The figure shows a beam of electrons leaving the electron gun and traveling stra

Question

The figure shows a beam of electrons leaving the electron gun and traveling straight through the deflection plates, because both plates are grounded there is no electric field between them. Let d be the vertical distance between the plates and let the horizontal length of the deflection plates be l (lower case L). Then let L be the distance from the center of the deflection plates to the screen and D the deflection or distance on the screen from central line up to the spot where the beam strikes.

The objective is to express D in terms of the accelerating voltage V_o (which is V_b + V_C of the electron gun), the voltage V applied to the deflection plates, the distance L to the screen and the distance d between the plates. From conservation of energy an electron at rest on the filament has potential energy e(V_b)

Show that the deflection on the spot on the screen measured from the central axis would be D where D=(lLV)/2dV_o

Explanation / Answer

v = speed of the electron when it enters the deflection plates
eV0 = mv2/2,

Inside the deflection region, electric field E = V/d, electric force E = eE, acceleration a = F/m = eV/(md)

time in the deflection region t = l/v

deflected distance inside the region y = at2/2

u = velocity of the electron just leaving the region

ux = v, uy = at

time outside the region t' = (L - l/2)/v

deflected distance outside the region y' = uyt' = at(L - l/2)/v

So the total deflected distance = y + y' = at2/2 + at(L - l/2)/v = at[t/2 + (L - l/2)/v]

= eV/(md) * l/v * [l/(2v) + (L - l/2)/v] = eVlL/(mv2d) = eVlL/(2eV0d) = VlL/(2V0d)

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