IBDP Physics 11.3 – Capacitance: IB style Question Bank HL Paper 2

Question

  1. A student makes a parallel-plate capacitor of capacitance 68 nF from aluminium foil and plastic film by inserting one sheet of plastic film between two sheets of aluminium foil.

    μ

    The aluminium foil and the plastic film are 450 mm wide.

    The plastic film has a thickness of 55 μm and a permittivity of 2.5 × 10–11 C2 N–1 m–2.

    1. (i) Calculate the total length of aluminium foil that the student will require. [3]

      (ii) The plastic film begins to conduct when the electric field strength in it exceeds 1.5 MN C–1. Calculate the maximum charge that can be stored on the capacitor. [2]

    2. The student uses a switch to charge and discharge the capacitor using the circuit shown. The ammeter is ideal.

                                       

      The emf of the battery is 12 V.

        1. The resistor R in the circuit has a resistance of 1.2kΩ. Calculate the time taken for the charge on the capacitor to fall to 50 % of its fully charged value. [3]

        2. The ammeter is replaced by a coil. Explain why there will be an induced emf in the coil while the capacitor is discharging. [2]

        3. Suggest one change to the discharge circuit, apart from changes to the coil, thawill increase the maximum induced emf in the coil. [2]

Answer/Explanation

Ans:(191)

a i length = \(\frac{d\times C}{width\times \epsilon }\) = 0.33 «m» so 0.66/0.67 «m» «as two lengths required»

a ii 1.5 ×106 ×55 ×10-6 = 83 «V»  q «= CV»= 5.6 ×10-6 «C» 8. b i 0.5 = e –\(\frac{1}{Rc}\) = e \(\frac{t}{1200\times 6.8\times 10^{-8}}\) t = «- »1200 × 6.8 × 10-8 ln0.5  5.7× 10−5 «s» OR use of t=\(\frac{1}{2}\) = RC × ln2 1200×6.8×10−8× 0.693  5.7× 10−5 «s»

b ii mention of Faraday’s law  indicating that changing current in discharge circuit leads to change in flux in coil/change in magnetic field «and induced emf»

b iii decrease/reduce  resistance (R) OR capacitance (C)

Question

A negatively charged thundercloud above the Earth’s surface may be modelled by a parallel plate capacitor.

The lower plate of the capacitor is the Earth’s surface and the upper plate is the base of the thundercloud.

The following data are available.

\[\begin{array}{*{20}{l}} {{\text{Area of thundercloud base}}}&{ = 1.2 \times {{10}^8}{\text{ }}{{\text{m}}^2}} \\ {{\text{Charge on thundercloud base}}}&{ = -25{\text{ C}}} \\ {{\text{Distance of thundercloud base from Earth’s surface}}}&{ = 1600{\text{ m}}} \\ {{\text{Permittivity of air}}}&{ = 8.8 \times {{10}^{ – 12}}{\text{ F }}{{\text{m}}^{ – 1}}} \end{array}\]

Lightning takes place when the capacitor discharges through the air between the thundercloud and the Earth’s surface. The time constant of the system is 32 ms. A lightning strike lasts for 18 ms.

a.

Show that the capacitance of this arrangement is C = 6.6 × 10–7 F.

[1]
b.i.

Calculate in V, the potential difference between the thundercloud and the Earth’s surface.

[2]
b.ii.

Calculate in J, the energy stored in the system.

[2]
c.i.

Show that about –11 C of charge is delivered to the Earth’s surface.

[3]
c.ii.

Calculate, in A, the average current during the discharge.

[1]
d.

State one assumption that needs to be made so that the Earth-thundercloud system may be modelled by a parallel plate capacitor.

[1]
 
Answer/Explanation

Markscheme

a.

C = «ε\(\frac{A}{d}\) =» 8.8 × 10–12 × \(\frac{{1.2 \times {{10}^8}}}{{1600}}\)

«C = 6.60 × 10–7 F»

[1 mark]

b.i.

V = «\(\frac{Q}{C}\) =» \(\frac{{25}}{{6.6 \times {{10}^{ – 7}}}}\)

V = 3.8 × 107 «V»

Award [2] for a bald correct answer

[2 marks]

b.ii.

ALTERNATIVE 1

E = «\(\frac{1}{2}\)QV =» \(\frac{1}{2}\) × 25 × 3.8 × 107

E = 4.7 × 108 «J»

ALTERNATIVE 2

E = «\(\frac{1}{2}\)CV2 =» \(\frac{1}{2}\) × 6.60 × 10–7 × (3.8 × 107)2

E = 4.7 × 108 «J» / 4.8 × 108 «J» if rounded value of V used

Award [2] for a bald correct answer

Allow ECF from (b)(i)

[2 marks]

c.i.

Q = «\({Q_0}{e^{ – \frac{t}{\tau }}}\) =» 25 × \({e^{ – \frac{{18}}{{32}}}}\)

Q = 14.2 «C»

charge delivered = Q = 25 – 14.2 = 10.8 «C»

«≈ –11 C»

Final answer must be given to at least 3 significant figures

[3 marks]

c.ii.

