Electric field slowing a proton from 3.5 × 10⁶ m/s to rest

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pat666
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Homework Statement


A proton traveling along the x-axis is slowed by a uniform electric field E. At x = 20 cm, the proton has a speed of 3.5 x 106 ms-1 and at x = 80 cm, its speed is zero. Find the magnitude and direction of electric field intensity.

Can someone please! check my procedure and if your really nice/helpful my answer-----thanks.


Homework Equations


W=Fx
W=Eqx
W=ΔK_E
ΔK_E=1/2 mv^2-1/2 mu^2



The Attempt at a Solution


1st i found the change in energy (work) to slow the proton to a stop. then that's equal to Eqx. I found E to be 1.06*10^5 N/C (positive direction) please check my answer if you have time - its worth marks.
 
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The magnitude of the electric field is correct. In what direction should the force on the proton be and why?
 
thanks kuruman - I thought positive (is that what's meant by direction??) because they need to be "like" to repel??
 
I am talking about the force on the proton. If it moves left to right (x is increasing), in what direction should the force be to stop it?
 
towards the origin - - but left or right would be dependent on the way the cc were defined? +x could be towards the left or right couldn't it?
 
pat666 said:
towards the origin - - but left or right would be dependent on the way the cc were defined? +x could be towards the left or right couldn't it?
Yes it could, so let me rephrase my question. Relative to the direction of the proton's velocity, in what direction should the force on it point? Same or opposite direction? The answer to this question has nothing to do with which way is positive and which way is negative.
 
Yes. Now how is the direction of the force on a positively charged particle related to the direction of the electric field? Same or opposite?
 
same- the direction of an electric field is the way a positively charged test particle would go
 
So put it together. Is the direction of the electric field the same as or opposite to the direction of the proton's velocity?
 
so the full answer is 1.06*10^5 N/C in the opposite direction of the protons velocity.