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=(F)/(m)=(6.717398*10^(-14))/(9.109*10^(-31))=7.37446*10^(16)(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(m)/(s)`