Fields (gravitational, electric, capacitors, magnetic fields, EM induction) – A-level Physics exam questions with mark scheme

1Core3 marks

Mars has a mass of 6.42 × 10$^{23}$ kg and a radius of 3.39 × 10$^6$ m.

(a) Calculate the gravitational field strength at the surface of Mars. [2]

(b) Calculate the gravitational potential at the surface of Mars. [1]

(a) $g = \frac{GM}{r^2} = \frac{6.67\times10^{-11} \times 6.42\times10^{23}}{(3.39\times10^6)^2}$(1)
(a) $g$ = 3.7 N kg$^{-1}$ (3.73) (allow m s$^{-2}$)(1)
(b) $V = -\frac{GM}{r}$ = −1.3 × 10$^7$ J kg$^{-1}$ (minus sign required)(1)
Answer: (a) 3.73 N kg$^{-1}$ (b) −1.26 × 10$^7$ J kg$^{-1}$
Examiner tip: Gravitational potential is always negative — the minus sign is part of the mark.

2Standard5 marks

A geostationary satellite orbits the Earth. The mass of the Earth is 5.97 × 10$^{24}$ kg.

(a) Show that the radius of a geostationary orbit is about 4.2 × 10$^7$ m. [3]

(b) State two features of a geostationary orbit, other than its radius. [2]

(a) $\frac{GMm}{r^2} = m\omega^2 r$ or $T^2 = \frac{4\pi^2}{GM}r^3$(1)
(a) $T$ = 24 h = 86 400 s used (allow 86 164 s, giving 4.21 × 10$^7$ m)(1)
(a) $r = \sqrt[3]{\frac{6.67\times10^{-11}\times5.97\times10^{24}\times86\,400^2}{4\pi^2}}$ = 4.22 × 10$^7$ m (at least 3 sf)(1)
(b) any one feature from: period of 24 hours (one day) / same angular speed as the Earth's rotation; orbits directly above the equator (in the equatorial plane); moves west to east, in the same direction as the Earth rotates; stays above the same point on the Earth's surface(1)
(b) a second, different feature from the list above ('period 24 h' and 'same angular speed as the Earth' count as the same feature; ignore uses such as 'used for communications/TV')(1)
Answer: (a) 4.22 × 10$^7$ m (b) any two of: period 24 h; above the equator; same direction as Earth's rotation; stays above the same point on the surface
Examiner tip: In a 'show that', give your answer to one more significant figure than the value quoted.

3Hard4 marks

A satellite of mass 500 kg is moved from the Earth's surface (radius 6.37 × 10$^6$ m) into a circular orbit of radius 7.00 × 10$^6$ m. The mass of the Earth is 5.97 × 10$^{24}$ kg.

(a) Calculate the increase in gravitational potential energy of the satellite. [2]

(b) Calculate the orbital speed of the satellite. [2]

(a) $\Delta E = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$ with correct substitution(1)
(a) $\Delta E$ = 2.8 × 10$^9$ J (2.81)(1)
(b) $\frac{GMm}{r^2} = \frac{mv^2}{r}$ so $v = \sqrt{\frac{GM}{r}}$(1)
(b) $v$ = 7.5 × 10$^3$ m s$^{-1}$ (7540)(1)
Answer: (a) 2.81 × 10$^9$ J (b) 7.54 × 10$^3$ m s$^{-1}$
Examiner tip: Use the change in potential between two radii — mgh is not valid here because g is not uniform over 630 km.

4Standard4 marks

Two point charges, +3.0 nC and −5.0 nC, are placed 0.12 m apart in air.

(a) Calculate the magnitude of the electrostatic force between the charges and state whether it is attractive or repulsive. [2]

(b) Calculate the electric potential at the point midway between the charges. [2]

(a) $F = \frac{Q_1Q_2}{4\pi\varepsilon_0 r^2} = \frac{3.0\times10^{-9}\times5.0\times10^{-9}}{4\pi \times 8.85\times10^{-12}\times0.12^2}$(1)
(a) $F$ = 9.4 × 10$^{-6}$ N, attractive(1)
(b) potentials add as scalars: $V = \frac{1}{4\pi\varepsilon_0 (0.060)}(3.0\times10^{-9} - 5.0\times10^{-9})$(1)
(b) $V$ = −300 V (sign required)(1)
Answer: (a) 9.4 × 10$^{-6}$ N, attractive (b) −300 V
Examiner tip: Potential is a scalar — add the two values with their signs; don't treat it like a vector.

5Hard5 marks

An electron enters the region between two parallel plates, midway between them, travelling at 2.5 × 10$^7$ m s$^{-1}$ parallel to the plates. The plates are 4.0 cm apart and 6.0 cm long, with a potential difference of 600 V between them.

