Capacitance and the parallel plate capacitorAQA A-Level Physics: Subtopic test

1. 1
   A 470 µF capacitor is connected across a 12 V d.c. supply until it is fully charged.

   (a)

   What is the charge stored on one plate of the capacitor?

   [1 mark]

   • A3.9 × 10⁻⁵ C
   • B2.6 × 10⁴ C
   • C5.6 × 10⁻³ C
   • D5.6 × 10³ C

   (b)

   Which of the following is equivalent to the farad?

   [1 mark]

   • AC V⁻¹
   • BV C⁻¹
   • CA s V
   • DJ C⁻¹

   (c)

   Define the capacitance of a capacitor and state what the charge Q in the definition refers to.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A parallel plate capacitor has two identical square plates, each of area 0.020 m², separated by an air gap of 0.50 mm. Treat the relative permittivity of air as 1.0. ε₀ = 8.85 × 10⁻¹² F m⁻¹.

   (a)

   Calculate the capacitance of the capacitor.

   [1 mark]

   • A3.5 × 10⁻¹³ F
   • B3.5 × 10⁻¹⁰ F
   • C8.9 × 10⁻¹⁷ F
   • D1.8 × 10⁻¹⁰ F

   (b)

   The plate separation is doubled and the plate area is kept the same. What happens to the capacitance?

   [1 mark]

   • AIt doubles
   • BIt stays the same
   • CIt falls to a quarter
   • DIt halves

   (c)

   A sheet of dielectric of relative permittivity 4.0 is inserted so that it completely fills the gap. Calculate the new capacitance.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   An air-filled parallel plate capacitor has a capacitance of 2.4 nF. It is charged to 160 V and then disconnected from the supply. A sheet of dielectric made of polar molecules is then slid between the plates so that it fills the gap, and the capacitance rises to 9.6 nF.

   (a)

   Explain how the polar molecules of the dielectric increase the capacitance of the capacitor.

   [3 marks]

   (b)

   Calculate the relative permittivity of the dielectric and the potential difference across the capacitor after the dielectric has been inserted.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A capacitive pressure sensor has two parallel plates, each of area 4.0 × 10⁻⁴ m², separated by a compressible foam that can be treated as having a relative permittivity of 1.0. With no load the plate separation is 0.20 mm. Under load the separation falls to 0.15 mm. The sensor is connected to a 5.0 V supply. ε₀ = 8.85 × 10⁻¹² F m⁻¹.

   (a)

   Calculate the capacitance of the sensor with no load and under load, and the percentage increase in capacitance caused by the load. Explain why the capacitance increases.

   [6 marks]

   (b)

   Calculate the extra charge that flows onto the plates when the load is applied. A designer proposes replacing the foam with a compressible material of relative permittivity 3.0. Evaluate this proposal.

   [6 marks]

   Total for question 4: 12 marks

End of questions