InterviewSolution
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InterviewSolution
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A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed `v_x` (like particle 1 in Figure). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is `(qEL^2)/(2m v_(x)^2)`. Compare this motion with motion of a projectile in gravitational field. |
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Answer» Charge on a particle of mass m = − q Velocity of the particle = `V_(X)` Length of the plates = L Magnitude of the uniform electric field between the plates = E Mechanical force, F = Mass (m) `xx` Acceleration (a) `a=(F)/(m)` However, electric force, `F=qE` Therefore, acceleration, `a=(qE)/(m)" "...(1)` Time taken by the particle to cross the fields of length L is given by, `t=("Length of the plate")/("Velocity of the particle")=(L)/(v_(x))" "...(2)` In the vertical direction, initial velocity, u = 0 According to the third equation of motion, vertical deflection s of the particle can be obtained as, `s=ut+(1)/(2)at^(2)` `s=0+(1)/(2)((qE)/(m))((L)/(v_(x)))^(2)` `s=(qEL^(2))/(2mv_(x)^(2))" "...(3)` Hence, vertical deflection of the particle at the far edge of the plate is `qEL^(2)//(2mv_(x)^(2))`. This is similar to the motion of horizontal projectiles under gravity. |
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