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23.1 Capacitance AQA A2 Physics P23 Capacitors Kerboodle Answers
1 a 5.0 μF
b 2.2 v
c 9.9 mC
d 1.4 μF
2
a the charge stored on the capacitor at 12.0 V,
264 μC
b the time taken.
106 s
3
a the charge stored,
27.5 μC
b the capacitance of the capacitor.
5.5 μF
4
a charge stored at a pd of 4.2 V,
0.91 μC
b the capacitance of the capacitor,
0.22 μF
c the extra charge stored at a current of 14
700 μC
d the pd after the extra charge was stored.
7.4 v
23.2 Energy stored in a charged capacitor AQA A2 Physics P23 Capacitors Kerboodle Answers : Page No. 383
1 a 3.OV,
30 μC, 45 μJ
b 6.OV.
60 μC, 180 μJ
2
a the charge and energy stored in the capacitor before discharge,
0.45 C, 2.0 J
b the average power supplied to the light bulb.
10 w
3
a the charge and energy
i stored in the capacitor,
6.6 μC. 9.9 μJ
ii supplied by the battery.
6.6 μC, 19.8 μJ
4
a the initial charge and energy stored in the 4.? UF capacitor,
56 μC, 338 μJ
b i the final charge stored by each capacitor,
2.2 μF; 18 μC, 4.7 μF; 38 ~ μC
ii the final pd across the two capacitors,
8.2 v
c the final energy stored in each capacitor.
Explain the loss of energy stored.
2.2 μF; 73 μJ, 4.7 μF; l 57 μJ
23.3 Charging and discharging a capacitor through a fixed resistor AQA A2 Physics P23 Capacitors Kerboodle Answers : Page No. 387
1
a i the charge stored in the capacitor immediately after it has been charged,
300 μC
ii the time constant of the circuit.
5.0 s
b i Estimate how long the capacitor would take to discharge to about 2 V.
5 s approx
ii Estimate the resistance of the resistor that you would use in place of the 100 k’) resistor if the discharge is to be 99% completed within about 5 s.
20 kQ
2
i the charge stored by the capacitor at a pd of 9.0 V,
0.61 mC
ii the initial discharge current.
0.45 mA
0.23 V, 11 μA
3 a the charge and energy stored in this capacitor at 6.0 V,
13 μC, 40 μJ
b the pd across the capacitor 0.5 s after the discharge started,
0.62 V, 0.42 μJ
4 a i initial charging current,
60 mA
ii the energy stored in the capacitor at 12.0 V,
0.34 mJ
b the time taken for the pd to fall from 12.0 V to 3.0 V,
1.4 s
c the energy lost by the capacitor in this time.
0.32 mJ
1 a C increases
ii the charge stored by the capacitor.
Q increases
b
The energy stored = 1/2QV, so the energy stored increases because Q increases and V is unchanged.
2
The capacitor is isolated from the battery, so the charge stored does not change.The capacitance decreases when the dielectric is removed. The energy stored , so it increases because C decreases and Q is unchanged. The increase of energy is due to the work done to overcome the electrostatic attraction between the charge on each dielectric surface and the opposite type of charge on the adjacent plate.
3 a 9.8 pf
b Calculate the energy stored by the capacitor in (a) when the pd across it is 15.0 V.
1100 pJ
4
a 0. 14μm
b 2.2 MJ m
-3
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Practice Questions: Page No. 392-395
1
(a)
(b)
(i) Complete the missing entries in Table 1.
(ii) Use the measurements to plot a graph of In Von the y-axis against r on the x-axis.
For correct labels on each axis
- for suitable scales
- for correctly plotted points
- for best-fit line.
(iii) Use your graph to determine the time constant of the discharge circuit.
(iv) The resistance R of the resistor was 68k’). Determine the capacitance C of the capacitor. (10 marks)
(c) (i) Discuss the reliability of the measurements.
The range of each set of readings is no more than 0.03 V except for the reading at t = 150 s which is 0.12 V. This exception is probably due to an anomalous reading.
The readings are therefore reliable because the range of each set of readings is very small compared with the mean value.
(ii) Estimate the accuracy of your value of capacitance, given the resistor value is accurate to within 1%.
Apart from the exception at t = 150 s, the uncertainty of the readings is therefore no more than ± 0.02 V.
or:
At t = 0 the values of ln V0 are 1.504, 1.506, and 1.504; a mean of 1.505 and an uncertainty of 0.001 (half the range).
Using the values at 300 s as an example, the values of ln V are −0.117, −0.105, and −0.139; a mean of −0.120 and an uncertainty of 0.017 (half the range).
In finding the gradient, the change in ln V is divided by the time taken. The uncertainty in the change in ln V from t = 0 to 300 s is about
0.017 + 0.001 = 0.018 and as a percentage
The error in t is not known, but taking the largest value of t = 300 s, this error is likely to be small (if timing was by hand then the uncertainty in t will be about 0.3 s or 0.1%) given R is accurate to 1%, the value of C is accurate to within 2.3% (= 1.3% + 1%) or about 2%.
2
(a) Calculate the charge on the capacitor. (2 marks)
Charge on capacitor Q = CV
= 2.0 × 10
-6 × 150
= 3.0 × 10
-4 C (300 μC)
(b) (i)
The required graph is of pd V on the vertical axis and charge Q on the horizontal axis.
(ii)
(c)
(i)
(ii)
3 (a) (i)
(ii)
Time constant = RC
= 470 × 10
3 × 330 × 10
-6 = 155 s
(iii)
(b) (i)
time constant = RC = 1.0 × 10
3 × 330 × 10
-6
= 0.33 s
Graphs to show:
- correct axes and labels on both
- exponential growth for pd
- exponential decay for current
- value dropped to 0.37 of maximum after RC
for current, and increased to 0.63 of maximum after RC for pd.
(ii)
4
(a) the initial charge stored, (2 marks)
Use of Q = CV gives
charge stored Q = 4.7 × 10
-6 × 6.0
= 2.82 × 10
-5 C or 2.8 × 10
-5 C
(b)
(c)
The time constant is the time taken for V to decrease from V0 to (V0/e)
∴V must fall to (6.0/e) = 2.2 V
From the graph, V = 2.2 V when t = 32 ms
∴ time constant = 32 ms
(The graph shows a value between 32 and 33 ms)
(d)
5
(b)
(c) (i)
(ii)
It is reasonable to assume that the capacitor has completely discharged in time T/2 because either
- 40.9 mV is only 0.34% of 12 V, or
- 1.25 ms is almost 6 time constants
6 (a)
A capacitor has a capacitance of 1 farad if it stores 1 coulomb of charge when the pd across it is 1 V.
(b) (i) the charge it stores.
Charge stored Q = CV
= 2.3 × 10
-11 × 6.0
= 1.38 × 10
-10 C or
1.4 × 10
-10 C
(ii) the energy it stores. (4 marks)
(c) (3 marks) AQA 2003
7
(a)
(i)
(ii)
Work done by battery = QV
= 2 × (energy stored by capacitor)
= 2 × 1.215 × 10
-3 = 2.43 × 10
-3 J
or 2.4 × 10
-3 J
(b) (i)
= 0.01 of the initial energy E0
∴ energy released by capacitor = 0.99 E0, which is almost all of the initial energy.
(ii)
(iii)
8 (a)
(b) (i)
(ii)
The energy is dissipated in the 680 Ω resistor.
This energy becomes internal energy of the resistor, and eventually internal (thermal) energy in the surroundings.
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