11th Class Physics Unit 9 Electrostatics and Current Electricity Numericals with Solutions

9.1.Two unequal point charges repel each other with a force of 0.4 N when they are 5.0 cm apart. Find the force which each charge exerts on the other when they are (a) 2.5 cm apart (b) 15.0 cm apart.

Given

Initial force F = 0.4 N

Initial distance r = 5 cm

Distance (a) r₁ = 2.5 cm

Distance (b) r₂ = 15 cm

Formula

Coulomb's inverse square law F ∝ 1r²

For part (a) FF₁ = r²r₁²

Working

Substituting values 0.4F₁ = 5²2.5² = 256.25

F₁ = 256.25 × 0.4

Result

Force at 2.5 cm F₁ = 1.6 N

For part (b) 0.4F₁ = 15²5² = 22525

Working

Substituting F₁ = 25225 × 0.4

Result

Force at 15 cm F₁ = 0.04 N

9.2.A particle of charge +20 μC is placed between two parallel plates, 10 cm apart and having a potential difference of 0.5 kV between them. Calculate the electric field between the plates, and the electric force exerted on the charged particle.

Given

Charge q = 20 μC = 20 × 10^-6 C

Potential difference V = 0.5 kV = 500 V

Distance between plates d = 10 cm = 0.1 m

Formula

Electric field E = Vd

Working

Substituting E = 5000.1

Result

Electric field E = 5000 V/m or 5 × 10³ V/m

Formula

Electric force F = qE

Working

Substituting F = (20 × 10^-6 C) × (5000 V/m)

Result

Electric force F = 0.1 N

9.3.The electron and proton in a hydrogen atom are separated (on the average) by a distance of approximately 5.3 × 10⁻¹¹ m. Find the ratio of the electric force and the gravitational force between the electron and proton in this state.

Given

Distance r = 5.3 × 10^-11 m

Charge on electron q₁ = -1.6 × 10^-19 C

Charge on proton q₂ = 1.6 × 10^-19 C

Mass of electron m₁ = 9.11 × 10^-31 kg

Mass of proton m₂ = 1.67 × 10^-27 kg

Formula

Electric force F_e = k q₁ q₂r²

Gravitational force F_g = G m₁ m₂r²

Ratio F_eF_g = k q₁ q₂G m₁ m₂

Working

Substituting F_eF_g = (9 × 10⁹) × (1.6 × 10^-19)²(6.67 × 10^-11) × (9.1 × 10^-31) × (1.67 × 10^-27)

Result

Ratio F_eF_g = 2.27 × 10³⁹

9.4.After a pleasant showering, a water droplet of mass 1.2 × 10⁻¹¹ kg is located in the air near the ground. An atmospheric electric field of magnitude 6.0 × 10³ NC⁻¹ points vertically downward in the vicinity of the water droplet. The droplet remains suspended at rest in the air. Find the electric charge on the droplet.

Given

Mass of droplet m = 1.2 × 10^-11 kg

Electric field E = 6 × 10³ NC^-1

Droplet suspended F_e = mg

Formula

Electric force F_e = qE

Balance condition qE = mg

Charge q = mgE

Working

Substituting q = frac{1.2 × 10^-11 kg × 9.8 ms^-2{6 × 10³ NC^-1

Result

Charge on droplet q = 1.96 × 10^-14 C

9.5.An electron enters the region of a uniform electric field, with vᵢ = 2.99 × 10⁶ m·s⁻¹ and E = 300 N C⁻¹. The horizontal length of the plates is 10.0 cm. Find the acceleration of the electron while it is in the electric field. How long will it take to pass through the field?

Given

Initial velocity v_i = 2.99 × 10⁶ m/s

Electric field E = 300 N C^-1

Length of plates L = 0.1 m

Charge on electron q = -1.6 × 10^-19 C

Mass of electron m = 9.11 × 10^-31 kg

Formula

Acceleration F = qE = ma

a = qEm

Working

Substituting a = frac{(-1.6 × 10^-19 C) × 300 N/C{9.11 × 10^-31 kg

Result

Acceleration a = -5.27 × 10¹³ ms^-2

Formula

Time to pass through t = Lv_i

Working

Substituting t = frac{0.1 m{2.99 × 10⁶ m/s

Result

Time t = 3.34 × 10^-8 s

9.6.A disc of 10 cm² area is placed in a vertical electric field E = 5 × 10⁵ N C⁻¹. If the plane of the disc makes an angle of 30° with the horizontal, determine the electric flux through the disc.

