Oxford Revise AQA A Level Physics | Chapter P16 answers

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P16: Capacitance

Questions

Answers

Extra information

Mark

AO

Spec reference

01.1

 \(
\begin{array}{lll}
Q & = & CV = 1000 \times 10^{-6} \times 12 = 1.2 \times 10^{-2}\textrm{ C}\\
I & = & \frac{V}{R} = \frac{12\ \rm{V}}{2400\ \Omega} = 5.0 \times 10^{-3}\textrm{ A}\\
\end{array}
\)

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3.7.4.2

01.2

Time for p.d. to drop to half its value

= RC ln2 = 2400 × 1000 × 10−6 × 0.693 = 1.67 s

Graph with scales/labelled axes extending to 6 seconds

Initial p.d. = 12 V and exponential shape by eye

Evidence for p.d. halving in 1.7 s

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3.7.4.4

01.3

Original time constant = RC = 2400 × 1000 × 10–6 = 2.4 s

New time constant = 1.8 × 2.4 = 4.32 s

Effective capacitance = \(\frac{{4.32}}{2400}\) = 1.8 × 10–3 F

Capacitances in parallel add so

Ctotal = C + 1000 × 10-6 F = 1.8 × 10–3 F

C = 0.8 × 10–3 F = 800 μ F

Calculation of new time constant/method involving time constant

Answer

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3.7.4.4

01.4

Assumption is that the voltmeter has infinite resistance

If the voltmeter has a large but finite resistance, this reduces the resistance of the circuit because there are now two resistances in parallel

The time constant will be smaller than it should be, so the unknown capacitance is larger than the value in 06.3

Effect of resistance of voltmeter on resistance of circuit

Effect on capacitance

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3.7.4.4

02.1

From the graph, the time for the p.d. to halve 0.12 ms

Time to halve = RC ln2

\(
\begin{array}{lll}
\textrm{Time constant } = RC & = & \frac{\textrm{(time to halve)}}{0.693}\\
& = & 1.73 \times 10^{-4}\textrm{ s}\\
\end{array}
\)

Use of graph to find time to halve

Answer

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3.7.4.4

02.2

 \(
\begin{array}{lll}
\textrm{C} & = & \frac{\textrm{time constant}}{R}\\
& = & \frac{2.89 \times 10^{-4}}{10 \times 10^3}\\
& = & 2.89 \times 10^{-8}\textrm{ F}\\
\end{array}
\)

Use of time constant to find C

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3.7.4.4

02.3

Curve that starts at half the p.d. on the y-axis, and has

T½ that is double the original value by eye

If the resistance doubles, the maximum current will halve, so the maximum p.d. will halve

If the resistance is doubled, the time constant is doubled, so the time to halve the p.d. is also doubled

Correct line on graph

Explanation of p.d. × \(\frac{1}{2}\)

Explanation of time to halve × 2

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3.7.4.4

02.4

Use the p.d. and resistance to work out the current using \(I = \frac{V}{R}\)

The area under the graph is the charge stored; work out the charge represented by each square using Q = It; count squares and multiply

Conversion of p.d. to current

How to find charge from area

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3.7.4.4

03.1

When the switch is closed, there is a potential difference across the resistor

A current flows, so the charge on the capacitor decreases

As the charge decreases, the p.d. decreases \((V = \frac{Q}{C}),\) so the current decreases in the same way

The rate of change of p.d. depends on the charge, and hence p.d., so the relationship is a negative exponential

Link between p.d. and current

Link to charge on capacitor

Explanation of exponential

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3.7.4.2

03.2

Time / min

p.d. / V

Time / s

ln V

0

2.50

0

0.92

10

1.62

600

0.482426

20

1.08

1200

0.076961

30

0.70

1800

–0.35667

40

0.46

2400

–0.77653

50

0.30

3000

–1.20397

60

0.20

3600

–1.60944

Calculations of t in seconds and lnV

do not penalise excessive significant figures

Graph starting at (0, 0), points plotted, linear line of best fit

Correct labels/units

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3.7.4.4

03.3

 \(
\begin{array}{lr}
V = {V_0}e^{-\;\frac{t}{RC}}\\
\ln \ V = \ln \ V_0 – \frac{t}{RC}\\
\textrm{So a graph of ln } V\textrm{ against } t\\
\textrm{gradient } = -\frac{1}{RC}\\
\end{array}
\)

