Oxford Revise AQA A Level Physics | Chapter P15 answers

P15: Electric fields

Question

Answers

Extra information

Mark

AO

Spec reference

01.1

Direction of arrow from centre of gold nucleus outwards

Judge by eye

1

3.7.1

AO1

01.2

\(
\begin{aligned}
& 6.2\text{ MeV }= 6.2 \times 10^{6} \times 1.6 \times 10^{-19}\mathrm{\ J}\\
& {E_k} = \frac{1}{2}m{v^2}\\
& {v^2} = \frac{2 \times E_{\mathrm{k}}}{m} = \frac{2 \times 6.2 \times 10^6 \times 1.6 \times 10^{19}}{6.64 \times 10^{-27}\mathrm{\ kg}}\\
& \textit{v} = 1.73 \times 10^7\text{ m s}^{-1}\\
\end{aligned}
\)

1

1

3.1.1

3.2.1.2

AO2

01.3

\(
\begin{aligned}
& \Delta\textit{W} = \textit{Q}\Delta\textit{V} so EPE = \textit{V} \times \textit{Q}\\
& \frac{1}{2}m{v^2} = \frac{Qq}{4\pi \varepsilon_0 r}\\
& \frac{1}{2}m{v^2} = \frac{Ze \times 2e}{4\pi \varepsilon _0r_c}\\
& r_c = \frac{Ze^2}{\pi \varepsilon _0mv^2}\\
\end{aligned}
\)

Must be clear how the 4 cancelled – watch for 2 disappearing

1

1

1

3.7.3.3

AO3

01.4

\(
\begin{aligned}
& Z = 79\\
& {r_c} = \frac{Ze^2}{\pi \varepsilon_0\ mv^2} = \frac{79 \times \left( 1.6 \times 10^{-19} \right)^2}{\pi \times 8.85 \times 10^{-12} \times 6.64 \times 10^{-27} \times \left( 1.73 \times 10^7 \right)^2}\\
& \textit{r}_{c} = 3.7 \times 10^{-14}\\
\end{aligned}
\)

allow e.c.f from 01.2

if Z = 197 is used, deduct one mark (then rc would = 9.13 × 10−14)

1

1

3.7.3.3

AO2

02.1

Lines leaving spheres perpendicular to surface

Arrows point away from positive

Suitable pattern between repelling spheres

1

1

1

3.7.3.2

AO1

02.2

One problem with

One related solution

e.g. difficulty of affecting the field using metal instruments

Use wooden/plastic ruler

Difficulty in measuring distances between curved objects

Set up ruler with set squares fixed or use light and measure distance between shadows

1

1

PS1.1 ATc

AO3

02.3

\(
\begin{aligned}
& F \propto \frac{1}{r^2}\text{ or }F = \frac{Q_1 Q_2}{4\pi \varepsilon_0 r^2}\\
& Fr^2 = \text{ constant or }mr^2 = \text{ constant}\\
\end{aligned}
\)

Data tested for at least 3 data sets and conclusion

\(
\begin{aligned}
& \text{e.g. }0.02^2 \times (0.0827 \times 9.81) = 3.2 \times 10^{-4} (\text{N m}^2)\\
& 0.025^2 \times 0.053 \times 9.81 = 3.2 \times 10^{-4}\\
& 0.030^2 \times 0.0368 \times 9.81 = 3.2 \times 10^{-4}\\
\end{aligned}
\)

Constant = 33 000 g mm2 (if you don’t change units)

1

1

1

3.7.3.1

AO2

02.4

\(
\begin{aligned}
& F = \frac{Q_1 Q_2}{4\pi \varepsilon_0 r^2}\\
& Q^2 = Fr^2 4\pi \epsilon_0\\
& Q^2 = 3.2 \times 10^{-4} \times 4\pi \times 8.85 \times 10^{-12}\\
& Q = 1.9 \times 10^{-7}\mathrm{\ C}\\
\end{aligned}
\)

Allow using a pair of values from table for full marks

1

1

3.7.3.1

AO2

03.1

The potential difference between the lines is constant but the distance is not

1

3.7.3.3

AO2

03.2

At least 4 lines drawn perpendicular to surface of the cable and equipotentials

Arrows pointing away from the cable

1

1

3.7.3.2

AO1

03.3

\(
\begin{aligned}
& V \propto \frac{1}{r}\\
& \textit{Vr} = \text{ constant}\\
& 300 \times 10 = 200 \times d\\
& d = 15\mathrm{\ cm}\\
\end{aligned}
\)