Given

Area of disc A = 10 cm² = 10 × 10^-4 m² = 10^-3 m²

Electric field E = 5 × 10⁵ N/C

Angle with horizontal Plane makes 30° with horizontal

Angle between E and normal varphi = 60°

Formula

Electric flux Phi = EAcos(varphi)

Working

Substituting Phi = (5 × 10⁵ N/C) × (10^-3 m²) × cos(60°)

Phi = (5 × 10⁵) × (10^-3) × 0.5

Result

Electric flux Phi = 250 Nm²/C

9.7.A circular copper rod is 50 cm long and has 1 cm diameter. Find the resistance across its ends. What should be the side of a square cross-section of a 50 cm long tungsten rod if its resistance is the same? [Resistivity of copper is 1.69×10⁻⁸ Ωm and that of tungsten is 5.0×10⁻⁸ Ωm.]

Given

Length of copper rod L = 50 cm = 0.5 m

Diameter of copper rod d = 1 cm = 0.01 m

Radius of copper rod r = d/2 = 0.005 m

Area of copper rod A = pi r² = pi(0.005)² = 7.85 × 10^-5 m²

Resistivity of copper rho = 1.68 × 10^-8 Ωm

Formula

Resistance formula R = rho LA

Working

For copper rod R = frac{(1.68 × 10^-8 Ωm) × 0.5 m{7.85 × 10^-5 m²}

Result

Resistance of copper rod R ≈ 1.07 × 10^-4 Ω

Given

Resistivity of tungsten rho_W = 5.51 × 10^-8 Ωm

For tungsten rod with same resistance 1.07 × 10^-4 = (5.51 × 10^-8) × 0.5A

Area of tungsten A = frac{5.51 × 10^-8 × 0.5}{1.07 × 10^-4 = 2.57 × 10^-4 m²

Side of square side = sqrt{A} = sqrt{2.57 × 10^-4

Result

Side of square cross-section side ≈ 0.016 m or 1.6 cm

9.8.The copper winding of an electric fan has a resistance of 50 Ω at 30 °C. After running for some time, the resistance becomes 52 Ω. How much is the increase in temperature of the winding? [For copper α = 0.0039 K⁻¹]

Given

Initial resistance R₁ = 50 Ω

Final resistance R₂ = 52 Ω

Change in resistance ΔR = R₂ - R₁ = 52 - 50 = 2 Ω

Temperature coefficient α = 0.0039 K^-1

Formula

Temperature change formula α = ΔRR₁ ΔT

Rearranging ΔT = ΔRR₁ α

Working

Substituting ΔT = 250 × 0.0039

ΔT = 20.195

Result

Temperature increase ΔT ≈ 10.26 ≈ 10.3 °C

9.9.During an experiment, a copper wire of 50 m long and 150 μm thick is hung vertically. Then a current of 1 A is passed across its ends for 50 s. Find the resistance of the wire and the heat dissipated during this process. [Resistivity of copper is 1.69 × 10⁻⁸ Ωm.]

Given

Length of wire L = 50 m

Diameter of wire d = 150 μm = 1.5 × 10^-4 m

Radius of wire r = d/2 = 0.75 × 10^-4 m

Area of wire A = pi r² = pi(0.75 × 10^-4)² = 1.767 × 10^-8 m²

Current I = 1 A

Time t = 50 s

Resistivity of copper rho = 1.68 × 10^-8 Ωm

Formula

Resistance formula R = rho LA

Working

Substituting R = frac{(1.68 × 10^-8 Ωm) × 50 m{1.767 × 10^-8 m²}

Result

Resistance R ≈ 47.55 Ω

Formula

Heat dissipated Q = I²Rt

Working

Substituting Q = (1)² × 47.55 × 50

Result

Heat dissipated Q = 2377.5 J

9.10.The e.m.f. of a battery is 12 V. It is connected to a 3.6 Ω resistor. If the internal resistance of the battery is 0.2 Ω, what will be the terminal voltage across the battery?

Given

e.m.f. E = 12 V

External resistance R = 3.6 Ω

Internal resistance r = 0.2 Ω

Formula

Total resistance R_{total = R + r

Working

Substituting R_{total = 3.6 + 0.2 = 3.8 Ω

Formula

Current I = frac{E}{R_{total

Working

Substituting I = frac{12 V{3.8 Ω = 3.16 A

Formula

Terminal voltage V = E - Ir

Working

Substituting V = 12 - (3.16 × 0.2)

V = 12 - 0.632

Result

Terminal voltage V ≈ 11.37 V

Unit 9: Electrostatics and Current Electricity — Numericals