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Question
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, me = 9.1 × 10−31 kg)
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Solution
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, me = 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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