Capacitance for Cambridge International AS & A Level Physics

Reading cue. 5.0 mm is the gap, not a plate length, and no field lines are drawn — the plus and minus signs on the plates are the only thing in the figure that fixes the field direction.

1. aState State the direction of the electric field in the gap in Fig. 10.1, and the direction of the force it exerts on an electron placed there.

   recall2 marks

   Check answer · 2 marks

   1. the field points vertically downwards, from the positive upper plate to the negative lower plate
   2. the force on an electron is vertically upwards, towards the positive plate

2. bCalculate Calculate the work done on an electron that moves from the lower plate to the upper plate, and the speed with which it arrives.

   routine3 marks

   Check answer · 3 marks

   1. W = eV = 1.60 × 10⁻¹⁹ × 2.0 × 10³ = 3.2 × 10⁻¹⁶ J
   2. ½mv² = 3.2 × 10⁻¹⁶ J
   3. v = 2.7 × 10⁷ m s⁻¹

3. cDetermine A charged dust particle of weight 1.28 × 10⁻¹³ N is held at rest midway between the plates. Determine the magnitude and sign of its charge, and the number of excess electrons it carries.

   demanding4 marks

   Check answer · 4 marks

   1. E = V/d = 2.0 × 10³ / 5.0 × 10⁻³ = 4.0 × 10⁵ V m⁻¹
   2. for equilibrium qE = weight, so q = 1.28 × 10⁻¹³ / 4.0 × 10⁵ = 3.2 × 10⁻¹⁹ C
   3. the electric force must act upwards while the field points downwards, so the charge is negative
   4. 3.2 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2 excess electrons

4. dDeduce The battery p.d. is suddenly reduced to 1.0 kV while the particle is still midway between the plates. Deduce the acceleration of the particle, and calculate the time it takes to reach a plate.

   top of the paper4 marks

   Check answer · 4 marks

   1. new field = 2.0 × 10⁵ V m⁻¹, so the electric force = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N upwards
   2. resultant = 1.28 × 10⁻¹³ − 6.4 × 10⁻¹⁴ = 6.4 × 10⁻¹⁴ N downwards
   3. mass = 1.28 × 10⁻¹³ / 9.81 = 1.30 × 10⁻¹⁴ kg, so a = 4.9 m s⁻² downwards, that is g/2
   4. falling the 2.5 mm to the lower plate: t = √(2 × 2.5 × 10⁻³ / 4.9) = 3.2 × 10⁻² s

Reading cue. The voltmeter is permanently across C, so it is part of the discharge path even before S closes; R joins the circuit only when S is closed.

1. aCalculate The capacitor in Fig. 12.1 has capacitance 470 μF. Calculate the time constant of the circuit after S is closed.

   recall2 marks

   Check answer · 2 marks

   1. τ = RC = 100 × 10³ × 470 × 10⁻⁶
   2. τ = 47 s

2. bDetermine The voltmeter reads 9.0 V at the instant S is closed. Determine its reading 60 s later.

   routine3 marks

   Check answer · 3 marks

   1. V = V₀e^(−t/RC), with t/RC = 60/47 = 1.28
   2. V = 9.0 × e^(−1.28)
   3. V = 2.5 V

3. cDetermine Determine the time taken for the voltmeter reading to halve, and state whether this time would differ if the reading at the instant of closing S were 4.5 V instead of 9.0 V.

   demanding3 marks

   Check answer · 3 marks

   1. ½ = e^(−t/RC), so t = RC ln 2
   2. t = 47 × 0.693 = 33 s
   3. the time to halve is independent of the starting p.d., so it would still be 33 s

4. dDeduce Deduce the energy transferred to R while the voltmeter reading falls from 9.0 V to 4.5 V, and state the assumption you make about the voltmeter.

   top of the paper4 marks

   Check answer · 4 marks

   1. energy stored at 9.0 V = ½CV² = ½ × 470 × 10⁻⁶ × 9.0² = 1.9 × 10⁻² J
   2. energy stored at 4.5 V = ½ × 470 × 10⁻⁶ × 4.5² = 4.8 × 10⁻³ J
   3. energy transferred to R = 1.9 × 10⁻² − 4.8 × 10⁻³ = 1.4 × 10⁻² J
   4. assumes no charge flows through the voltmeter branch, i.e. its resistance is effectively infinite