**NCERT Class 12 Physics Chapter 1 Electric Charges and Fields Solutions**,

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**NCERT Class 12 Physics Chapter 1 Electric Charges and Fields**

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**Electric Charges and Fields**

**Electric Charges and Fields****Chapter: 1**

Part – I |

**EXERCISE**

**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?**

Ans:

**Given:**

q_{1} = 2 × 10^{-7} C

q_{2} = 3 × 10^{-7} C

r = 30 cm = 0.3 m

F = 8.99 × 10^{9} Nm^{2}/C^{2} × | 2 x 10^{-7} C × 3 × 10^{-7} C | / (0.3 m)^{2}

F = 8.99 × 10^{9} × 6 × 10^{-14} / 0.09

F = 53.94 × 10^{-5} / 0.09

= 59.93 × 10^{-5} N

F = 5.99 × 10^{-4} N.

**2. The electrostatic force on a small sphere of charge 0.4 mC due to another small sphere of charge –0.8 mC in air is 0.2 N.**

**(a) What is the distance between the two spheres?**

Ans: Here,

**Rearranging coulomb’s law to solve for r:**

r = 3.79 m.

**(b) What is the force on the second sphere due to the first?**

Ans: The force on the second sphere due to the first is also 0.2 N and is attractive in nature.

**3. Check that the ratio ke2 /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?**

Ans:

K_{e}^{2} / Gm_{e}m_{p} =

K_{e}^{2} / Gm_{e}m_{p} = 2.29824 × 10^{-28} / 1.013448 × 10^{-67 }

= 2.27 × 10^{39}.

The factor Ke^{2} / Gm_{e}m_{p}, represents the ratio of the electrostatic force to the gravitational force between an electron and a proton. The large value of this ratio indicates that the electrostatic force is immensely stronger than the gravitational force. This disparity highlights why the electrostatic force is dominant in atomic and molecular interactions, while gravity becomes significant only at much larger scales, such as in planetary and astronomical phenomena.

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

Ans: Charge is quantized, meaning it exists in discrete units. This unit is the elementary charge, which is the charge of a proton or electron.

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

Ans: At the macroscopic level, dealing with large amounts of charge, the individual units of elementary charge become so small compared to the total charge that they appear continuous. Therefore, we can treat charge as a continuous quantity without noticeable error.

**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.**

Ans: When a glass rod is rubbed with a silk cloth, opposite charges appear on both, because electrons are transferred from glass to silk. The phenomenon of charging by friction is in complete harmony with the law of conservation of charge. When two neutral objects are rubbed in such a phenomenon, both objects get charged. Before friction, both objects are neutral, that is, their total charge is zero.

**6. Four point charges q**_{A}** = 2 μC, q**_{B}** = –5 μC, q**_{C}** = 2 mC, and q**_{D}** = –5 μC are located at the corners of a square ABCD of 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?**

Ans:

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. Due to the symmetry of the arrangement and the equal and opposite charges, the net force on the central charge is zero.

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

Ans: An electrostatic field line is a continuous curve because it represents the path along which a positive test charge would move under the influence of the electric field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to another.

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

Ans: This is not possible. Hence, two field lines never cross each other.Crossing would imply two different directions at the same location, which is not possible.

**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?**

Ans: Electric field at the midpoint O,

r = 0.2 m / 2

= 0.1 m

E = k × |q| / r^{2}

= E_{A} = 8.99 × 10^{9} × 3 × 10^{-6} / (0.1)^{2}

E_{A} = 2.697 × 10^{4} / 0.01

= 2.697 × 10^{6} N/C

**Electric field due to q**_{B }**at O:**

E_{B} = 8.99 × 10^{9} × 3 × 10^{-6} / (0.1)^{2}

E_{B} = 2.697 × 10^{6} N/C.

**Total electric field at O:**

Etotal = E_{A} + E_{B}

= 2.697 × 10^{6 }+ 2.697 × 10^{6}

= 5.394 × 10^{6} N/C.

