Question

Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 170 V. An electron enters the region between them at a speed of 2.5 × 105 m/s by passing through a small hole in the negative plate and continues moving toward the positive plate. Assume the electric field between the plates is uniform. what is the electron's speed, in meters per second, when it is 0.1 cm from the negative plate?

Answer 1

To solve this problem it is necessary to apply the concepts related to energy as a function of voltage, load and force, and the definition of Force given by Newton in his second law.

By definition we know that force is equal to

F= ma

Where,

m = mass (at this case of an electron)

a = Acceleration

But we also know that the Energy of an electric object is given by two similar definitions.

$1) E= \frac{F}{q}$

Where,

F= Force

q = Charge of proton/electron

$2) E = \frac{V}{d}$

V = Voltage

d = Distance

Equating and rearrange for F,

$\frac{F}{q} = \frac{V}{d}$

$F = \frac{Vq}{d}$

The two concepts of force can be related to each other, then

$ma = \frac{Vq}{d}$

Acceleration would be,

$a = \frac{Vd}{dm}$

Replacing with our values we have that the acceleration is

$a = \frac{Vq}{dm}$

$a = \frac{(170)(1.6*10^{-19})}{(2*10^{-2})(9.1*10^{-31})}$

$a = 1.49*10^{15}m/s^2$

Now through the cinematic equations of motion we know that,

$V_f^2-V_i^2 = 2ax$

Where,

$V_f =$ Final velocity

$V_i =$ Initial velocity

a = Acceleration

x = Displacement

Re-arrange to find v_f,

$v_f = \sqrt{v_i^2+2ax}$

$v_f = \sqrt{2.5*10^5+2*(1.49*10^{15})(0.1*10^{-2})}$

$v_f = 1.726*10^6 m/s$

Therefore the electron's speed when it is 0.1 cm from the negative plate is $1.726*10^6 m/s$