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Question: Take the particle in question number 52 electron projected with velocity \(v_{x} = 4 \times 10^{6}m^...

Take the particle in question number 52 electron projected with velocity vx=4×106ms−1v_{x} = 4 \times 10^{6}m^{s - 1}. If electric field between the plates separated by 1 cm is 8.2×102NC−1,8.2 \times 10^{2}NC^{- 1}, the the electron will strike the upper plate at (Take me=9.1×10−31kgm_{e} = 9.1 \times 10^{- 31}kg)

A

2.14 cm

B

39 cm

C

1.23 cm

D

3.3 cm

Answer

3.3 cm

Explanation

Solution

: Given vx=4×106ms−1,d=1cm=1×10−2mv_{x} = 4 \times 10^{6}ms^{- 1},d = 1cm = 1 \times 10^{- 2}m

E=8.2×102NC−1E = 8.2 \times 10^{2}NC^{- 1}

q=e=1.6×10−19C,me=9.1×10−31kgq = e = 1.6 \times 10^{- 19}C,m_{e} = 9.1 \times 10^{- 31}kg

The electron will strike the upper plate at its other end of x = L as soon as its deflection.

Or y=dZ=10−22m=5×10−3my = \frac{d}{Z} = \frac{10^{- 2}}{2}m = 5 \times 10^{- 3}m

From (iii).

L=2mevx2yqE=2×9.1×10−31×(4×106)2×5×10−31.6×10−19×8.2×102=3.33×10−2mL = \sqrt{\frac{2m_{e}v_{x}^{2}y}{qE} =}\sqrt{\frac{2 \times 9.1 \times 10^{- 31} \times (4 \times 10^{6})^{2} \times 5 \times 10^{- 3}}{1.6 \times 10^{- 19} \times 8.2 \times 10^{2}}} = 3.33 \times 10^{- 2}m

L=3.33cmL = 3.33cm

The electrons will strike the upper palte at its other end, if length of plate is 3.33 cm