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Sunday, April 8, 2018

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.}

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