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Particle in a uniform electric field

  • Thread starter Bryon
  • Start date
  • #1
99
0

Homework Statement


Throughout space there is a uniform electric field in the -y direction of strength E = 450 N/C. There is no gravity. At t = 0, a particle with mass m = 4 g and charge q = -19 µC is at the origin moving with a velocity v0 = 40 m/s at an angle θ = 25° above the x-axis.

(a) What is the magnitude of the force acting on this particle?

F = 0.00855N

(b) At t = 5.7 s, what are the x- and y-coordinates of the position of the particle?

x = 206.6381754m
y = ?

Homework Equations



y = y0 + vy0*t + 0.5*ay*t2

vx = v*cos(25)
vy = v*sin(25)

a = m/F

The Attempt at a Solution



a = 0.004/0.00855 = 0.467836257

vy = 40*sin(25) = 16.90473047

y = 16.90473047*5.7 + 0.5*(-0.467836257)*5.7^2 = 88.75696368

Im not sure where I messed up here. Any thoughts?

Thanks!
 

Answers and Replies

  • #2
911
18
hi

your calculation for 'a' is wrong

[tex] a=\frac{F}{m} = \frac{0.00855}{0.004} = 2.1375 m/s^2 [/tex]

now electric field is downwards , and charge is negative so the electric force on the charge will be in +y direction, so your acceleration should be positive.

I am getting [itex] y = 131.08 \, \, m [/itex]


:smile:
 
  • #3
gneill
Mentor
20,793
2,773
Make sure of the direction of the force on the particle.
 
  • #4
99
0
Ah thanks...I didnt realize I flipped mass and force for acceleration, and I should have realized the correct direction. Ill punch in the numbers and see what I get. Thanks!
 

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