1

1

3.7.3.3

AO2

03.4

P is at a distance of 12.5 cm

300 × 10 = 12.5 × V

V = 240 V

1

1

3.7.3.3

AO2

04.1

At least 6 lines drawn – equidistant

Arrows pointing down

ignore field outside/near edge of plates

1

1

3.7.3.2

AO1

04.2

Path deflected upwards

ignore size of deflection

1

3.7.3.2

AO1

04.3

\(
\begin{aligned}
& \text{Use of }E = \frac{F}{Q} = \frac{V}{d}\text{ or }F = ma\\
& F = \frac{VQ}{d}\\
& ma = \frac{VQ}{d}\\
& a = \frac{VQ}{md}\\
& a = \frac{1500 \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31} \times 0.025} = 1.1 \times 10^{16} \text{ m s}^{-2}\\
\end{aligned}
\)

1

1

1

3.7.3.2

3.4.1.5

AO2

04.4

\(
\begin{aligned}
& \text{Time between plates }= \frac{\text{length of plates}}{\text{speed of electrons}}\\
& t = \frac{0.04}{3 \times 10^7} = 1.3 \times 10^{-9}\mathrm{\ s}\\
\end{aligned}
\)

Use of suvat for vertical displacement

\(
\begin{aligned}
& s = ut + \frac{1}{2}\ at^2 \textbf{ and } u = 0\\
& s = \frac{1}{2} \times 1.1 \times 10^{16} \times (1.3 \times 10^{-9})^2\\
& s = 0.01\mathrm{\ m\ }= 10\text{ mm or }0.0098\mathrm{\ m }= 9.8\mathrm{\ mm}\\
\end{aligned}
\)

Distance from top plate = 12.5 mm – 10 mm = 2.5 mm or 2.7 mm)

Use of rounded numbers gives s = 8.5 mm and so final answer = 4 mm

1

1

1

1

3.4.1.4

AO2

05.1

\(
\begin{aligned}
& \text{Use of C }= \frac{Q}{V}\\
& V = \frac{Q}{4\pi \varepsilon_0 R}\\
& C = Q \times \frac{4\pi \varepsilon_0 R}{Q} = 4\pi \varepsilon_0 R\\
\end{aligned}
\)

Clear substitution seen for second mark

1

1

3.7.4.1

3.7.3.3

AO2

05.2

C = 4πε0R = 4 × π × 8.85 × 10–12 × 0.20 = 2.2 × 10–11

F (Farads)

1

1

3.7.4.1

AO1

05.3

\(
\begin{aligned}
& E = \frac{V}{r}\\
& V = Er = 3 \times 10^6 \times 0.20 = 6 \times 10^5\mathrm{\ V}\\
\end{aligned}
\)

1

1

3.7.3.2

AO2

05.4

\(
\begin{aligned}
& \text{Use of } Q = VC = 2.2 \times 10^{-11} \times 6 \times 10^5 = 1.3 \times 10^{-5}\mathrm{\ C}\\
& \text{Number of excess charges } = \frac{1.3 \times 10^{-5} C}{1.6 \times 10^{-19}\mathrm{\ C}}\ = 8.3 \times 10^{13}\\
\end{aligned}
\)

Be aware of possible e.c.f from answer to 05.2 and 05.3

Could also use \(V = \frac{Q}{4\pi \varepsilon_0 R}\)

1

1

3.7.4.1

AO2

06.1

\(
\begin{aligned}
& F = \frac{Q_1 Q_2}{4\pi \varepsilon_0 R^2}\\
& F = \frac{\left( 1.6 \times 10^{-19} \right)^2}{4\pi \times 8.85 \times 10^{-12} \times \left( 5.3 \times 10^{-11} \right)^2}\\
& F = 8.2 \times 10^{-8}\mathrm{\ N}\\
\end{aligned}
\)

1

1

3.7.3.1

AO2

06.2

8.2 × 10–8 N

e.c.f same as 06.1

ignore minus sign

1

3.7.3.1

AO1

06.3

\(
\begin{aligned}
& F = ma\\
& a = \frac{F}{m} = \frac{8.2 \times 10^{-8}\mathrm{\ N}}{9.11 \times 10^{-31}\mathrm{\ kg}} = 9.0 \times 10^{22}\text{ m s}^{-2}
\end{aligned}
\)

1

3.4.1.5

AO2

06.4

\(
\begin{aligned}
& \text{Total energy}, E = E_{\mathrm{k}} + E_{\mathrm{p}}\\
& \Delta W = Q \Delta V\\
& E = \frac{1}{2} mv^2 – \frac{e^2}{4\pi \varepsilon_0 r}\\
& \text{since } \frac{mv^2}{r} = \frac{e^2}{4\pi \varepsilon_0 r^2}\\
& mv^2 = \frac{e^2}{4\pi \varepsilon_0 r}\\
& E = \frac{e^2}{2 \times 4\pi \varepsilon_0 r} – \frac{e^2}{4\pi \varepsilon_0 r} = -\frac{e^2}{8\pi \varepsilon_0 r}\\
& E = \frac{\left( 1.6 \times 10^{-19} \right)^2}{4\pi \times 8.85 \times 10^{-12} \times 5.3 \times 10^{-11}}\\
& E = 2.2 \times 10^{-18}\mathrm{\ J}\\
& E = \frac{2.2 \times 10^{-18}\mathrm{\ J}}{1.6 \times 10^{-19}\mathrm{\ J}} = 13.57\mathrm{\ eV}\\
\end{aligned}
\)