I «= \(\frac{{\Delta Q}}{{\Delta t}} = \frac{{11}}{{18 \times {{10}^{ – 3}}}}\)» ≈ 610 «A»

Accept an answer in the range 597 611 «A»

[1 mark]

d.

the base of the thundercloud must be parallel to the Earth surface

OR

the base of the thundercloud must be flat

OR

the base of the cloud must be very long «compared with the distance from the surface»

[1 mark]

Question

A capacitor consists of two parallel square plates separated by a vacuum. The plates are 2.5 cm × 2.5 cm squares. The capacitance of the capacitor is 4.3 pF.

a.

Calculate the distance between the plates.

[1]
b.

The capacitor is connected to a 16 V cell as shown.

Calculate the magnitude and the sign of the charge on plate A when the capacitor is fully charged.

[2]
c.

The capacitor is fully charged and the space between the plates is then filled with a dielectric of permittivity ε = 3.0ε0.

Explain whether the magnitude of the charge on plate A increases, decreases or stays constant.

[2]
d.

In a different circuit, a transformer is connected to an alternating current (ac) supply.

The transformer has 100 turns in the primary coil and 1200 turns in the secondary coil. The peak value of the voltage of the ac supply is 220 V. Determine the root mean square (rms) value of the output voltage.

[3]
e.

Describe the use of transformers in electrical power distribution.

[3]
 
Answer/Explanation

Markscheme

a.

d = «\(\frac{{8.85 \times {{10}^{ – 12}} \times {{0.025}^2}}}{{4.3 \times {{10}^{ – 12}}}}\) =» 1.3 × 10–3 «m»

[1 mark]

b.

6.9 × 10–11 «C»

negative charge/sign

[2 marks]

c.

charge increases

because capacitance increases AND pd remains the same.

[2 marks]

d.

ALTERNATIVE 1

εs = \(\frac{{1200}}{{100}}\) × 220

= 2640 «V»

Vrms = \(\frac{{2640}}{{\sqrt 2 }}\) = 1870 «V»

ALTERNATIVE 2

(Primary) Vrms = \(\frac{{220}}{{\sqrt 2 }}\) = 156 «V»

(Secondary) Vrms = \(\frac{{156 \times 1200}}{{100}}\)

Vrms = 1870 «V»

Allow ECF from MP1 and MP2.

Award [2] max for 12.96 V (reversing Np and Ns).

[3 marks]

e.

step-up transformers increase voltage/step-down transformers decrease voltage

(step-up transformers increase voltage) from plants to transmission lines / (step-down transformers decrease voltage) from transmission lines to final utilizers

this decreases current (in transmission lines)

to minimize energy/power losses in transmission

[3 marks]

Question

The electrical circuit shown is used to investigate the temperature change in a wire that is wrapped around a mercury-in-glass thermometer.

A power supply of emf (electromotive force) 24 V and of negligible internal resistance is connected to a capacitor and to a coil of resistance wire using an arrangement of two switches. Switch S1 is closed and, a few seconds later, opened. Then switch S2 is closed.

a.

The capacitance of the capacitor is 22 mF. Calculate the energy stored in the capacitor when it is fully charged.

[1]
b.

The resistance of the wire is 8.0 Ω. Determine the time taken for the capacitor to discharge through the resistance wire. Assume that the capacitor is completely discharged when the potential difference across it has fallen to 0.24 V.

[3]
c.i.

The mass of the resistance wire is 0.61 g and its observed temperature rise is 28 K. Estimate the specific heat capacity of the wire. Include an appropriate unit for your answer.

[2]
c.ii.

Suggest one other energy loss in the experiment and the effect it will have on the value for the specific heat capacity of the wire.

[2]
Answer/Explanation

Markscheme

a.

«\(\frac{1}{2}C{V^2} = \frac{1}{2} \times 0.22 \times {24^2}\)» = «J»

b.

\(\frac{1}{{100}} = {e^{ – \frac{t}{{8.0 \times 0.022}}}}\)

\(\ln 0.01 = – \frac{t}{{8.0 \times 0.022}}\)

0.81 «s»

c.i.

c = \(\frac{Q}{{m \times \Delta T}}\)

OR

\(\frac{{6.3}}{{0.00061 \times 28}}\)

370 J kg–1 K–1

Allow ECF from 3(a) for energy transferred.

Correct answer only to include correct unit that matches answer power of ten.

Allow use of g and kJ in unit but must match numerical answer, eg: 0.37 J kg–1 K–1 receives [1]

c.ii.

ALTERNATIVE 1

some thermal energy will be transferred to surroundings/along connecting wires/to
thermometer

estimate «of specific heat capacity by student» will be larger «than accepted value»

ALTERNATIVE 2

not all energy transferred as capacitor did not fully discharge

so estimate «of specific heat capacity by student» will be larger «than accepted value»