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Electron traveling between two plates

  • Thread starter talaroue
  • Start date
303
0
1. 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


2. Homework Equations

I thought that

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


3. 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?
 
303
0
So then my equation should be (Kq1q2/R^2)/qVB
 
1,180
24
So then my equation should be (Kq1q2/R^2)/qVB
You mean (Kq1q2/R^2) = qVB :wink:

Yes that's right, but it's better to use Eq = qVB
Find E, then solve for B
 
303
0
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?
 
1,180
24
For two parallel charged plates, E = V/d, where V = voltage across the parallel plate and d = distance between the plate
 
303
0
oh so instead of using E=KQ1Q1/R^2 I should use E=V/D?
 
1,180
24
Yups
 
303
0
For some reason it didn't worl
 
1,180
24
Why? Have you converted 1.47 cm to m?
 
303
0
Here is my work....
Fe+Fm=0
Fe=-Fm
Voltage/s(distance between plates)=qV(velocity)B

B=Voltage/(s*q*V)
 
1,180
24
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
 
303
0
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.
 

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