Electron traveling between two plates

[talaroue]

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?

 

[songoku]

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

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

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

Fe = E x q

 

[talaroue]

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

 

[songoku]

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

 

[talaroue]

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?

 

[songoku]

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

 

[talaroue]

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

 

[songoku]

Yups

 

[talaroue]

For some reason it didn't worl

 

[songoku]

Why? Have you converted 1.47 cm to m?

 

[talaroue]

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

B=Voltage/(s*q*V)

 

[songoku]

The charge (q) has canceled 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

 

[talaroue]

right that's 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.