Also credit for full marks use of \(\frac{1}{2}m{v^2}\) and

\(V = \frac{Q}{4\pi \varepsilon_0 r} (E_\mathrm{p} = Q\frac{Q}{4\pi \varepsilon_0r})\)

1

1

1

1

3.7.3.3

3.1.1

AO3

07.1

Lines drawn at \(\frac{1}{4}, \frac{1}{2},\text{ and }\frac{3}{4}\) points and correctly labelled

1

3.7.3.3

AO1

07.2

\(
\begin{aligned}
& E = \frac{V}{d}\\
& = \frac{40}{0.01} = 400\text{ V m}^{-1}\\
\end{aligned}
\)

1

1

3.7.3.2

AO1

07.3

ΔW = QΔV

= 1.6 × 10–19 × 40 V = 6.4 × 10–18 J

1

1

3.7.3.3

AO2

07.4

Max 3 from:

Electron is attracted by B/repelled by A/experiences force to the right

Electron decelerates (initially)

Electron does not reach A/stops/reverses direction

Stops at half way point (20 eV)

When it returns it has 20 eV

max 3

3.7.3.3

AO3

08.1

Is the work done per unit positive charge when it is moved from infinity to that point

Must include positive

1

3.7.3.3

AO1

08.2

\(
\begin{aligned}
& V \propto \frac{1}{r}\\
& Vr = \text{ constant}\\
\end{aligned}
\)

Data checked at least three times and conclusion,

\(
\begin{aligned}
& \text{e.g. }1800 \times 0.01 = 18\\
& 600 \times 0.03 = 18\\
& 300 \times 0.06 = 18\\
\end{aligned}
\)

1

1

1

3.7.3.3

AO2

08.3

\(
\begin{aligned}
& V = \frac{Q}{4\pi \varepsilon_0 r}\\
& Q = V \times 4\pi \varepsilon_0 r = 18 \times 4 \times \pi \times 8.85 \times 10^{-12}\\
& Q = 2.0 \times 10^{-9}\mathrm{\ C}\\
& Q = 2\mathrm{\ nC}\\
\end{aligned}
\)

1

1

3.7.3.3

3.1.1

AO2

08.4

Draw a tangent to the curve at 3 cm

Calculate the gradient of the tangent,

\(
\begin{aligned}
& \text{e.g. }\frac{1180}{0.068} = 1.7 \times 10^{4}\text{ V m}^{-1}\\
& \pm 0.3 \times 10^4 V m^{-1}\\
\end{aligned}
\)

Allow 170 if units quoted as V cm–1

1

1

1

3.7.3.3

AO3

08.5

V at 6 cm = 300 V

ΔW = QΔV = 4 × 10–9 × 300 = 1.2 × 10–6 J

1

1

3.7.3.3

AO2

Skills box answers

Question

Answer

1

\(
\begin{aligned}
& F = \frac{1}{4\pi \varepsilon_0 r} \frac{Qq}{r^2}\\
& Q = +25 \times 10^{-6}\mathrm{\ C}; q = +100 \times 10^{-6}\mathrm{\ C}\\
& \varepsilon_0 = 8.85 \times 10^{-12} F m^{-1}; r = 60 \times 10^{-3}\mathrm{\ m}\\
& F = \frac{2.5 \times 10^{-5} \left( 1.00 \times 10^{-4} \right)}{\left( 6.0 \times 10^{-2} \right)^2}\\
& F = 5.5 \times 10^4\mathrm{\ N}\\
\end{aligned}
\)

2(a)

The force is attractive because the charges have opposite signs.

2(b)

\(
\begin{aligned}
& F = \frac{1}{4\pi \times 8.85 \times 10^{-12}} \frac{4.0 \times 10^{-9} \times \left( -8.0 \times 10^{-9} \right)}{\left( 80 \times 10^{-3} \right)^2}\\
& F = 4.5 \times 10^{-5}\mathrm{\ N}\\
\end{aligned}
\)

3

\(F \propto \frac{1}{r^2}\) so if the distance doubles, the force will decrease by \(\frac{1}{\left( 2 \right)^2}\). The new force will be 10 N.

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