An electron is travelling in a straight line through a vacuum with a constant speed of 1.5 × 107 m s-1

Question 5

An electron is travelling in a straight line through a vacuum with a constant speed of 1.5 × 10⁷ m s^-1.

The electron enters a uniform electric field at point A, as shown in Fig. 5.1.

Fig. 5.1

The electron continues to move in the same direction until it is brought to rest by the electric field at point B. Distance AB is 2.0 cm.

(a) State the direction of the electric field. [1]

(b) Calculate the magnitude of the deceleration of the electron in the field. [2]

(c) Calculate the electric field strength. [3]

(d) The electron is at point A at time t = 0.

On Fig. 5.2, sketch the variation with time t of the velocity v of the electron until it reaches point B. Numerical values of v and t do not need to be shown.

Fig. 5.2

[1]

[Total: 7]

Reference: Past Exam Paper – March 2017 Paper 22 Q5

Solution:

(a) to the right / from the left / from A to B / in the same direction as electron velocity

{Electric field is from positive to negative.

Since the electron is brought to rest, the force on it should oppose the direction of motion. So, the positive should be at A (so that it attracts the electron) and the negative should be at B (so that it repels and cause the electron to come to rest).}

(b)

v² = u² + 2as

a = (1.5 × 10⁷)² / (2 × 2.0 × 10^-2)

a = 5.6 × 10¹⁵ m s^-2     

(c)

E = F / Q

E = (9.1 × 10^-31 × 5.6 × 10¹⁵) / 1.6 × 10^-19

E = 3.2 × 10⁴ V m^-1

(d)

straight line with negative gradient starting at an intercept on the v-axis and ending at an intercept on the t-axis.