Charged Particle Travelling Through Electric Field

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Homework Help Overview

The discussion revolves around a charged particle traveling through a uniform electric field at the Earth's surface. The problem involves determining the direction and magnitude of the vertical deflection of the particle as it exits the electric field region. Key parameters include the charge of the particle, its mass, velocity, and the strength of the electric field.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the electric field's units and how they relate to the forces acting on the charged particle. There are questions about the inclusion of gravitational force and how to properly account for it in the calculations. Some participants explore the relationship between vertical acceleration and displacement, while others question the initial conditions for vertical motion.

Discussion Status

The discussion is ongoing, with participants sharing their interpretations and calculations. Some have provided insights into the forces acting on the particle, while others are seeking clarification on the correct approach to incorporate gravity. There is no explicit consensus yet, but various lines of reasoning are being explored.

Contextual Notes

Participants note a potential typo in the homework regarding the units of the electric field, which has led to some confusion. The necessity of clarifying the role of gravity in the overall force acting on the particle is also a point of contention.

pdeco1
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Homework Statement


A uniform electric field is non zero in a 20x10m region at the surface of the earth. Determine the direction and magnitude of the total vertical deflection of the incoming charged particle by the time it exits the region. The charged particle is traveling horizontal through the electric field which is pointing down.

q= -10^-4 C
v= 1000m/s
m= .01 kg
g= 10m/s^2

E = 500N/m downward.

Homework Equations


F= qE + mg
a = qE/m
x = vt

The Attempt at a Solution



E, given in Newtons per meter is giving me problems.

I know the horizontal velocity is unchanged due to the perpendicular forces.
Vertical velocity and displacement can be found through kinematics when E is N/C.
 
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pdeco1 said:

Homework Statement


A uniform electric field is non zero in a 20x10m region at the surface of the earth. Determine the direction and magnitude of the total vertical deflection of the incoming charged particle by the time it exits the region. The charged particle is traveling horizontal through the electric field which is pointing down.

q= -10^-4 C
v= 1000m/s
m= .01 kg
g= 10m/s^2

E = 500N/m downward.


Homework Equations


F= qE + mg
a = qE/m
x = vt
You should never write formulas without stating what the parameter represent.

The Attempt at a Solution



E, given in Newtons per meter is giving me problems. [/quote]
It certainly should! "qE" as a force only makes sense when E is in "Newtons per coulomb"
or "Newtons per electron volt". Who told you that E was in "Newtons per meter"?

I know the horizontal velocity is unchanged due to the perpendicular forces.
Vertical velocity and displacement can be found through kinematics when E is N/C.
 
The question states E in Newtons/meter. I emailed my professor but he has yet to get back to me.

Edit: Was a typo on the homework. It is in N/C.

I am not sure where to include gravity. Is the Force in the vertical direction (F = qE -mg) Since the electric field creates an upward force on the particle but gravity creates a downward force?

I can find the time by x/v=t

I can find the vertical acceleration by a=qE/m or do I include gravity somehow?

Then I can find the vertical displacement by: y=vt + .5at^2 with vertical velocity initial = 0?

Is the vertical deflection just y?
 
Last edited:
Bump!
 
I would love some expert input.

The particle holds a (-) charge, so the electric field force would move the particle upwards, but the gravitational force will push the particle down.

Using the formula F = qE + m(-g) the Force = -.05N, the - indicating an overall Force in the downward direction.

Using a =qE/m I found the acceleration to be 5m/s(squared)

using y= .5at^2 = .5 x 5 x 4*10^-4 = .001

Thanks!
 
Last edited:

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