**(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?**

Ans:** Force Experienced by the Test Charge:**

F = E × qt

F = 5.394 × 10^{6} × 1.5 × 10^{-9}

F = 8.091 × 10^{-3} N.

**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 moments of the system?**

Ans: Hence total charge = q_{A} = q_{B} = 0

**Dipole moment:**

q = q_{A} = q_{B} = 2.5 × 10-7 C

2a = 30 cm = 0.3 m.

Therefore,

**10. An electric dipole with a 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.**

Ans: T = p × E sin (θ)

**Given:**

p = 4 × 10^{-9} cm

E = 5 × 10^{4} N/C

θ = 30°

T = (4 × 10^{-9}) × (5 × 10^{4}) × sin (30°)

T = (4 × 10^{-9}) × (5 × 10^{4}) × 0.5

T = (4 × 5 × 0.5) × 10^{-9} × 10^{4}

T = 10 × 10^{-5}

T = 1 × 10^{-4} Nm.

**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?)**

Ans: n = q/e

**Where:**

e is the elementary charge, e = 1.6 × 10^{-19} C.

q = 3 × 10^{-7} C

n = 3 × 10^{-7 }/ 1.6 × 10^{-19}

n = 1.875 × 10^{12}

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

Ans: Yes, there is a transfer of mass from wool to polythene because electrons have mass. The mass transferred Δm = n (m_{e}).

= 1.875 × 10^{12} × 9.1 × 10^{-31} kg

= 1.7 × 10^{-18} kg = 2 × 10^{-18} kg

However, this mass transfer is negligible and cannot be detected experimentally.

**12. (a) 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. **

Ans:

F = 8.99 × 10^{9} × 169 × 10^{-14}

F = 1.52 × 10^{-2} N.

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

Ans: q^{/}_{1} = q^{/}_{2} = 2 × 6.5 × 10^{-7} = 1.3 × 10^{-6}C.

r^{/} = 0.5 / 2

= 0.25m.

F = 8.99 × 10^{9} × 27.04 × 10^{-12}

F’ = 2.43 × 10^{-1} N.

**13. Figure 1.30 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?**

Ans: Particles (1) and (2) are deviating towards the positive plate, therefore the sign of their charges is negative. The particle (3) is deviating towards the negative plate, therefore the sign of its charge is positive.

Therefore, the deflection of the particle,

y = ut + ½ at^{2}

= q/m ∝y

Therefore, the particle (3) having maximum deflection (y), has highest ratio of charge to mass.

**14. Consider a uniform electric field E = 3 × 103 î 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?**

Ans: **Given:**

Φ_{E} = E × A = EA cos θ

A = (0.1 m )^{2} = 0.01 m^{2}

= 3 × 10^{3} N/C × 0.01 m^{2} × 1

= 30 N m^{2}/N

**(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?**

Ans: Φ_{E} = E × A × cos 60°

= 3 × 10^{3} N/C × 0.01 m^{2} × 0.5

= 15 N m^{2}/C.

**15. What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?**

Ans:

Due to the field’s direction along the x-axis, faces in the x-y and y-z planes have zero flux. For faces in the y-z plane, the flux is equal but opposite, resulting in a net flux of zero.

**16. 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}** Nm**^{2}** /C. **

**(a) What is the net charge inside the box?**

Ans: Φ_{E} = q_{inside} / Σ_{0}

= qinside = Φ_{E} × Σ_{0}

= 8.0 × 10^{3} Nm^{2}/C × 8.854 × 10^{-12} C^{2}/ Nm^{2}

= 7.08 × 10^{-8} C

**(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?**

Ans: 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. There could be equal amounts of positive and negative charges inside the box that cancel each other out, resulting in zero net charge and hence zero net flux.

