Question bank › A-level Physics › Fields and their consequences

Electric fields and electric potential

Answer on paper first, then reveal the mark scheme one mark at a time.

1Standard4 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.

2Hard5 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.

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