Question

Two 2 cm-diameter disks spaced 2.028 mm apart form a parallel-plate capacitor. The electric field between the disks is 419,837.408 V/m. An electron is launched from the negative plate. It strikes the positive plate at a speed of 14,672,627.552 m/s. What was the electron's speed as it left the negative plate?

Answer 1

Firstly, the force a charged particle feels when submited to an eletric field is:

$\vec{F}=q.\vec{E}$

And, in our case, F will be:

$F=1.6*10^{-19}*419837.408=6.717398*10^{-14}N$

By using Newton's second law, we know that

$\vec{F}=m.\vec{a}$

Then, we can find out our acceleration, which will be

$a=\frac{F}{m}=\frac{6.717398*10^{-14}}{9.109*10^{-31}}=7.37446*10^{16}\frac{m}{s^2}$

We now have the acceleration the electron would feel. We also know the distance it'll travel, which is 2.028mm. With this, we can apply Torricelli's equation, which tells us that:

$v^2=v_0^2+2.a.\Delta s$

By isolating v0 we get

$v_0=\sqrt[2]{v^2-2.a.\Delta s}$

Finally, replacing our values, we get:

$v_0=\sqrt[2]{14672627.552^2-2*7.37446*10^{16}*2.028*10^{-3}}$

Then

$v_0=9155440\frac{m}{s}$