Capacitance

Question 1

State what is meant by a capacitance of 370 μF

Answer 1

(Refers to a capacitor that) stores/holds/changes by 370 μC of charge ✔
For every (1) volt/volt change (of pd across its plates) ✔
OR
Reference to charge to pd OR charge to voltage ratio ✔ includes units C or coulomb and V or volt ✔

Question 2

for a capacitor charging. what is the current against time graphs shape

Answer 2

decreasing curve. No pd opposing the current at first so it starts off high. As the capacitor charges the pd across the the capacitor gets bigger, the pd across the resistor gets smaller and therefore the current drops.

Question 3

An air-filled parallel-plate capacitor is charged from a source of emf. The electric field has a strength E between the plates. The capacitor is disconnected from the source of emf and the separation between the isolated plates is doubled.
What is the final electric field between the plates?

A
2E

B
E

C
E/2

D
E/4

Answer 3

B

Question 4

A parallel-plate capacitor has square plates of length l separated by distance d and is filled with a dielectric.
A second capacitor has square plates of length 2l separated by distance 2d and has air as its dielectric.
Both capacitors have the same capacitance.
What is the relative permittivity of the dielectric in the first capacitor?

A
1/2

B
1

C
2

D
8

Answer 4

C

Question 5

Explain what is meant by a capacitance of 120 µF.

Answer 5

the capacitor stores 120 µC of charge for each one volt of pd between the plates ✔

Question 6

Explain why there is an increase in the energy stored by the capacitor when the polythene sheet is pulled out from between the plates.

Answer 6

Energy increases because:
In the polar dielectric molecules align in the field with positive charged end toward the negative plate (or WTTE).✓
Work is done on the capacitor separating the positively charged surface of the dielectric from the negatively charged plate (or vice versa).✓

Question 7

The student intends to measure the potential difference (pd) across the capacitor at 10 s intervals while adjusting the variable resistor to keep the charging current constant.
The power supply has an emf of 6.0 V and negligible internal resistance. The capacitor has a capacitance of 680 µF. The variable resistor has a maximum resistance of 100 kΩ.
(a) The student chooses a digital voltmeter for the experiment. A digital voltmeter has a very high resistance.
Explain why it is important to use a voltmeter with very high resistance.

Answer 7

Capacitor must not lose charge through the meter ✓

Question 8

Suggest one advantage of using an analogue ammeter rather than a digital ammeter for this experiment.

Answer 8

Position on scale can be marked / easier to read quickly etc ✓

Question 9

The student decides to confirm the value of the capacitance by first determining the time constant of the circuit when the capacitor discharges through a fixed resistor.
Describe an experiment to do this. Include in your answer:
• a circuit diagram
• an outline of a procedure
• an explanation of how you would use the data to determine the time constant.

Answer 9

Suitable circuit drawn ✓
Charge C then discharge through R and record V or I at 5 or 10 s intervals ✓
Plot ln V or ln I versus time ✓
gradient is 1 / RC ✓
OR
Suitable circuit drawn ✓
Charge C then discharge through R and record V or I at 5 or 10 s intervals✓
Use V or I versus time data to deduce half-time to discharge ✓
1 / RC = ln 2 / t½ quoted ✓
OR
Suitable circuit drawn ✓
Charge C then discharge through R and record V or I at 5 or 10 s intervals ✓
Plot V or I against t and find time T for V or I to fall to 0.37 of initial value ✓
T = CR ✓

Question 10

Explain how the initial charge on the capacitor could be determined from a graph of current against time.

Answer 10

determine area under the graph

[or determine area between line and time axis] ✓

Question 11

Explain why the rate of change of pd between the capacitor plates decreases as the capacitor discharges.

Answer 11

Current decreases (I = V / R) / describes rate of flow of electrons decreasing / rate of flow of charge decreases
M1
Charge lost more slowly so pd falls more slowly because V∝Q or Q=CV where C is constant

Question 12

The discs are replaced by ones of larger area placed at the same separation, to give a larger capacitance.
State and explain what effect this increased capacitance will have on:
(i) the time between consecutive discharges,

Answer 12

time increases
(larger C means) more charge required (to reach breakdown pd)
Mark sequentially i.e. no explanation mark if effect is wrong. or time ∝ capacitance