NCEA Level 2 Physics 91173 Electricity free practice questions with answers

Question 1 (9 marks) [Electric fields]

(a) Two parallel metal plates 0.0400 m apart have a potential difference of 250 V between them, giving a uniform electric field. An electron is released from rest at the surface of the plate that repels it, and accelerates across the gap. (i) Calculate the electric field strength between the plates. (ii) Calculate the force on the particle and its acceleration. (iii) Calculate the speed with which it reaches the other plate. (Charge of the particle 1.60 × 10⁻¹⁹ C; mass 9.11 × 10⁻³¹ kg.)

(a) Answer: (i) 6250 V m⁻¹; (ii) 1.00 × 10⁻¹⁵ N and 1.10 × 10¹⁵ m s⁻²; (iii) 9.37 × 10⁶ m s⁻¹

• Evidence level: Excellence (link field, force, acceleration and energy; use energy conservation for the final speed).

• (i) E = V/d = 250 ÷ 0.0400 = 6250 V m⁻¹.

• (ii) F = Eq = 6250 × 1.60 × 10⁻¹⁹ = 1.00 × 10⁻¹⁵ N. a = F/m = 1.00 × 10⁻¹⁵ ÷ 9.11 × 10⁻³¹ = 1.10 × 10¹⁵ m s⁻².

• (iii) The work done on the charge by the field, qV = 1.60 × 10⁻¹⁹ × 250 = 4.00 × 10⁻¹⁷ J, becomes kinetic energy: ½mv² = qV.

• v = √(2qV/m) = √(2 × 4.00 × 10⁻¹⁷ ÷ 9.11 × 10⁻³¹) = 9.37 × 10⁶ m s⁻¹.

(b) A small sphere carrying a charge of 3.00 × 10⁻⁶ C and a mass of 2.00 × 10⁻³ kg is in a uniform electric field of strength 4000 V m⁻¹. It is released from rest and moves 0.120 m along the field direction.

(i) Calculate the force on the sphere. (ii) Calculate the work done on the charge by the field, and the gain in kinetic energy. (iii) Calculate the speed of the sphere after moving 0.120 m. Ignore gravity.

(b) Answer: (i) 0.0120 N; (ii) 1.44 × 10⁻³ J; (iii) 1.20 m s⁻¹

• Evidence level: Merit (use F = Eq, work = Fd (or Ep = Eqd) and energy conservation).

• (i) F = Eq = 4000 × 3.00 × 10⁻⁶ = 0.0120 N.

• (ii) W = Fd = 0.0120 × 0.120 = 1.44 × 10⁻³ J, which equals the gain in kinetic energy.

• (iii) ½mv² = 1.44 × 10⁻³ J, so v = √(2 × 1.44 × 10⁻³ ÷ 2.00 × 10⁻³) = 1.20 m s⁻¹.