, we have,

0 = E₁ - E₂ (Just inside the hollow cavity of conductor)

Or E₁ = E₂

i.e. Electric field due to hole and rest of conductor are equal in magnitude and in same direction outside the conductor and equal in magnitude and opposite in direction in the hollow cavity.

Now using, E₁ = E₂ and E = E₁ + E₂

We have, 2E₁ = E and 2E₂ = E

Or, E₁ = E/2 and E₂ = E/2

Also we know E = (/)

i.e. the electric field due to the hole only is

E₁ = (/2)

And electric field due to rest of surface when hole is made is also

E₂ = (/2)

i.e. the electric field in the hole of a charged hollow conductor is (/2)

Where  is a unit vector in the outward normal direction.

Question 6.

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:

Let AB be a long thin wire of uniform linear charge density λ.

Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.

The charge on a small length dx on the line AB is q which is given as q = λdx.

So, according to Coulomb’s law, electric field at P due to this length dx is

where, ϵ₀ is the vacuum permittivity of the medium

But 

⇒ 

This electric field at P can be resolved into two components as dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only dEcosθ component.

So, the net electric field at P due to dx is

dE' = dE cosθ

⇒  ………………….(1)

In ΔPOC,

⇒ x = h tan θ

Differentiating both sides w.r.t. θ,

⇒ dx = h sec²θ dθ …………………….(2)

Also, h² + x² = h² + h²tan²θ

⇒ h² + x² = h²(1+ tan²θ)

⇒ h² + x² = h² sec²θ ………….(3)

(Using the trigonometric identity, 1+ tan²θ = sec²θ)

Using equations (2) and (3) in equation (1),

⇒,

The wire extends from  to  since it is very long.

Integrating both sides,

⇒ 

⇒ 

⇒ 

⇒ 

This is the net electric field due to a long wire with linear charge density λ at a distance h from it.

NOTE: In the given case, it has been assumed that the length of the wire tends to infinity. In case of wires of finite lengths, the sin(θ) components cancel out only along the perpendicular bisector of the wire. At any other point, the net electric field will have both sin(θ) and cos(θ) components.

Question 7.

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:

Let there be n up quarks.

Number of down quarks = 3-n (Since there are 3 quarks in a proton)

Charge on n up quarks =  where e is the charge of a proton

Charge on 3-n down quarks = 

In case of a proton:

Since the charge on a proton is e,

⇒ 

⇒ 

⇒ 

⇒ n = 2

So, 3-n = 3-2 = 1

Hence, there can be two up quarks ⇒, 2u and one down quark ⇒, 1d in a proton.

In case of a neutron:

Since a neutron is electrically neutral,

⇒ 

⇒ 

⇒ 

⇒ n = 1

So, 3-n = 3-1 = 2

Hence, there can be one up quark i.e. 1u and two down quarks i.e. 2d in a neutron.

Question 8.

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (⇒, 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) At a null point, the net electrostatic field is zero. If a small test charge is placed at the null point, it experiences no electrostatic force. But if it is displaced from the null point, it tends to move in the direction which is tangential to the field lines ⇒, towards the negative charge and away from the positive charge. So, if the test charge is displaced from the equilibrium position at the null point, it tends to move away from the equilibrium position. Hence, a null point is necessarily unstable.

(b) Let us consider the case of two charges of equal magnitude and opposite sign. The null point occurs at the midpoint of the line joining the two charges. If the test charge is moved away from this null point, it experiences a force towards the negative charge and away from the positive charge. This results in the movement of the charge away from the null point. Hence, the null point is an unstable equilibrium position.

NOTE: A test charge is a positive charge of unit magnitude. A null point or neutral point is a point where magnitude of electric field is zero. A pair of charges of equal magnitude and opposite sign is known as a dipole.

Question 9.

A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed v_x(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 

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Answer:

Given,

Mass of the particle = m

Charge of the particle = -q

Velocity of the particle = v_x

Length of the plate = L

Electric field between the plates = E

From Newton’s second law of motion,

Where, F = Force on the particle

m = mass of the particle

a = acceleration of the particle

⇒ 

⇒  (Since F = qE)

Time taken by the particle to cover the distance L is given by

 (Time = Displacement/Velocity)

For the movement in vertical direction, initial velocity (u) is zero.

According to Newton’s second equation of motion,

⇒

⇒ 

This is the vertical displacement of the particle at the far edge of the plate.

This motion is very similar to the motion of a projectile in a gravitational field. In a gravitational field, the force acting on the particle is mg and in the given case it is qE. The trajectory followed by the object will be similar in both the cases.

NOTE: A projectile is any object thrown into space by the exertion of a force. The path followed by a projectile is known as its trajectory.

Question 10.

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity v_x= 2.0 × 10⁶ m s^–1. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e| = 1.6 × 10^–19 C, m_e= 9.1 × 10^–31 kg.)

Answer:

Given,

Velocity of the electron, v_x = 2.0 × 10⁶ m s^–1

Separation of plates, l = 0.5cm = 0.5 × 10^-2 m

Electric field between the plates, E = 9.1 × 10² N C^-1

Charge on the electron, q = |e| = 1.6 × 10^-19 C

Mass of the electron, m_e = 9.1 × 10^-31 kg

The vertical deflection of the particle (s) is given by

⇒ 

⇒ 

⇒ L = √(2.5 × 10^-4) m

hence, L = 1.58 × 10^-2m = 1.58cm

Hence, the electron will strike the upper plate after travelling 1.58cm.