Lesson 3.7.4.2
3.7.4.2 Parallel plate capacitor Quiz: AQA Physics, Unit 7
20 questions
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Lesson 3.7.4.2, Parallel plate capacitor: 20 multiple choice questions for the AQA Physics (7408), Unit 7: Fields and their consequences (A-level only), written with Revision Ninja.
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The 20 questions
-
Which expression gives the capacitance of a parallel-plate capacitor?
- C = epsilon0 x epsilonr x A/d, where A is plate area and d is plate separation
- C = epsilon0 x A/(epsilonr x d), where A is plate area and d is plate separation
- C = epsilon0 x epsilonr x A x d, where A is plate area and d is plate separation
- C = epsilon0 x epsilonr x d/A, where A is plate area and d is plate separation
-
What is the relative permittivity of a dielectric?
- The ratio of the plate area to the plate separation of the capacitor
- The ratio of the capacitance with the dielectric to the capacitance with a vacuum between the same plates
- The product of the permittivity of free space and the plate separation
- The charge per unit area on each plate when a voltage is applied
-
A parallel-plate capacitor in air has plate area 0.020 m^2 and plate separation 2.0 mm. Taking epsilon0 = 8.85 x 10^-12 F m^-1, what is its capacitance?
- 8.85 pF
- 88.5 pF
- 885 pF
- 8.85 nF
-
The same capacitor, with the same geometry, is now filled with a dielectric of relative permittivity 4. What is its capacitance?
- 1420 pF
- 88.5 pF
- 22.1 pF
- 354 pF
-
What is the value of the permittivity of free space, epsilon0?
- 8.85 x 10^12 F m^-1
- 1.26 x 10^-6 F m^-1
- 4.00 x 10^-7 F m^-1
- 8.85 x 10^-12 F m^-1
-
What happens to the polar molecules of a dielectric when it is placed in an electric field?
- They rotate so that their dipoles line up with the field, partly cancelling it inside the dielectric
- They expand in the field, increasing the separation between the plates
- Their electrons are stripped off, creating free charges that carry a current through the dielectric
- Their ions migrate to the plates and neutralise the charge completely
-
A parallel-plate capacitor has its plate area doubled while the separation is unchanged. What happens to its capacitance?
- It doubles, since capacitance is proportional to plate area
- It halves, since capacitance is inversely proportional to plate area
- It is unchanged, since only the separation affects capacitance
- It quadruples, since capacitance is proportional to area squared
-
A parallel-plate capacitor has its plate area doubled and its plate separation halved. By what factor does its capacitance change, with air between the plates?
- 1 time, so unchanged
- 8 times
- 2 times
- 4 times
-
A capacitor has capacitance 10 pF in air. The gap is then filled with a dielectric of relative permittivity 2.5, geometry unchanged. What is the new capacitance?
- 12.5 pF
- 4 pF
- 25 pF
- 40 pF
-
A student measures the capacitance of a parallel-plate capacitor for several plate separations. What graph shows the expected relationship?
- A horizontal line of C against d, since capacitance does not depend on separation
- A straight line of C against d that does not pass through the origin
- A straight line of C against 1/d through the origin
- A curve of C against d that falls to zero as d increases
-
A parallel-plate capacitor in air has plate area 100 cm^2 and separation 0.50 mm. What is its capacitance?
- 1.77 nF
- 354 pF
- 17.7 pF
- 177 pF
-
A capacitor is connected to a fixed-voltage supply. A dielectric with higher relative permittivity replaces the air between the plates. What happens to the charge stored?
- It becomes zero, because the dielectric blocks charge flow
- It decreases, because the stronger field reduces the charge
- It stays the same, because the charge depends only on the supply
- It increases, because capacitance rises while the pd is fixed
-
A parallel-plate capacitor has plate area 0.05 m^2, separation 1.0 mm and dielectric relative permittivity 3. It is charged to 100 V. What charge is stored?
- 1.33 x 10^-9 C
- 4.43 x 10^-10 C
- 4.43 x 10^-8 C
- 1.33 x 10^-7 C
-
A 300 V supply is connected across an air-filled parallel-plate capacitor with a 3.0 mm gap. What is the electric field strength between the plates?
- 1.0 x 10^6 V m^-1
- 1.0 x 10^5 V m^-1
- 9.0 x 10^4 V m^-1
- 1.0 x 10^4 V m^-1
-
A capacitor is charged in air, then a dielectric of relative permittivity 4 fills the gap. The capacitor is disconnected from the supply first, so its charge stays the same. What happens to the pd?
- It rises to four times its original value
- It stays the same, since the charge is unchanged
- It falls to one half of its original value
- It falls to one quarter of its original value
-
A 1.0 m^2 parallel-plate capacitor in air has plates 1.0 mm apart. Which value is closest to its capacitance?
- about 9 pF
- about 9 nF
- about 90 nF
- about 9 uF
-
A parallel-plate capacitor with air between its plates has a 1.0 cm gap. The dielectric strength of air is 3.0 x 10^6 V m^-1. What is the maximum pd before breakdown?
- 30 kV
- 300 V
- 300 kV
- 3 kV
-
A capacitor is kept connected to a supply of fixed pd while a dielectric of relative permittivity 6 fills it. By what factor does its stored energy change?
- unchanged, since the pd is fixed
- 6 times
- 36 times
- 1/6 times
-
A 50 pF air-filled capacitor has plate separation 0.40 mm. What separation gives a capacitance of 100 pF with the same plate area?
- 0.80 mm
- 0.40 mm
- 0.20 mm
- 0.10 mm
-
Which change gives the largest capacitance for the stated changes to a parallel-plate capacitor?
- Area doubled, separation unchanged, dielectric of relative permittivity 2, giving a factor of 4
- Area unchanged, separation halved, dielectric of relative permittivity 3, giving a factor of 6
- Area and separation unchanged, dielectric of relative permittivity 5, giving a factor of 5
- Area doubled, separation halved, air between the plates, giving a factor of 4
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