**17. A point charge +10 mC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. 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.) **

Ans: Since the charge is at the centre of the cube, it’s enclosed by the cube. Using Gauss’s law, we find the total flux through the cube. Then, we divide this by 6 (the number of faces) to get the flux through the square.

**Therefore, electric flux through the square:**

Φ_{c} =

= 1.88 × 10^{5 }Nm^{2} C^{-1}.

**18. A point charge of 2.0 C is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?**

Ans: Φ_{E} = Qenc / Σ_{0}

**We know that:**

Σ_{0} is the vacuum permittivity, approximately 8.85 × 10^{−12} C^{2}/N × m^{2}

**Applying Gauss’s Law:**

Φ_{E} = 2.0 C / (8.85 × 10^{-12} C^{2} / N × m^{2})

Φ_{E} = 2.26 × 10^{11} N × m^{2} / C.

**19. 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 centred on the charge.**

**(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?**

Ans: The electric flux depends only on the charge present in the Gaussian surface. Therefore, through the of double the radius will be same, i.e.,

Ø = – 1.0 × 10^{3} Nm^{2} C^{-1}

= – 10^{3} Nm^{2} C^{-1}.

**(b) What is the value of the point charge?**

Ans: Ø_{E} = Qenc / Σ_{0}

Qenc = Ø_{E} × Σ_{0}

Qenc = (-1.0 × 10^{3}) × (8.85 × 10^{-12}) C

= – 8.85 × 10^{-9} C.

**20. 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?**

Ans: **Given:**

E = kQ / r^{2}

Here,

E = 1.5 × 10^{3} N/C

k = 8.99 × 10^{9} Nm^{2} / C^{2}

Q = ?

r = 20 cm = 0.2 m.

Q = Er^{2} / k

= (1.5 × 10^{3} N/C) × (0.2m)^{2} / (8.99 × 10^{9} Nm^{2}/C^{2})

= 6.67 × 10^{-9} C.

As electric field points radially inward, it means that the charge is negative i.e., – 6.67 × 10^{-9} C.

**21. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 C/m**^{2}**. **

**(a) Find the charge on the sphere.**

Ans: d = 2.4 m

Σ = 80.0 C/m^{2}

**Radius:**

r = d/2 = 2.4 m /2 = 1.2 m.

Surface of the Area A = 4𝜋r^{2}

A = 4𝜋 (1.2)^{2}

= 4𝜋 × 1.44

= 18.1 m2

Q = Σ × A

= 80.0 C/m2 × 18.1 m2

= 1448 C.

**(b) What is the total electric flux leaving the surface of the sphere?**

Ans: Φ_{E} = Qenc / Σ_{0}

Σ_{0} (Approximately) = 8.85 × 10^{-12} C^{2} /N

Φ_{E} = 1448 / 8.85 × 10^{-12}

= 1.64 × 10^{14} N × m^{2}/ C.

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

Ans: E = 9 × 10^{4 }NC^{-1}

= r = 2 cm = 0.02 m

= 10^{-7} Cm^{-1}.

**23. 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?**

Ans: **Given:**

Σ_{0} (Approximately) = 8.85 × 10^{-12} C^{2} /N

σ = 17.0 × 10^{-22} C/m^{2}.

**(a) Electric Field in the Outer Region of the First Plate:**

E_{1} = σ / 2Σ_{0}

= 17.0 × 10^{-22} / 2 × 8.85 × 10^{-12} N/C

= 17.0 × 10^{-22} / 1.77 × 10^{-11} N/C

= 9.6 × 10^{-12} N/C.

**(b) Electric Field in the Outer Region of the Second Plate:**

E_{2} = σ / 2Σ_{0} = E_{1} = 9.6 × 10^{-12} N/C

**(c) Electric Field Between the Plates:**

E_{between} = σ / 2Σ_{0}

= 17.0 × 10^{-22} / 8.85 × 10^{-12} N/C

= 1.92 × 10^{-11} N/C.