Two vertical metal plates in a vacuum have a separation of 4.0 cm. A potential difference of 2.0 × 102 V is applied between the plates. Fig. 5.1 shows a side view of this arrangement.

Question 3

(a) Define the coulomb. [1]

(b) Two vertical metal plates in a vacuum have a separation of 4.0 cm. A potential difference of 2.0 × 10² V is applied between the plates. Fig. 5.1 shows a side view of this arrangement.

Fig. 5.1

A smoke particle is in the uniform electric field between the plates. The particle has weight 3.9 × 10^-15 N and charge –8.0 × 10^-19 C.

(i) Show that the electric force acting on the particle is 4.0 × 10^-15 N. [2]

(ii) On Fig. 5.1, draw labelled arrows to show the directions of the two forces acting on the smoke particle. [1]

(iii) The resultant force acting on the particle is F.

Determine

1. the magnitude of F,

2. the angle of F to the horizontal. [3]

(c) The electric field in (b) is switched on at time t = 0 when the particle is at a horizontal

displacement s = 2.0 cm from the left-hand plate. At time t = 0 the horizontal velocity of the particle is zero. The particle is then moved by the electric field until it hits a plate at time t = T.

On Fig. 5.2, sketch the variation with time t of the horizontal displacement s of the particle from the left-hand plate.

Fig. 5.2

[2]

[Total: 9]

Reference: Past Exam Paper – November 2017 Paper 22 Q5

Solution:

(a) 1 coulomb is the amount of charges that passes when a current of 1 ampere flows for 1 second.

(b)

(i)

E = V / d          or E = F / Q

F = VQ / d

F = (2.0 × 10² × 8.0 × 10^-19) / 4.0 × 10^-2 = 4.0 × 10^-15 N

(ii)

arrow pointing to the left labelled ‘electric force’ and

arrow pointing downwards labelled ‘weight’   

(iii)

1.

{The resultant force F of the 2 forces mentioned above can be obtained from Pythagoras’ theorem.

F² = (weight)² + (electric force)²}

resultant force = √ [(3.9 × 10^-15)² + (4.0 × 10^-15)²]     

= 5.6 × 10^-15 N                                   

2. angle = tan^-1 (3.9 × 10^-15 / 4.0 × 10^-15) = 44°         

(c)

downward sloping line from (0, 2.0)

magnitude of gradient of line increases with time and line ends at (T, 0)

{The horizontal displacement is affected by the electric force. The particle accelerates towards the left. So, the velocity increases. From the s-t graph, the velocity is obtained by the gradient. Thus, the gradient increases.}