Taking natural logs of both sides of equation

Gradient correct

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3.7.4.4

03.4

ln(1.5) = 0.405

Reading from graph when ln V = 0.4, t = 700 s

= 11 minutes and 40s

Correct deduction from graph

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3.7.4.4

04.1

 \(
\begin{array}{lll}
E & = & \frac{1}{2}CV^2\\
& = & \frac{1}{2} \times 470 \times 10^{-6} \times (6.0)^{2}\\
& = & 8.46 \times 10^{-3}\textrm{ J} = 8.5 \times 10^{-3}\textrm{ J}\\
\end{array}
\)

Answer

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3.7.4.3

04.2

 \(\textrm{Resistance of lamp } = \frac{6.0}{0.3} = 20\ \Omega\)

Time to discharge to 37% = RC = 470 × 10-6 × 20 = 9.4 × 10–3 s

energy transferred = \(8.46 \times {10^{-3}} \times {0.37^2} = 1.158 \times 10^{-3}\ \rm{J}\)

\(
\textrm{power} = \frac{\textrm{energy}}{\textrm{time}} = \frac{1.158 \times 10^{-3}}{9.4 \times 10^{-3}} = 0.12\ \textrm{W}\)

You may not see this as it is less than a tenth of the bulbs normal power rating

power = 6 V × 0.3 A = 1.8 W

Calculation of resistance

Explicit use of RC as time for p.d. to reduce to 37%

energy transferred from capacitor ∝ V2

Calculation of power

Appropriate comment with numerical comparison

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3.7.4.3

04.3

The energy stored would be multiplied by 4 as energy stored depends on V2

The time is the same

Power would be multiplied by 4; this would definitely be observable but still dim

Reference to E proportional to V2

Effect on what is observed / appropriate comment

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3.7.4.3

04.4

(use of \(E = \frac{1}{2}C{V^2}\))

\(\rm{470\ \mu F\ capacitor\ would\ need\ a\ p.d.\ of\ }V = \sqrt { \frac{2E}{C} } = 650\ \textrm{V}\)

a 6 V p.d. would need a capacitance of \(C = \frac{E}{{2{V^2}}} = 1.4\textrm{ F}\)

1.4 F is a very large capacitor, so the energy stored is achieve by increasing the p.d.

Comment on size of capacitance

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3.7.4.3

05.1

Initially there is no charge on the capacitor, so zero p.d.; as the capacitor charges the p.d. increases as \(V = \frac{Q}{C}\)

And increases at a decreasing rate

Exponential growth by eye

Asymptotic to 12 V

Only a sketch needed, so no values needed on x-axis

Correct description of V proportional to charge

Comment about shape

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3.7.4.4

05.2

Reducing the resistance; as the current normally decreases as the capacitor charges, a smaller resistance is needed to maintain the current at a constant value

The graph will be a straight line through (0, 0) as the p.d. increases at a constant rate

The graph will be horizontal when the capacitor is fully charged

Answer and reason needed for mark

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3.7.4.4

05.3

Procedure described, for example:

  • note the capacitance of the capacitor
  • open the switch and short circuit the capacitor to ensure that it is uncharged
  • close the switch and reduce the resistance of the variable resistor to maintain the current at a constant value
  • when the graph on the computer is horizontal, open the switch
  • use the graph to find the time it took to charge the capacitor from time the p.d. started to rise until the time the p.d. was constant
  • multiply the current by the time to work out the charge
  • replace the capacitor with one of a different capacitance, and repeat
  • repeat for a range of capacitors
  • repeat the experiment three times for each capacitor, and calculate the mean charge stored
  • plot a graph of charge against capacitance

Sufficient detail

Using the graph to find the time

Calculating charge from current and time

Repetition/finding mean

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3.7.4.4

05.4

Appropriate suggestion and solution, for example:

The reading on the ammeter will not be constant as it will be difficult to change the resistance to exactly match the exponential decay of current

Repeating the experiment many more times will give a more accurate measurement

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3.7.4.4

06.1

The water molecule aligns with the electric field between the plates so that the positive side of the molecule (H+) is attracted to the negative plate, and the negative side (O–) is attracted to the positive plate

