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Motion of a Charged Particle in an Electric Field

by zandbera
Tags: charged, electric, field, motion, particle
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zandbera
#1
Mar31-09, 02:40 AM
P: 18
1. The problem statement, all variables and given/known data
A proton moves at 4.50 x 108 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 x 103 N/C. Ignoring any gravitational effects, find (a) the time interval required for the proton to move 5.00 cm horizontally, (b) its vertical displacement during that time interval, and (c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.


2. Relevant equations
[tex]\vec{a}[/tex] = q[tex]\vec{E}[/tex] / m


3. The attempt at a solution
(a) Since it's moving at constant horizontal velocity, and the only force its encountering is a vertical force, would the particle continue to move at the constant horizontal speed? Which would mean that the time interval to move 5.00 cm is just its velocity / 5.00 cm??

(b) I don't know how to relate the electric field to the velocity. But I know that the initial vertical velocity is 0, right?

(c) If i was right in (a) then its horizontal component will be the same
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Doc Al
#2
Mar31-09, 07:39 AM
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Quote Quote by zandbera View Post
(a) Since it's moving at constant horizontal velocity, and the only force its encountering is a vertical force, would the particle continue to move at the constant horizontal speed?
Right.
Which would mean that the time interval to move 5.00 cm is just its velocity / 5.00 cm??
You have that fraction upside down. Distance = velocity*time.

(b) I don't know how to relate the electric field to the velocity.
But you know how to find the acceleration caused by the electric field. The rest is kinematics.
But I know that the initial vertical velocity is 0, right?
Right.

(c) If i was right in (a) then its horizontal component will be the same
That's true.
zandbera
#3
Apr1-09, 12:16 AM
P: 18
(b) From the equation and my given data, is qE = 9.60 x 10^3 N/C or is it just E = 9.60 x 10^3?

If it's E = 9.60 x 10^3, I would just multiply by (+e) to get qE and then divide by the mass of a proton, right?

zandbera
#4
Apr1-09, 12:46 AM
P: 18
Motion of a Charged Particle in an Electric Field

Okay yeah for (b) I did what I thought, I used E = 9.60 x 10^3, then multiplied by +e, divided by mass of proton to get a, then used delta x = Vi t + 1/2 a t^2 with Vi = 0, and i got 5.68 mm, the correct answer.

So then for (c), the horizontal component is just the initial horizontal velocity, and then i can use the time interval from (a) and the acceleration from (b) to find the vertical velocity, so would those be my two components?
Doc Al
#5
Apr1-09, 07:23 AM
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Quote Quote by zandbera View Post
So then for (c), the horizontal component is just the initial horizontal velocity, and then i can use the time interval from (a) and the acceleration from (b) to find the vertical velocity, so would those be my two components?
Right!


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