Proton is shot away from inf. charged plane

  • Thread starter megr_ftw
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  • #1
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Homework Statement


The surface charge density on an infinite charged plane is -2.6*10^-6 C/m^2. A proton is shot straight away from the plane at 2.2*10^6 m/s. How far does the proton travel before reaching its turning point?


Homework Equations


Im very unsure of what equations to use, the only one I know is the equation for density charge. n=Q/A


The Attempt at a Solution


Im trying to figure out how to incorporate kinematic equations with the electric field equations. I think I may need to use the force equation for the electric field somehow.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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What is the expression of electric field near an infinity charged plane?
What is th KE of the proton near the plate?
What is the force on the proton due to this field?
If the proton moves a distance x before it comes to rest, what is the work done?
 
  • #3
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i derived an equation for the distance.
x= (-m*v*eps0)/(qn) n= surface charge density eps0= 8.85*10^-12

this seemed to work, but thanks for the questions they got me to thinking!
 
  • #4
rl.bhat
Homework Helper
4,433
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i derived an equation for the distance.
x= (-m*v*eps0)/(qn) n= surface charge density eps0= 8.85*10^-12

this seemed to work, but thanks for the questions they got me to thinking!
It is not clear how you got the above equation.
The KE of proton = 1/2*m*v^2.
 
  • #5
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i used the equations
(v-final)^2= (v-initial)^2 + 2ax
F=qE=ma

a= (qE)/m = -(v-initial)^2/(2x) => x= (-m(v-initial)^2)/(2qE)

The electric field of a plane charge with surface charge density n is E=n/2(e_0)
e_0= 8.85*10^-12

x= -m(v-initial)^2(e_0)/qn
 
  • #6
rl.bhat
Homework Helper
4,433
7
i used the equations
(v-final)^2= (v-initial)^2 + 2ax
F=qE=ma

a= (qE)/m = -(v-initial)^2/(2x) => x= (-m(v-initial)^2)/(2qE)

The electric field of a plane charge with surface charge density n is E=n/2(e_0)
e_0= 8.85*10^-12

x= -m(v-initial)^2(e_0)/qn
That is correct. Previously square was missing.
 

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