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Electric Fields Question Uniform Field 
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#1
Jan3110, 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 µC is at the origin moving with a velocity v0 = 25 m/s at an angle θ = 25° above the xaxis. (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 ycoordinates of the position of the particle? x = ? y = ? HELP: Recall and apply the kinematic expressions for 2D 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^2vo^2 = 2a(xxo) 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? 


#2
Jan3110, 09:15 PM

P: 26

Hi
acceleration is Force divided by mass Best regards 


#3
Jan3110, 09:17 PM

P: 621

whoops, answered above. 


#4
Jan3110, 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.



#5
Feb110, 06:13 AM

P: 26

v0 is a vector doing 25º
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 


#6
Feb110, 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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