Magnitude of elec field to balance electron?

[Brad_1234]

Hi all,

Im doing a question in the homework, its #13 and I am struggling on how to get started.

The question is: what are the magnitude and direction of the electric field that will balance the weight of (a) an electron and (b) a proton?

There is an answer in the book, for the electron "55.8 pN/C down"

This is day 2 of my trying to figure this one out, since the electron has a simple single charge of "e" how to get the force to balance -1.602 E-19 C ? I am searching the book to find out how to convert this into N/C

Ive used the examples on the books companion site, move the charge around and it shows the vector and resulting charge, neat. So I am guessing since the weight is so small, one only has to make a field that counters one charge, but 55.8 pN/C??

I already found a typo in the book, it was showing a right triangle, the 2 sides = .1 and the long side was square root of 2 times .1? so now I don't trust the book, and cannot begin to see how they got the 55 number there, I was going to answer just -1.602 E-19 because one electron should balance the other?? no that's probably not it?

well thanks in advance

 

[emptymaximum]

The force needed to balance the particle can be found by coulomb's law.
The electric field is $E = F/q_0$. This will give units of N/C.

 

[Brad_1234]

ok I found out that the direction of the E-field must be vertically downward. This will generate an upward electrical force since the electron has a negative charge -e.

For the magnitude of the E field, solve

m g = e E

E = g m /e where g is the acceleration of gravity. now the number comes out correctly.