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Deflection of Plates

  • Thread starter anyone1979
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[SOLVED] Deflection of Plates

sqs5km.jpg

Answer choices: A) 63V B) 112V C) 520V D) 642V E) 882V

y = .5cm = 5 x 10^-3 m
x = 3cm = 3 x 10 ^-2 m
angle = 10 degrees


acceleration(a) = ((1.602 x 10^-19) (15000))/(9.11 x 10^-31) = 2.6 x 10^15 m/s^2
Force(F) = 2.6 x 10 ^15(9.11 x 10^-31) = 2.4 x 10^15 N
E = F/e = (2.4 x 10 ^-15)/(1.602 x 10^-19) = 1.5 x 10^4 N/C
velocity(v) = sqrt(2 x (2.6 x 10^15) x (.005)) = 5.1 x 10^6 m/s
y = (1/2) x ((1.602 x 10^-19)/(9.11 x 10^-31)) x ((1.5 x 10^4)/(5.1 x 10^6)^2) x (.03) = 1.5 m
displacement = y tan 10 = (1.5) x (tan 10) = .26

Voltage(V) = ?
I am lost here on how to calculate the voltage applied tp to plates.
V = E x .005 = 75....But the answer is not one of the choices
 

Answers and Replies

454
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your equation for acceleration is wrong the acceleration is just qE/m the energy of the particle doesn't belong in there.

you aren't given the charge or the mass of the particle, so you just assumed it was an electron. That is OK. It might be better to use q and m in all calculations, they should
get eliminated at the end.
You should compute the velocity of the particle next. You know its kinetic energy if you know q.
The speed will give you the time between the plates, and the velocity change to get a deflection of ten degrees. For the deflection angle you just need the speed in the x
and y direction, not the displacement.
 
38
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Ok this is what I have so far.

(15000)(1.602 x 10^-19) = 2.4 x 10^-15 J

KE = (1/2)(m)(v^2) = 2.4 x 10^-15 J

v = sqrt(2 x (2.4 x 10^-15))/(9.11 x 10^-31) = 7.3 x 10^7 m/s

x = 3 x 10^-2 m
d = 0.0025 m, y = .0025 m

x = vt
t = (3 x 10^-2)/(7.3 x 10^7) = 4.1 x 10^-10 sec

a = (2(0.0025))/(4.1 x 10^-10)^2 = 3 x 10^16 m/s^2

F = ma = (9.11 x 10^-31)(3 x 10^16) = 2.7 x 10^14 N

E = F/e = (2.7 x 10^14)/(1.602 x 10^-19) = 1.7 x 10^5 N/C

This voltage calculated is still not right.....HELP
Voltage(V) = ?
V = E(d) = (1.7 x 10^5)(.005) = 850V
 
454
0
a = (2(0.0025))/(4.1 x 10^-10)^2 = 3 x 10^16 m/s^2
why are you doing this? you only need the change in velocity to get the deflection angle, NOT the distance that the particle moved.
 
38
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why are you doing this? you only need the change in velocity to get the deflection angle, NOT the distance that the particle moved.
I needed it to calculate the electric force and electric field.
But when I finished solving the problem, I did not need the electric force or the electric field. I got the answer by finding the displacement and then finding the voltage applied to the plates based on the deflection angle.
 
38
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your equation for acceleration is wrong the acceleration is just qE/m the energy of the particle doesn't belong in there.
Another reason why I found the acceleration was because of what you stated above.
 
454
0
I needed it to calculate the electric force and electric field.
But when I finished solving the problem, I did not need the electric force or the electric field. I got the answer by finding the displacement and then finding the voltage applied to the plates based on the deflection angle.
the deflection is the angle between the velocity vectors, before and after going through the condensator. you don't need the displacement for that.
 
38
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the deflection is the angle between the velocity vectors, before and after going through the condensator. you don't need the displacement for that.
After doing solving the problem, I came up with 882V.
I went over it every possible way I can think of.
 
38
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Thank you for your help.
 

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