Find vertical position and exit angle of electron.

Question

Electrons move initially at 2.4 x 10^6 m/s horizontally. They penetrate a region between two 2 cm long plates and have an acceleration of 4 x 10^14 m/s^2 vertically and directed upwards. Determine a) the vertical position of the electrons when they leave the region situated between the plates and b) their exit angle.

Much appreciated!

Answer 1

Consider just one electron -- they all behave the same.

Let

v = 2.4×10⁶ m/s be the electron's horizontal velocity,

a = 4×10¹⁴ m/s² be its upward acceleration,

L = 2cm = 0.02 m be the length of the plates,

t₀ be the time instance the electron enters the region between the plates,

t₁ be the time instance the electron leaves the region between the plates,

x(t) be the electron's horizontal position at time t,

y(t) be the electron's vertical position at time t,

the origin of time and space coordinates be set so that t₀ = 0 and x(0) = y(0) = 0.

The electron's horizontal position is unaffected by the plates, so

x(t) = vt (1)

The electron's vertical position is affected only by the acceleration a, so

y(t) = at²/2 (2)

The time t₁ is defined by

x(t₁) = L (3)

(a)

We are to calculate y(t₁).

From (1) express

t = x(t)/v

and plug it into (2).

y(t) = ax²(t)/2v²

Using (3)

y(t₁) = ax²(t₁)/2v² = aL²/2v²

Substituting actual numbers

y(t₁) = 4×10¹⁴×0.02²/(2×(2.4×10⁶)²) = 0.014 m

(b)

The tangent of the exit angle is the derivative dy/dx at the point x = L.

dy/dx = 2ax/2v² = ax/v²

Thus the tangent is aL/v² = 4×10¹⁴×0.02/(2.4×10⁶)² = 1.39

Thus the angle is about 54.3°.