Simple electrostatic fields questions

[lovelyrwwr]

Imagine nine vertical and parallel lines drawn side by side. They represent infinite planes of uniform voltage. The first line has 0 V. The 2nd line has 1 V. The 3rd has 2 V and so on, with the 9th line having 8 V. The distance between each line is 1 cm, which is 0.01 meter.

There is an electron located on the vertical line with 2 V. This electron has a q = -1.6E-19 C and a mass = 9.11E-31 kg.

What will be the acceleration of the electron assuming there are no other forces?

MY SOLUTION
d = 0.01 meter
q = -1.6E-19 C
mass = m =9.11e-31 kg

E = delta V / delta d = 1 V / 0.01 m= 100 V/m

F = Eq = ma
F = (100 V/m)(-1.6E-19 C) = (9.11E-31 kg) a
a = 1.76E13 m/s^2

Is this correct? My friend is getting something like 6.86E50 m/s^2 or 8.68E50 m/s^s.

 

[astarael7]

I'm going to use a simplified version of the problem. Consider the case where there is only two plates: the ground plate and the plate at 1 V. The electron is released from the grounded plate and arrives the second plate having gained 1 eV (1.6e-19 J) of energy (electron-volt: literally the energy gained by an electron moving through a potential of 1 V). Note that the actual distance between the plates was irrelevant.

Next consider the same grounded plate, but now the second plate is twice as far away as it used to be and has twice the potential. Therefore, when the electron arrives at that plate its gain in energy will be 2 eV.

That's as far as I will take it. Calculating the value of the potential halfway between the second set of plates might be a good way to figure out the solution.

 

[rude man]

I see nothing wrong with what you did.

As for your friends' computations - looks like the electron will have to shed quite a bit of rest mass before achieving those kinds of acceleration!

 

[lovelyrwwr]

thank you so much rude man! 1000x hugs :)

 

[Nickie]

Your solution is correct. The acceleration of the electron would be 1.76E13 m/s^2. This is due to the fact that the electric field is constant between each line and the force acting on the electron is directly proportional to the electric field. Your friend's solution seems to be incorrect as it is several orders of magnitude higher than the expected value. It is important to double check calculations and units to ensure accuracy in scientific calculations.