An electron is projected with an initial speed v_0 = 1.30 Times 10^6 m/s into th

Question

An electron is projected with an initial speed v_0 = 1.30 Times 10^6 m/s into the uniform field between the parallel plates in the following figure. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. (Assume d_1 = 3.50 cm and d_2 = 7.00 cm.) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Explanation / Answer

As it moves into the electric field, the electron accelerates towards the upper plate.

Now, the acceleration of the electron, a = F/m = Eq/m

So, in the vertical direction,

using the equation of motion:

s = ut + 0.5*at^2

u = initital vertical speed = 0

a = Eq/m

s = d1/2 = 1.75 cm = 0.0175 m

So, 0.0175 = 0 + 0.5*(E*1.6*10^-19/(9.1*10^-31))*t^2

Now, t = d2/v0 = 0.07/(1.3*10^6) = 5.38*10^-8 s

So, 0.0175 = 0 + 0.5*(E*1.6*10^-19/(9.1*10^-31))*(5.38*10^-8)^2

So, E = 68.8 N/C <------answer

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