A uniform electric field E=(8m)/e V m⁻¹ is created between two parallel plates of length 1 m (where m is the mass of the electron and e the magnitude of its charge). An electron enters the field symmetrically between the plates with a speed of 2 m s⁻¹. The angle of deviation θ of the path of the el…

Question

A uniform electric field E=(8m)/e V m⁻¹ is created between two parallel plates of length 1 m (where m is the mass of the electron and e the magnitude of its charge). An electron enters the field symmetrically between the plates with a speed of 2 m s⁻¹. The angle of deviation θ of the path of the electron as it comes out of the field will be

Accepted Answer

(2) tan⁻¹(2). The electric force on the electron is transverse, so the horizontal speed stays vₓ=2 m s⁻¹. Transverse acceleration a=(eE)/m=e/m·(8m)/e=8 m s⁻². Time inside the plates: t=L/(vₓ)=1/2=0.5 s. Transverse speed on exit: v_y=at=8×0.5=4 m s⁻¹. tan θ=(v_y)/(vₓ)=4/2=2, so θ=tan⁻¹(2).

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