Question

Two 2 cm-diameter disks spaced 1.926 mm apart form a parallel-plate capacitor. The electric field between the disks is 576,184.575 V/m. An electron is launched from the negative plate. It strikes the positive plate at a speed of 31,724,040.891 m/s. What was the electron's speed as it left the negative plate?

Answer 1

If we know the electric field and the charge of a particle, we can find out the force it feels. This can be written as:

$\vec{F}\text{ = }q\vec{E}=1.6*10^{-19}*576185.575=9.2189692*10^{-14}N$

Then, knowing the force, we can find out its acceleration by Newton's second law, thus:

$a=\frac{F}{m}=\frac{9.2189692*10^{-14}}{9.1093837*10^{-31}}=1.012029958*10^{17}\frac{m}{s}$

We can then rearrange Torricelli's formula in order to obtain the initial velocity:

$v_0=\sqrt[\placeholder{⬚}]{(v_f)^2-2a\Delta s}$

Then, replacing our values we get:

$v_0=\sqrt[\placeholder{⬚}]{(31724040.891)^2-2*1.012*10^{17}}=24831045.7\frac{m}{s}$

Then, our final answer is v0=24831045.7 m/s