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Deflection of an electron

  • #1

Homework Statement



An electron with an initial velocity of 4.7 x 10^7 m/s in the x-direction moves along a trajectory that takes it directly between the center of two 20 cm x 20 cm plates separated by 1 cm. The plates are connected to a high voltage power supply so that the potential difference between them is 500V. What is the magnitude of the y-component of the electron’s velocity when it leaves the plates? (Answer in units of ANSWER x 10^7 m/s.)


Homework Equations


E = V/d


The Attempt at a Solution



I think I found the electric field of the plates to be:
E = 500V / .01m E = 50,000 V/m

From there I have no idea where to go...Any help would be appreciated!
 

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Answers and Replies

  • #2
288
0
The electric field is gonna apply a force to the electron

Do you know the equation that relates electric field, charge, and force?
 
  • #3
just started EM stuff now, sorry cant help. but blochwave is right, there is an equation that relates to electric field, charge and force, the only one i can think of uses two particles and a separation between them, would not be useful for in this instance. best of luck. lol im curious actually...
 
  • #4
288
0
Well that's the equation FOR an electric field, which you can't quite use here exactly in that form, but he already used the equation for parallel plate capacitors and now KNOWS the electric field, and needs to find the force on a particle with known charge within that (conveniently constant) electric field
 
Last edited:
  • #5
rock.freak667
Homework Helper
6,230
31
If F=EQ, and E=V/d, can you find F? What does Newton's 2nd law say now?

(Hint: the velocity in the x-direction stays constant)
 
  • #6
So the force would be F = (50000 V)*(1.602 x10^-19 C)
F = 8.01 x10^-15 N
 
  • #7
rock.freak667
Homework Helper
6,230
31
So the force would be F = (50000 V)*(1.602 x10^-19 C)
F = 8.01 x10^-15 N
Yes that is the force. Note that the units of E are either [itex]Vm^{-1} or NC^{-1}[/itex]
 
  • #8
Kinetic + potential = Emech
(1/2m(Vi^2)) - q(E)*(Xi) = (1/2m(Vf^2)) - q(E)*(Xf)
Then we changed the equation to this:
qE(20cm) = 1/2(mass of electron)*((Vf^2) + (Vi^2))
Then we solved for Vf:
Vf = 3.61697 x10^7 m/s
 

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