# Electron traveling between two plates

## Homework Statement

An electron travels with speed 4.91×107 m/s between the two parallel charged plates. The plates are separated by s=1.47 cm and are charged by a 203 V battery. What magnetic field strength will allow the electron to pass between the plates without being deflected?
----------+------- < postive plate
>>>>O < electron
----------(-)------- < negative plate

## Homework Equations

I thought that

Fnet=Fe+Fm
Fe= electric force
Fm=magnetic force

## The Attempt at a Solution

Set it equal to zero o=Fe+Fm
Fe=Fm
(Kq1)/(r^2)=qVB

solve for B....that didn't work what am I doing wrong?

## Answers and Replies

Fe=Fm
(Kq1)/(r^2)=qVB

Fe is not (Kq1)/(r^2)

(Kq1)/(r^2) is E ( electric field )

Fe = E x q

So then my equation should be (Kq1q2/R^2)/qVB

So then my equation should be (Kq1q2/R^2)/qVB

You mean (Kq1q2/R^2) = qVB

Yes that's right, but it's better to use Eq = qVB
Find E, then solve for B

What charges do I use, I can tell from that equation that the charge of the plates are going to cancel out. So then....
E=VB
Kq/r^2=VB
B=KQ/(R^2*V)

do i assume is half of the distance between the plates?

For two parallel charged plates, E = V/d, where V = voltage across the parallel plate and d = distance between the plate

oh so instead of using E=KQ1Q1/R^2 I should use E=V/D?

Yups

For some reason it didn't worl

Why? Have you converted 1.47 cm to m?

Here is my work....
Fe+Fm=0
Fe=-Fm
Voltage/s(distance between plates)=qV(velocity)B

B=Voltage/(s*q*V)

The charge (q) has cancelled out and even if they haven't, you can subs it with the charge of an electron which is 1.6 x 10^(-19) C

right thats is what q I used.....My professor had his c++ language wrong online so it wouldn't take my answer but he fixed it and my answer is right now.