(a) Calculate the electric field strength between the plates. [1]

(b) Calculate the acceleration of the electron between the plates. [1]

(a) $E = \frac{V}{d} = \frac{600}{0.040}$ = 1.5 × 10$^4$ V m$^{-1}$ (allow N C$^{-1}$)(1)
(b) $a = \frac{eE}{m} = \frac{1.60\times10^{-19}\times1.5\times10^4}{9.11\times10^{-31}}$ = 2.6 × 10$^{15}$ m s$^{-2}$(1)
(c) time between plates $t = \frac{0.060}{2.5\times10^7}$ = 2.4 × 10$^{-9}$ s (horizontal velocity constant) (alternative method: time to deflect 2.0 cm $t = \sqrt{\frac{2\times0.020}{a}}$ = 3.9 × 10$^{-9}$ s)(1)
(c) vertical deflection $s = \frac{1}{2}at^2$ = 7.6 × 10$^{-3}$ m (allow 7.5 × 10$^{-3}$ m from rounded $a$ = 2.6 × 10$^{15}$; ecf from (b); alternative method: horizontal distance travelled in 3.9 × 10$^{-9}$ s = 2.5 × 10$^7$ × 3.9 × 10$^{-9}$ = 0.097 m (9.7 cm))(1)
(c) 7.6 mm is less than 2.0 cm (half the separation), so the electron leaves the plates (deflected towards the positive plate) (alternative: 9.7 cm is greater than the 6.0 cm plate length, so it leaves before reaching a plate; ecf a consistent comparison)(1)
Answer: (a) 1.5 × 10$^4$ V m$^{-1}$ (b) 2.6 × 10$^{15}$ m s$^{-2}$ (c) deflection 7.6 mm < 20 mm, so it emerges
Examiner tip: Compare the deflection with half the plate separation, since the electron starts midway.

6Standard6 marks

A 470 μF capacitor is charged to 12 V and then discharged through a 22 kΩ resistor.

(a) Calculate the initial charge on the capacitor. [1]

(b) Calculate the initial energy stored by the capacitor. [1]

(d) Calculate the pd across the capacitor 15 s after the discharge starts. [2]

(e) Calculate the time taken for the pd to fall to 3.0 V. [1]

(a) $Q = CV = 470\times10^{-6}\times12$ = 5.6 × 10$^{-3}$ C(1)
(b) $E = \frac{1}{2}CV^2$ = 0.034 J (3.38 × 10$^{-2}$)(1)
(c) $RC = 22\times10^3 \times 470\times10^{-6}$ = 10.3 s (allow 10 s)(1)
(d) $V = V_0 e^{-t/RC} = 12e^{-15/10.34}$(1)
(d) $V$ = 2.8 V (allow 2.7 V from $RC$ = 10 s; ecf from (c))(1)
(e) $t = RC\ln\frac{12}{3.0} = 10.34 \ln 4$ = 14 s (14.3) (accept two half-lives with $T_{1/2} = 0.69RC$)(1)
Answer: (a) 5.6 × 10$^{-3}$ C (b) 0.034 J (c) 10.3 s (d) 2.8 V (e) 14 s
Examiner tip: Check your powers of ten: μF is 10⁻⁶ and kΩ is 10³ — the time constant should come out in seconds.

7Hard5 marks

A proton travelling at 3.0 × 10$^6$ m s$^{-1}$ enters a uniform magnetic field of flux density 0.25 T. Its velocity is perpendicular to the field.

(a) Explain why the proton moves in a circular path. [2]

(b) Calculate the radius of the path. [2]

(a) the magnetic force ($BQv$) is always perpendicular to the velocity (Fleming's left-hand rule)(1)
(a) so no work is done / speed is constant and the force (constant magnitude) acts towards a centre, providing the centripetal force(1)
(b) $BQv = \frac{mv^2}{r}$ so $r = \frac{mv}{BQ}$(1)
(b) $r = \frac{1.673\times10^{-27}\times3.0\times10^6}{0.25\times1.60\times10^{-19}}$ = 0.13 m (0.125)(1)
(c) $t = \frac{\pi r}{v}$ (or $\frac{\pi m}{BQ}$) = 1.3 × 10$^{-7}$ s (allow 1.4 × 10$^{-7}$ s from $r$ = 0.13 m; ecf from (b))(1)
Answer: (b) 0.13 m (c) 1.3 × 10$^{-7}$ s
Examiner tip: For the 'explain', say the force is perpendicular to the velocity — 'the force is towards the centre' alone just restates what you are explaining.

8Hard7 marks

A simple ac generator has a rectangular coil of 150 turns, each of area 4.0 × 10$^{-3}$ m$^2$. The coil rotates at 50 revolutions per second in a uniform magnetic field of flux density 0.080 T.

(a) Calculate the peak emf induced in the coil. [2]

(b) The rotation frequency is doubled. State and explain the effect on the peak emf. [2]

(c) A step-down transformer has 1150 turns on its primary coil. The primary coil is connected to a 230 V supply and the output pd of the secondary coil is 12 V. Assume the transformer is ideal for part (i).

(i) Calculate the number of turns on the secondary coil. [1]

(ii) Explain why the core of the transformer is laminated. [2]

(a) $\varepsilon_0 = BAN\omega$ with $\omega = 2\pi \times 50$(1)
(a) $\varepsilon_0 = 0.080\times4.0\times10^{-3}\times150\times314$ = 15 V (15.1)(1)
(b) peak emf doubles (to 30 V)(1)
(b) because the rate of change of flux linkage doubles (Faraday's law) (accept $\varepsilon_0 \propto \omega$)(1)
(c)(i) $N_s = N_p\frac{V_s}{V_p} = 1150\times\frac{12}{230}$ = 60(1)
(c)(ii) laminations (thin layers insulated from each other) reduce eddy currents in the core (accept 'increase resistance to eddy currents')(1)
(c)(ii) so less energy is wasted heating the core / transformer is more efficient(1)
Answer: (a) 15 V (b) doubles to 30 V (c)(i) 60 turns
Examiner tip: In (b), name the cause: rate of change of flux linkage — 'it spins faster' alone doesn't score.

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