NP16.1.A particle carrying a charge of 0.1 μC starts from rest in a uniform electric field of intensity 60 N C⁻¹. Find the kinetic energy acquired by the particle when it has moved 2 m. (Ans: 1.2 × 10⁻³ J)
NP16.2.What should be the potential difference through which an electron falls to acquire a speed of 2 × 10⁶ m s⁻¹? (Ans: 11.4 V)
Data
v = 2 × 10⁶ m s⁻¹, me = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C. The loss in potential energy equals the gain in kinetic energy: eV = ½mv², so V = mv²/(2e). Substituting: mv² = (9.1 × 10⁻³¹ kg)(2 × 10⁶ m s⁻¹)² = 9.1 × 10⁻³¹ × 4 × 10¹² = 3.64 × 10⁻¹⁸ J. Then V = 3.64 × 10⁻¹⁸ J / (2 × 1.6 × 10⁻¹⁹ C) = 3.64 × 10⁻¹⁸ / 3.2 × 10⁻¹⁹ ≈ 11.4 V.
NP16.3.The potential difference between points A and B is V_A − V_B = 84 V, while the potential difference between points C and B is V_C − V_B = 24 V. What should be the work done on moving a 40 μC charge from C to A? (Ans: 2.4 × 10⁻³ J)
Data
VA − VB = 84 V, VC − VB = 24 V, q = 40 μC = 4 × 10⁻⁵ C. First find VA − VC = (VA − VB) − (VC − VB) = 84 V − 24 V = 60 V. Work done in moving the charge from C to A: W = q(VA − VC) = (4 × 10⁻⁵ C)(60 V) = 2.4 × 10⁻³ J.
NP16.4.A parallel plate capacitor has a capacitance of 6 μF when filled with a dielectric. The area of each plate is 1.2 m² and the separation between the plates is 9 × 10⁻⁶ m. What is the dielectric constant? (Ans: 5.1)
Data
C = 6 μF = 6 × 10⁻⁶ F, A = 1.2 m², d = 9 × 10⁻⁶ m, ε₀ = 8.85 × 10⁻¹² F m⁻¹. Using C = ε₀εᵣA/d, the dielectric constant is εᵣ = Cd/(ε₀A). Substituting: εᵣ = (6 × 10⁻⁶ F)(9 × 10⁻⁶ m) / [(8.85 × 10⁻¹² F m⁻¹)(1.2 m²)] = 5.4 × 10⁻¹¹ / 1.062 × 10⁻¹¹ ≈ 5.1.
NP16.5.A 15 μF capacitor is charged till the potential difference across its plates becomes 300 V. Find the energy stored in it. (Ans: 0.675 J)
Data
C = 15 μF = 15 × 10⁻⁶ F, V = 300 V. Energy stored: W = ½CV² = ½(15 × 10⁻⁶ F)(300 V)² = ½(15 × 10⁻⁶)(9 × 10⁴) = 0.675 J.
NP16.6.Two capacitors of 2.0 μF and 8.0 μF capacitances are connected in series and a potential difference of 300 volts is applied. Find the charge and potential difference for each capacitor. (Ans: Q₁ = Q₂ = 4.8 × 10⁻⁴ C, V₁ = 240 V, V₂ = 60 V)
Data
C₁ = 2.0 μF, C₂ = 8.0 μF, V = 300 V. Equivalent series capacitance: 1/Cs = 1/C₁ + 1/C₂ = 1/2 + 1/8 (in μF⁻¹) = 5/8, so Cs = 8/5 = 1.6 μF = 1.6 × 10⁻⁶ F. Since the charge is the same on each capacitor in series: Q = CsV = (1.6 × 10⁻⁶ F)(300 V) = 4.8 × 10⁻⁴ C, so Q₁ = Q₂ = 4.8 × 10⁻⁴ C. Potential difference across each: V₁ = Q/C₁ = 4.8 × 10⁻⁴ C / 2.0 × 10⁻⁶ F = 240 V; V₂ = Q/C₂ = 4.8 × 10⁻⁴ C / 8.0 × 10⁻⁶ F = 60 V. (Check: V₁ + V₂ = 300 V ✓)
NP16.7.A capacitor has a capacitance of 3 μF. It is charged by a battery. When the potential difference between the plates is 240 volts, how many electrons have been removed from one plate and placed on the other plate? (Ans: 3.0 × 10¹⁵)
NP16.8.Two positive point charges are held at two points 1.2 m apart. They are then moved towards each other so that their electric potential energy becomes double of its previous value. How much are they apart at their new positions? (Ans: 0.6 m)
Data
initial separation r₁ = 1.2 m, charges unchanged. Since U = (1/4πε₀)(q₁q₂/r), for fixed charges U ∝ 1/r. If U doubles, r must be halved: r₂ = r₁/2 = 1.2 m / 2 = 0.6 m.
NP16.9.A 100 μF capacitor is initially charged to 8 V. It is allowed to discharge through a 500 Ω resistor. Find the voltage across the capacitor after 0.5 s. (Ans: 3.6 × 10⁻⁴ V)
Data
C = 100 μF = 100 × 10⁻⁶ F, R = 500 Ω, V₀ = 8 V, t = 0.5 s. Time constant: τ = RC = 500 Ω × 100 × 10⁻⁶ F = 0.05 s. Using the exponential decay equation V = V₀e^(−t/τ): t/τ = 0.5/0.05 = 10, so V = 8 × e⁻¹⁰ = 8 × (4.54 × 10⁻⁵) ≈ 3.6 × 10⁻⁴ V.
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