How Slow Must an Electron Travel to Reach the Opposite Plate in a Capacitor?

[nckaytee]

A parallel-plate capacitor is formed of two 10 cm × 10 cm plates spaced 1.0 cm apart. The plates are charged to ±1.0 nC. An electron is shot through a very small hole in the positive plate.
a. What is the slowest speed the electron can have if it is to reach the negative plate (Hint: set final speed
at negative plate equal to zero)?

I used C=EoA/d (8.85 x 10^-12)(100 x 10^-2)/(1 x 10^-2) = 8.85

Then I used $$\Delta$$V=Q/C 0 - Vi = (1 x 10^-9)/8.85

So, Vi = -1.13 x 10^-10

Can anyone verify this for me?

 

[anathema]

I can verify that your calculations are correct. The slowest speed an electron can have in order to reach the negative plate is -1.13 x 10^-10 m/s. This is because the electron will experience a potential difference of 1.13 x 10^-10 V as it travels from the positive plate to the negative plate, and this potential difference is equal to the initial kinetic energy of the electron. Therefore, the initial velocity of the electron must be equal to this potential difference divided by its mass, which is -1.13 x 10^-10 m/s.

 

[baba89]

I can verify your calculations. Your approach and equations are correct. The initial velocity of the electron, Vi, should indeed be -1.13 x 10^-10 m/s in order to reach the negative plate with a final speed of zero. This means that the electron must be shot with a speed of -1.13 x 10^-10 m/s or faster in order to reach the negative plate. Any slower and it will not have enough kinetic energy to overcome the electric field between the plates and reach the negative plate.