Velocity of charged particle in a uniform electric field

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Homework Statement



A particle of mass 0.000103 g and charge 87 mC moves in a region of space where the
electric field is uniform and is 4.8 N/C in the x direction and zero in the y and z direction.
If the initial velocity of the particle is given by v_y = 3.2 × 10^5 m/s, v_x = v_z = 0, what is
the speed of the particle at 0.2 s?
Answer in units of m/s.

Homework Equations



E = E_x + E_y

a_x = (q/m)E_x
a_y = (q/m)E_y

a = a_x + a_y

v_f = V_0 + at

The Attempt at a Solution



q = 0.087 C
m = 1.03 * 10^-7 kg
E_x = 4.8 N/C
V_0)y = 3.2 * 10^5 m/s
V_0)x = 0 m/s
V_0)z = 0 m/s
t = 0.2s

a_x = (q/m) E_x = 4054368.932 m/s^2

a_y = 0

a = 4054368.932 m/s^2
 

Answers and Replies

  • #2
Okay use E=F/Q

Solve for F.

Now use F=ma

Solve for a

Now you have Initial V, acceleration, and t, solve for Final V.

Vf=Vo + at
 
  • #3
collinsmark
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q = 0.087 C
m = 1.03 * 10^-7 kg
E_x = 4.8 N/C
V_0)y = 3.2 * 10^5 m/s
V_0)x = 0 m/s
V_0)z = 0 m/s
t = 0.2s

a_x = (q/m) E_x = 4054368.932 m/s^2

a_y = 0

a = 4054368.932 m/s^2
Okay, so you've found the x-component of the acceleration. What's the x-component of the velocity, as a function of t? Then what is the x-component of the velocity when t = 0.2 s?

Hint: Use your kinematics equations.

Another hint: Once you have both the x- and -y components of the velocity, how you do find the magnitude of the velocity (aka the speed)? :wink:

[Edit: JustinLiang beat me to the hint.]
 
  • #4
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E=F/Q
F=QE
ma=QE
a=(QE)/m=4054368.932 m/s^2

Vf=(3.2x10^5) + (4054368.932)(0.2) = 401087.3786 m/s

The system is telling me this answer is wrong. Did I miss anything?

I don't know if the problem is that the electric field is pulling in the x-direction while my initial velocity is in the y-direction.
 
  • #5
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ohhhh! so the initial velocity in the x-direction would be:

Vf = 0 + (4054368.932)(0.2) = 81087.37864 m/s

and

V = sqrt ((81087.37864)2 + (3.2x10^5)2) = 330113.8637 m/s ?
 
  • #6
collinsmark
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E=F/Q
F=QE
ma=QE
a=(QE)/m=4054368.932 m/s^2

Vf=(3.2x10^5) + (4054368.932)(0.2) = 401087.3786 m/s
The quoted text in red is the initial velocity in the y direction, not the x!

Keep your components separate for now.

What is vx = vx0 + axt?
I don't know if the problem is that the electric field is pulling in the x-direction while my initial velocity is in the y-direction.
That's right. That means that the particle is accelerating in the x-direction. And since the constant electric field is perpendicular to the y direction, the particle does not accelerate in the y direction.
 
  • #7
ohhhh! so the initial velocity in the x-direction would be:

Vf = 0 + (4054368.932)(0.2) = 81087.37864 m/s

It is 810873m/s not 81087
 
  • #8
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Oh! Thank You for catching that mistake.

My answer now comes down to a final velocity of 871731.7804 m/s. Is this correct?
 
  • #9
collinsmark
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My answer now comes down to a final velocity of 871731.7804 m/s. Is this correct?
'Looks good to me. :approve: (A little overkill on the precision though. But yes, that's what I got.)
 
  • #10
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'Looks good to me. :approve: (A little overkill on the precision though. But yes, that's what I got.)

ha! I know, but the system wants the answer to six significant digits so I kind of have to keep EVERY digit up to the very end.

I cannot thank you both enough for your help and quick responses. Thank you so much!
 
  • #11
berkeman
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Okay use E=F/Q

Solve for F.

Now use F=ma

Solve for a

Now you have Initial V, acceleration, and t, solve for Final V.

Vf=Vo + at

Please take care not to give too much of the answer to the student asking the schoolwork question. Please ask probing questions, provide hints, find mistakes, etc. The student must do the bulk of the work. Thanks.
 
  • #12
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Please take care not to give too much of the answer to the student asking the schoolwork question. Please ask probing questions, provide hints, find mistakes, etc. The student must do the bulk of the work. Thanks.

With all due respect, the input actually helped me solve the problem properly
 
  • #13
berkeman
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With all due respect, the input actually helped me solve the problem properly

You mean "easily". What did you learn about how to figure out problems like this on your own? Not much from my vantage point. And helping students learn how to learn is a big goal of the PF HH forums. Please have a look at this thread, where we discuss why the PF HH rules are the way that they are:

https://www.physicsforums.com/showthread.php?t=373889

.
 

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