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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Suppose that the Particle in Exercise in 1.33 is an Electron Projected with Velocity Vx= 2.0 × 10^6 ms^−1. If E Between the Plates Separated by 0.5 cm is 9.1 × 10^2 N/C, - Physics

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प्रश्न

Suppose that the particle is an electron projected with velocity vx = 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e| = 1.6 × 10−19 C, m= 9.1 × 10−31 kg)

संख्यात्मक
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उत्तर

Velocity of the particle, vx = 2.0 × 106 m/s

Separation of the two plates, d = 0.5 cm = 0.005 m

Electric field between the two plates, E = 9.1 × 102 N/C

Charge on an electron, q = 1.6 × 10−19 C

Mass of an electron, m= 9.1 × 10−31 kg

Let the electron strike the upper plate at the end of plate L, when deflection is s.

Therefore,

`"s" = ("qEL"^2)/(2"mv"_"x"^2)`

`"L" = sqrt((2"dmv"_"x"^2)/("qE"))`

= `sqrt((2 xx 0.005 xx 9.1 xx 10^-31 xx (2.0 xx 10^6)^2)/(1.6 xx 10^-10 xx 9.1 xx 10^2))`

= `sqrt(0.025 xx 10^-2)`

= `sqrt(2.5 xx 10^-4)`

= `1.6xx10^-2` m

= 1.6 cm

Therefore, the electron will strike the upper plate after travelling 1.6 cm.

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