Question 1

Q1: 2.1 Two charges 5 × 10⁻⁸ C and –3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zer...

Accepted Answer

Let the two charges be q₁ = 5 × 10⁻⁸ C and q₂ = –3 × 10⁻⁸ C. The distance between them is d = 16 cm = 0.16 m. Let us place the charge q₁ at the origin (x = 0). Then q₂ is located at x = 0.16 m. Let P be the point on the line where the total electric potential is zero. The potential at P is the sum o...

Question 2

Q10: 2.10 A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Accepted Answer

Given:    Capacitance, C = 12 pF = 12 × 10⁻¹² F    Voltage, V = 50 V The electrostatic energy (U) stored in a capacitor is given by the formula: U = (1/2)CV² Substituting the given values: U = (1/2) × (12 × 10⁻¹² F) × (50 V)² U = (6 × 10⁻¹² F) × (2500 V²) U = 15000 × 10⁻¹² J U = 1.5 × 10⁻⁸ J Therefo...

Question 3

Q11: 2.11 A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor....

Accepted Answer

Given:    Capacitance of the first capacitor, C₁ = 600 pF = 600 × 10⁻¹² F    Initial voltage, V₁ = 200 V *   Capacitance of the second capacitor, C₂ = 600 pF = 600 × 10⁻¹² F Step 1: Calculate the initial energy stored. The initial energy (U_initial) is stored only in the first capacitor. U_initial =...

Question 4

Q2: 2.2 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.

Accepted Answer

Given:    Charge at each vertex, q = 5 μC = 5 × 10⁻⁶ C    Side of the regular hexagon, a = 10 cm = 0.1 m In a regular hexagon, the distance from the centre to each vertex is equal to the side length. So, the distance of the centre from each charge is r = a = 0.1 m. The electric potential at a point...

Question 5

Q3: 2.3 Two charges 2 μC and –2 μC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction...

Accepted Answer

The system consists of two equal and opposite charges separated by a distance, which constitutes an electric dipole. (a) Equipotential Surface: An equipotential surface is a surface where the electric potential is the same at every point. For an electric dipole, the potential at any point on the equ...

Question 6

Q4: 2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field (a) inside the...

Accepted Answer

Given:    Radius of the spherical conductor, R = 12 cm = 0.12 m    Charge on the conductor, Q = 1.6 × 10⁻⁷ C (a) Inside the sphere: The electrostatic field inside a charged conductor is always zero. This is because the charges reside on the surface and rearrange themselves in such a way that the net...

Question 7

Q5: 2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = 10⁻¹² F). What will be the capacitance if the distance be...

Accepted Answer

Let the initial capacitance of the parallel plate capacitor be C₀. Given, C₀ = 8 pF = 8 × 10⁻¹² F. The capacitance of a parallel plate capacitor with air (or vacuum) as the medium is given by: C₀ = ε₀A / d where A is the area of the plates and d is the distance between them. According to the problem...

Question 8

Q6: 2.6 Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential...

Accepted Answer

Given:    Capacitance of each capacitor, C₁ = C₂ = C₃ = 9 pF = 9 × 10⁻¹² F    Supply voltage, V = 120 V (a) Total capacitance of the combination: When capacitors are connected in series, the equivalent capacitance (C_eq) is given by the formula: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ 1/C_eq = 1/9 + 1/9 + 1/9 1...

Question 9

Q7: 2.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determin...

Accepted Answer

Given:    Capacitances: C₁ = 2 pF, C₂ = 3 pF, C₃ = 4 pF    Supply voltage, V = 100 V (a) Total capacitance of the combination: When capacitors are connected in parallel, the equivalent capacitance (C_eq) is the sum of the individual capacitances. C_eq = C₁ + C₂ + C₃ C_eq = 2 pF + 3 pF + 4 pF C_eq =...

Question 10

Q8: 2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ m² and the distance between the plates is 3 mm. Calc...

Accepted Answer

Given:    Area of each plate, A = 6 × 10⁻³ m²    Distance between the plates, d = 3 mm = 3 × 10⁻³ m    Permittivity of free space, ε₀ ≈ 8.85 × 10⁻¹² F/m    Supply voltage, V = 100 V 1. Calculation of Capacitance: The capacitance (C) of a parallel plate capacitor with air as the dielectric is given b...

Question 11

Q9: 2.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between th...

Accepted Answer

From Exercise 2.8, we have:    Initial capacitance with air, C₀ = 17.7 pF    Initial voltage, V₀ = 100 V    Initial charge, Q₀ = 1.77 × 10⁻⁹ C A mica sheet of dielectric constant K = 6 is inserted, completely filling the space between the plates. The new capacitance (C) will be: C = K × C₀ = 6 × 17....

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