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Electric Fields Question- Uniform Field

by r34racer01
Tags: electric, field, fields, uniform
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r34racer01
#1
Jan31-10, 09:09 PM
P: 63


Throughout space there is a uniform electric field in the -y direction of strength E = 540 N/C. There is no gravity. At t = 0, a particle with mass m = 3 g and charge q = -17 ÁC is at the origin moving with a velocity v0 = 25 m/s at an angle θ = 25░ above the x-axis.

(a) What is the magnitude of the force acting on this particle?
F = 0.00918N

(b) At t = 6.5 s, what are the x- and y-coordinates of the position of the particle?
x = ? y = ?
HELP: Recall and apply the kinematic expressions for 2-D projectile motion from mechanics.


E = F/Q = (KQ)/r^2
Range = (vo^2*sin(2Θ))/g
Trajectory = x*tan(Θ)-(1/2)((g*x^2)/(vo^2*cos^2(Θ))
v = vo+at
x = xo + v0t+(1/2)at^2
v^2-vo^2 = 2a(x-xo)


Ok pt a.) was really easy but I'm completely stuck on part b. It says to use kinematics equations but I don't see how that can work. We're not given acceleration and we're told gravity is not acting on the particle, and we don't know the speed at t=6.5s. I thought maybe I could set the force from pt a equal to m*a and that gave me 3.06 but the coordinates I got were not correct. Am I suppose to assume that the acceleration is zero since the electric force field is constant?
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Zhivago
#2
Jan31-10, 09:15 PM
P: 26
Hi

acceleration is Force divided by mass

Best regards
pgardn
#3
Jan31-10, 09:17 PM
P: 621
Quote Quote by r34racer01 View Post


Throughout space there is a uniform electric field in the -y direction of strength E = 540 N/C. There is no gravity. At t = 0, a particle with mass m = 3 g and charge q = -17 ÁC is at the origin moving with a velocity v0 = 25 m/s at an angle θ = 25░ above the x-axis.

(a) What is the magnitude of the force acting on this particle?
F = 0.00918N

(b) At t = 6.5 s, what are the x- and y-coordinates of the position of the particle?
x = ? y = ?
HELP: Recall and apply the kinematic expressions for 2-D projectile motion from mechanics.


E = F/Q = (KQ)/r^2
Range = (vo^2*sin(2Θ))/g
Trajectory = x*tan(Θ)-(1/2)((g*x^2)/(vo^2*cos^2(Θ))
v = vo+at
x = xo + v0t+(1/2)at^2
v^2-vo^2 = 2a(x-xo)


Ok pt a.) was really easy but I'm completely stuck on part b. It says to use kinematics equations but I don't see how that can work. We're not given acceleration and we're told gravity is not acting on the particle, and we don't know the speed at t=6.5s. I thought maybe I could set the force from pt a equal to m*a and that gave me 3.06 but the coordinates I got were not correct. Am I suppose to assume that the acceleration is zero since the electric force field is constant?
You calculated a Force. YOu have the mass of the particle, so F = ma... Just gotta get the direction of the net force, thus the acceleration correct. Then use your kinematics. But dont use g, use the a you found from F=ma..

whoops, answered above.

r34racer01
#4
Jan31-10, 09:41 PM
P: 63
Electric Fields Question- Uniform Field

Ok well like I said before I tried F = ma => 0.00918 = 0.003kg * a => a = 3.06 m/s^2. I then did x = x0+v0*t + 0.5*a*t^2 ==> x = 0 + 25(6.5) + 0.5(3.06)(6.5)^2 = 227.1425. I then resolved this into x and y components and got x = 205.86 and y = 95.9. These didn't work so I'm still lost.
Zhivago
#5
Feb1-10, 06:13 AM
P: 26
v0 is a vector doing 25║
E (and therefore F) is a vector going in the y direction.

You must find the velocity in the x and y component, the force in the x and y component (in this case the x component of the force is 0), and apply kinematic's equation on each component.
Hope this helps
emanuel_hr
#6
Feb1-10, 07:26 AM
P: 10
The only difference form the classical projectile problem is that the gravitational field is replaced by the electrostatic field, they obey the same laws, only the expression of the force is different(G = mg for gravity, F = qE in this case, F plays the role of G).


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