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Question

# A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig.). The length of plate is L and an uniform electric field E is maintained between the plates.Suppose that the particle in Exercise is an electron projected with velocity vx=2.0×106ms−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|=1.6×10−19C,me=9.1×10−31kg.)

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Solution

## Velocity of the electron vx=2×106 m/sDistance covered d=0.005 mElectric field between plates E=9.1×102 N/Ccharge of electron is q and the mass is m.Acceleration of the electron a=qEm Time taken by the electron t=L/vxUsing d=ut+12at2where initial speed of the electron along y-axis u=0 m/s∴ d=0+12×qEm(Lvx)2so, we get d=qEL22mv2x⟹L=√2dmv2xqE = ⎷2(0.005)(9.1×10−31)(2×106)2(1.6×10−19)(9.1×102) =1.6×10−2 m=1.6 cm

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