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A particle having a charge of 1.6 × 10–19C enters between the plates of a parallel plate capacitor. The initial velocity of the particle is parallel to the plates. A potential difference of 300v is applied to the capacitor plates. If the length of the capacitor plates is 10cm and they are separated by 2cm, Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is 12 × 10–24kg . 

(A) 104(m/s) 

(B) 102(m/s) 

(C) 10–1(m/s) 

(D) 103(m/s)

1 Answer

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Best answer

The correct option (A) 104(m/s)   

Explanation:

Electric field = E = (v/d) = [(Potential difference)/(separation between plates)]

= [(300v)/(2 × 10–2)]

= 15 × 103v/m

also y = (1/2)at2

where a = (F/m) = (qE/m) ------- Force = mass × acceleration & time = [(distance)/(velocity)]

and t = (x/u)

∴  y = (1/2) (qE/m) (x/u)2

∴  u2 = (1/2) [(qEx2)/(my)]

= (1/2) × [{1.6 × 10–19 × 15 × 103 × (0.1)2}/(12 × 10–24 × 0.01)]

u2 = 108

u = 104(m/s)  

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