Movement of molecule to align with field

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3.7.4.2

06.2

The greater the humidity, the greater the capacitance

The water molecules effectively reduce the distance between the plates of the capacitor, and \(C = \frac{{A{\varepsilon_0}{\varepsilon_r}A}}{d},\)

C is inversely proportional to d, so as d decreases, C increases

Relationship

Effect of water molecules on distance

Link to capacitance

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3.7.4.2

06.3

When the humidity is zero, the capacitance is 282 µF, and when it is 100%, the capacitance is 330 mF,

\(
\begin{array}{lr}
C_{100} = \frac{\varepsilon_0 (\varepsilon_{\rm{r}} + \varepsilon_{\rm{w}}) A}{d}\textrm{ and }C_0 = \frac{\varepsilon_0 \varepsilon_{\rm{r}} A}{d}\\
\frac{C_{100}}{C_0} = \frac{\varepsilon_0 (\varepsilon_r + \varepsilon_w)}{d} \times \frac{d}{\varepsilon_0 \varepsilon_{\rm{r}}}\\
\frac{C_{100}}{C_0} = \frac{\varepsilon_{\rm{r}} – \varepsilon_{\rm{w}}}{\varepsilon_{\rm{r}}} = \frac{330}{285} = 1.17\\
\varepsilon_{\rm{r}} – \varepsilon_{\rm{w}} = 1.17\ \varepsilon_{\rm{r}}\\
\varepsilon_{\rm{w}} = 0.17\ \varepsilon_{\rm{r}}\\
\varepsilon_{\rm{r}} = \frac{80}{0.17} = 471\\
\end{array}
\)

Graph starting at (0, 0), points plotted, linear line of best fit

Correct labels/units

Values between 280 mF and 285 mF acceptable

Use of ratios of capacitances

Method

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3.7.4.2

06.4

 \(
\begin{array}{lll}
C_0 & = & \frac{\varepsilon_0 \varepsilon_{\rm{r}} A}{d}\\
d & = & \frac{\varepsilon_0 \varepsilon_{\textrm{r}} A}{C_0}\\
& = & \frac{8.85 \times 10^{-12} \times 470 \times \left( 10.8 \times 10^{-3} \times 3.81 \times 10^{-3} \right)}{282 \times 10^{-6}}\\
& = & 1.7(1) \times 10^{-7}\textrm{ m which is about }0.2 \times 10^{-6}\textrm{ m}\\
\end{array}
\)

Use of equation

Answer/comparison

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06.5

 \({\rm{Field\ strength\ }}\left( {{\rm{assuming\ parallel\ plates}}} \right) = \frac{V}{d}\)

So V = field strength × d = 74 000 × 1.7(1) × 10–7 m = 1.26 × 10–2 V

\(
\begin{array}{lr}
R = \frac{\rho l}{A} = \frac{10^{12} \times 1.7 \times 10^{-7}} {10.8 \times 10^{-3} \times 3.81 \times 10^3} = 4.13 \times 10^{6}\ \Omega\\
I = \frac{V}{R} \frac{1.26 \times 10^{-2}\ {\rm{V}}}{4.13 \times 10^6\ \Omega} = 3.05 \times 10^{-9}\textrm{ A}\\
\end{array}
\)

This is an extremely small current that would be very difficult to measure

Answer = 1.48 × 10–2 V if 0.2 mm used

Answer

Answer

Comment

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3.7.3.3

3.5.1.3

07.1

 \(
\begin{array}{lll}
C & = & \frac{\varepsilon_0 \varepsilon_{\rm{r}} A}{d},\ \varepsilon_\textrm{r} = 1\\
& = & \frac{{8.85 \times {{10}^{-12}} \times 150 \times {{10}^{-4}}}}{{0.1}} = 1.33 \times 10^{-12}\textrm{ V}\\
\end{array}
\)

Q = CV = 1.33 × 10–12 × 4000 = 5.31 × 10–9 C

Calculation of capacitance

Charge

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3.7.4.2

07.2

When the ball touches the plate, electrons are transferred to it, giving the ball a net negative charge and it is attracted to the other plate/repelled from the negative plate

