Electron traveling between two plates

  • Thread starter talaroue
  • Start date
  • #1
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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

  • #2
1,871
203
Fe=Fm
(Kq1)/(r^2)=qVB

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

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

Fe = E x q
 
  • #3
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So then my equation should be (Kq1q2/R^2)/qVB
 
  • #4
1,871
203
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
 
  • #5
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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?
 
  • #6
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203
For two parallel charged plates, E = V/d, where V = voltage across the parallel plate and d = distance between the plate
 
  • #7
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oh so instead of using E=KQ1Q1/R^2 I should use E=V/D?
 
  • #8
1,871
203
Yups
 
  • #9
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For some reason it didn't worl
 
  • #10
1,871
203
Why? Have you converted 1.47 cm to m?
 
  • #11
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Here is my work....
Fe+Fm=0
Fe=-Fm
Voltage/s(distance between plates)=qV(velocity)B

B=Voltage/(s*q*V)
 
  • #12
1,871
203
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
 
  • #13
303
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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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