An electron traveling 1000 km s-1 goes through holes on two parallel, narrowly-s

Question

An electron traveling 1000 km s-1 goes through holes on two parallel, narrowly-spaced plates (assume they have infinite extent). In the plane perpendicular to the plates, the electron comes in from the bottom left, making an angle of 45° to the horizontal (the normal to the plates). The first plate encountered is grounded (i.e., uncharged). The second plate encountered is positively charged such that the electric potential to the right of the plates is 1 V. The potential energy of the electron effectively changes discontinuously when the electron passes through the plates. Use conservation of energy to determine the velocity of the electron to the right of the plates. Use Snell's law (which follows from the Hamilton-Jacobi equation) to determine the angle the trajectory of the electron on the right makes with the horizontal. The velocity (in km s-1) and angle (in degrees) are ?

Explanation / Answer

As electron is accelerated through this potential difference, its KE increass.

Work done by this potential = change in KE

- q V = m ( vf^2 - vi^2) /2

1.6 x 10^-19 x 1 = (9.109 x 10^-31) (vf^2 - (1000 x 10^3 m/s)^2) /2

vf = 1.162 x 10^6 m/s

Speed right to the plate is 1.162 x 10^6 m/s.

in betweem, field will be perpendicular to the plates (horizontal) only.

hence velocity will only change in horizontal direction onyl.

vf_y = 1000 x 10^3 m/s sin45 = 0.707 x 10^6 m/s

and vf_x^2 + vf_y^2 = vf^2

(vf_x)^2 + (0.707 x 10^6)^2 = (1.162 x 10^6)^2

vf_x = 0.922 x 10^6 m/s

angle = tan^-1(vfy / vfx) = 40 deg

so 1162 km/s at an angle of 40 deg .

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