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InterviewSolution
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Suppose that the particle in the above question is an electron projected with velocity v_x = 2.0 xx 10^(6) m s^(–1). If E between the plates separated by 0.5 cm is 9.1 xx 10^(2) N//C, where will be the electron strike the upper plate? (|e|=1.6 xx 10^(-19) C, m_e = 9.1 xx 10^(–31) kg). |
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Answer» Solution :Velocity of the particle, `V_(X)=2.0xx10^(6)m//s` Separation of the TWO plates, d = 0.5 cm = 0.005 m Electric FIELD between the two plates, E = `9.1xx10^(2)N//C` Charge on an electron, q = `1.6xx10^(-19)C` Mass of an electron, `m_(e)=9.1xx10^(-31)` kg LET the electron STRIKE the upper plate at the end of plate L, when deflection is s. Therefore, `s=(qEL^(2))/(2mv_(x)^(2))` `L=sqrt((2dmv_(x)^(2))/(qE))` `=sqrt((2xx0.005xx9.1xx10^(-31)xx(2.0xx10^(6))^(2))/(1.6xx10^(-10)xx9.1xx10^(2)))` `=1.6xx10^(-2)` = 1.6 cm Therefore, the electron will strike the upper plate after travelling 1.6 cm. |
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