When it touches the positive plate, the electrons are transferred to the plate, so it is repelled from the plate

Transfer of electrons used

Correct attraction/repulsion

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3.5.1.1

07.3

 \(
\begin{array}{lr}
T = 2\,\pi \sqrt{\frac{l}{g}}\\
T = 2\,\pi \sqrt{\frac{{0.3}}{{9.8}}} = 1.1\textrm{ s}\\
\end{array}
\)

So it would take about 0.5 s to travel between the plates

Use of time period

Answer

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3.6.1.3

07.4

 \(\textrm{Current } = \frac{{{\rm{change}}}}{{{\rm{time}}}} = \frac{{0.1 \times 5.31 \times {{10}^{-9}}}}{{0.5}} = 1.1 \times 10^{-10}\textrm{ A}\)

Answer

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3.5.1.1

07.5

The p.d. would decrease

\(C = \frac{{{\varepsilon_0}{\varepsilon_{\rm{r}}}A}}{d}\)

And Q = CV where Q is constant

\(Q = \frac{{V{\varepsilon_0}{\varepsilon_{\rm{r}}}A}}{d}\) \(Qd = V{\varepsilon_0}{\varepsilon_r}A\)

p.d. is proportional to d

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3.7.4.2

08.1

The dielectric would break down/the capacitor will conduct

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3.7.4.2

08.2

 \(V = {V_0}{e^{-\frac{t}{RC}}}\) \(
\begin{array}{lr}
C = \frac{t}{R\left( \ln V_0 – \ln V \right)}\\
t = 7200 s, V = 1.5\textrm{ V},\\
V_0 = 3\textrm{ V}; C = \frac{7200}{{10\,000\left( {\ln \,3 – \ln \,1.5} \right)}} = 1.0\textrm{ F}\\
V_0 = 6\textrm{ V}; C = \frac{7200}{{10\,000\left( {\ln \,6 – \ln \,1.5} \right)}} = 0.51\textrm{ F}\\
\end{array}
\)

3.3 F capacitor

The operating p.d. for the 0.5 F and 1.3 F capacitor is only 3 V

Expression for C, explicit or implied

Values of C for both initial p.d.s

Conclusion with reason

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3.7.4.4

08.3

Q = It = 1400 × 10–3 × 3600 = 5040 C

E = QV = 5040 × 3 = 1.5 × 103 J

E =\(\frac{1}{2}\)CV 2 = 0.5 × 0.5 × 32 = 2.25 J

E =\(\frac{1}{2}\)CV 2 = 0.5 × 1.3 × 32 = 5.85 J

The energy is much less than that stored in the battery by a factor of 500

Calculation of energy

Calculations of energy

Comment

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3.5.1.1

3.7.4.3

08.4

The battery has an internal resistance, r, so if a current flows the p.d. will be reduced by a p.d. of Ir, V = ɛ – Ir

\(
\begin{array}{lr}
\textrm{Current in circuit }I = \frac{\varepsilon }{{R + r}}\\
\textrm{Terminal pd} = V = \varepsilon – Ir\\
V = \varepsilon – \frac{\varepsilon}{R + r}r\\
\textrm{So }\frac{r}{{R + r}} = \frac{1}{2}\\
\end{array}
\)

R is equal to the internal resistance of the battery

Explanation involving internal resistance

Use of equation

Answer

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3.5.1.6

Skills box answers

Question

Answer

1(a)

Graph will look similar to the one in the worked example column, with a negative gradient and an intercept on the y-axis at 2.33.

1(b)

Gradient of graph should be −0.02. This is equal to \( – \frac{1}{{RC}},\) so RC = 50 s.

1(c)

 \(\textrm{Capacitance } = \frac{50}{R} = \frac{50}{{\left( {330 \times {{10}^3}} \right)}}\Omega = 1.5 \times {10^{-4}}\textrm{ F}.\)

2

Electrolytic capacitors must be connected using the polarity marked on the capacitor. Otherwise the capacitor will overheat and be damaged.

3

The value for RC using this capacitor and resistor is 2.9 s, so the voltage will decay very quickly and be difficult to measure using a multimeter and stopclock. However, a data logger could be set to take 10 measurements per second and so could be used.
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