Charged Particle Travelling Through Electric Field

AI Thread Summary
A charged particle with a negative charge is traveling horizontally through a uniform electric field of 500 N/C directed downward. The force acting on the particle combines the electric force, which acts upward due to its negative charge, and the downward gravitational force. The net force is calculated as F = qE - mg, resulting in a downward force of -0.05 N. The acceleration is determined to be 5 m/s², and using kinematic equations, the vertical displacement is calculated to be 0.001 m by the time the particle exits the field. The discussion clarifies the correct interpretation of the electric field and the inclusion of gravitational effects in the calculations.
pdeco1
Messages
13
Reaction score
0

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.
 
Physics news on Phys.org
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:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top