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A proton is placed at rest near the positive side of two oppositely charged paralell...

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andrewmezzo | Student, College Freshman | Valedictorian

Posted February 11, 2013 at 3:53 AM via web

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A proton is placed at rest near the positive side of two oppositely charged paralell plates. There is a potential difference of 12V.

Calculate the final speed of proton just before it strikes plate

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted February 11, 2013 at 5:34 AM (Answer #1)

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Imagine a proton is a moving object. Since it is an object, it has energy. In this case, it is kinetic energy.

Kinetic Energy

= `1/2 mv^2`      e1

and energy can also be interpreted as:

`E = qV`      e2

Since we are getting the speed of the molecule, we can arrange e1 to be:

`K.E. = 1/2 mv^2`

`2K.E. = mv^2`

`(2K.E)/(m) = v^2`

`v = sqrt((2K.E)/(m))` 

Since qV = E = K.E, then:

`v = sqrt((2qV)/(m))`

q = charge of the proton;1.6x10^-19 C

V = 12 V

m = mass of the proton; 1.67x10^-27 kg

And we are ready to solve the problem.

`v = sqrt((2qV)/(m))`

`v = sqrt((2(1.6x10^(-19))(12))/(1.67x10^(-27)))`

 

v = 47,952 m/s

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