Suppose that the particle in Exercise in 1.33 is an electron projectedwith velocity vx = 2.0 × 106 m s–1. If E between the plates separatedby 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upperplate? (|e|=1.6 × 10–19 C, me = 9.1 × 10–31 kg.)

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Solution

The given velocity of the particle is, v x =2.0× 10 6 m/s.

The distance between the plates, d=0.5 cm=0.005 m.

Electric field E is 9.1× 10 −19 C.

Charge on an electron is 1.6× 10 −19 C and mass of the electron is 9.1× 10 −31 kg.

The formula for the deflection at the end of the plate is,

s= qE L 2 2m ( v x ) 2
Substitute the values in the above formula and find the value of L.

L= 2sm ( v x ) 2 qE = 2×0.005×9.1× 10 −31 1.6× 10 −19 ×9.1×10×2 = 0.00025 =0.016 m or 1.6 cm
Thus, the distance travelled by the electron before striking the plate is 1.6 cm.

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Similar questions

v_{x} = 2.0 \times 10^{6} m s^{-}

^{1}

9.1 \times 10^{2} N / C

(q_{e} = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g .)

v_{x}

v_{x} = 2.0 \times 10^{6} m s^{- 1}

9. 1 \times 10^{2}

(| e | = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g .)

1.6 \times 10^{- 19} C

9.1 \times 10^{- 31} k g

3 \times 10^{10} m s^{- 1}

0.2 T

e = 1.6 \times 10^{- 19} C

m_{e} = 9.1 \times 10^{- 31} k g

(| e | = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g, h = 6.6 \times 10^{- 34} J s) .

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