Solving Problems in Projectile Motion, SHM of Charge Ring & Electric Quadrapole

In summary: Q}{ma^3}}which is half of the frequency of the SHM. So the full frequency would bef = \frac{1}{2\pi}\sqrt{\frac{kqQ}{ma^3}} = \frac{1}{2\pi}\sqrt{\frac{kQq}{ma^3}}.
  • #1
Gale
684
2
i have a few problems that i was struggling with.

Projectile motion:
An electron is fired in an electric field E=-720j. Its launched with a velocity of 9550 m/s. if you want the x distance it travels to be 1.27mm, what are the two values of theta that will work? how long does it take?

so, like i do all projectile problems, i found t using the x components, (which i left in terms of theta,) because there's no force in x. then i plugged t into y_f= 1/2at^2 +v_ot. my problem was that i was left with cos and sin's and i didn't know what to do with them. i couldn't easily solve for theta. and at any rate, i wasn't really sure how to use the electric field. when i thought about gravity fields, i thought that i should be using energy or something, and i just got confused.

SHM of charge ring:
i forget the wording of this one, but its just like, show that the frequency of a particle moving on the axis of a ring is [tex]f=\frac{1}{2\pi}*(\frac{kQq}{ma^3})^1/2[/tex]

i tried starting with the electric field, and getting the force from that. when my prof did an example with the spring force in class, he did this really weird thing i can't follow now. he was like f=1/s and so if F=kx=ma then k/m*x=d^2x/dt^2, and what function looks like its second derivative? sin. and then we have the constant k/m, which must be frequency squares since sin(ax)'s second deriv is a^2sin(ax). and that's how he found the frequency for the spring. i have NO idea how to do that with the electric force. i tried working backwards too, but i couldn't really make sense of it because there's x's in the denominator of the force and not the frequency, and yeah, I'm stuck.

The last problem I'm stuck on is this-
Electric quadrapole:
whats the electric field along the y-axis if there's a quadrapole on the x axis, with charge q at (-a,0), -2q at (0,0) and q at (a,0)?

so i started with writing the field generated by each. knowing that the x components of the two q charges would cancel, and their y's would be equal. so i got two terms, and i simplified and used the binomial theorem, and i couldn't get it right. but i think I'm going to try this one again.

thanks!
 
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  • #2
i really do need help on the harmonic motion one. I'm super stuck. I'm doing the first one with work-energy theorem, and i think I'm just screwing up the math in the third one, but the second one I'm really clueless on.
 
  • #3
Gale said:
i have a few problems that i was struggling with.

Projectile motion:
An electron is fired in an electric field E=-720j. Its launched with a velocity of 9550 m/s. if you want the x distance it travels to be 1.27mm, what are the two values of theta that will work? how long does it take?

so, like i do all projectile problems, i found t using the x components, (which i left in terms of theta,) because there's no force in x. then i plugged t into y_f= 1/2at^2 +v_ot. my problem was that i was left with cos and sin's and i didn't know what to do with them. i couldn't easily solve for theta. and at any rate, i wasn't really sure how to use the electric field. when i thought about gravity fields, i thought that i should be using energy or something, and i just got confused.

Rewrite everything in terms of [tex] tan (\theta) [/tex] (you will need to use [tex] {1 \over cos^2 \theta} = 1 + tan^2 (\theta) [/tex]). Then you wilh have a quadratic formula for [tex] tan (\theta) [/tex] . Solve and then take the inverse tan of the two roots.


SHM of charge ring:
i forget the wording of this one, but its just like, show that the frequency of a particle moving on the axis of a ring is [tex]f=\frac{1}{2\pi}*(\frac{kQq}{ma^3})^1/2[/tex]

i tried starting with the electric field, and getting the force from that. when my prof did an example with the spring force in class, he did this really weird thing i can't follow now. he was like f=1/s and so if F=kx=ma then k/m*x=d^2x/dt^2, and what function looks like its second derivative? sin. and then we have the constant k/m, which must be frequency squares since sin(ax)'s second deriv is a^2sin(ax). and that's how he found the frequency for the spring. i have NO idea how to do that with the electric force. i tried working backwards too, but i couldn't really make sense of it because there's x's in the denominator of the force and not the frequency, and yeah, I'm stuck.


That's right. All you have to do is to show that the electric force on the point charge has the form [tex] - C~ y [/tex] where C is a constant. Then, y(t) has to follow simple harmonic motion with [tex] \omega [/tex] given by [tex] \omega = {\sqrt {C \over mass}} [/tex]. So all you have to know is the electric field of a ring along its axis and that times the charge of the point charge will give you the force.
 
  • #4
For your first question:
[tex]\vec{a} = \vec{F}/m_e = \frac{-e(-720\vec{j})}{m_e}[/tex]

[tex]at = v_0 \sin(\theta)[/tex]
[tex]\Delta x = v_0 \cos(\theta) t[/tex]
substituting t with other stuff
[tex]\Delta x = v_0 \cos(\theta) v_0 \sin(\theta) / a[/tex]

I think what you needed was trigonometic identity [tex]2 \sin(\theta) \cos(\theta) = \sin(2 \theta)[/tex]

Use it and you have

[tex]\sin(2 \theta)/2 = \frac{\Delta x a}{v_0^2}[/tex]
which you can easily solve for the angle.

For the second one, I guess you mean an axis passing through the middle of a ring, perpendicular to the surface of ring. The force for it is:
[tex]\frac{kqQx}{(x^2 + a^2)^{3/2}}[/tex]
where a is the radius of circle, Q is the charge of ring, and q is our hero. It appears that you've assumed a>>r, so the force would be simplified into
[tex]\frac{kqQx}{a^3}[/tex]. Plugging this to 2nd law of Newton
[tex]\frac{md^2}{dt^2} = \frac{kqQx}{a^3}[/tex]
It's solution Asin(wt) + Bcos(wt) where w is [tex]\sqrt{\frac{kqQx}{ma^3}}[/tex]. Since [tex]\omega = 2 \pi f[/tex]...

And finally, or the third: E at (0,y) should be the superposition of all three. For q charges, it's [tex](kq/{r^2}) \sin(\theta) = (kq/{r^2}) \frac{y}{r}[/tex], and [tex]r=\sqrt{y^2+a^2}[/tex]. There're 2 of them. And for -2q, it's simply [tex]k(-2q)/{y^2}[/tex]. Just add the 3 then.
 

1. What is projectile motion and how is it solved?

Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity. It can be solved using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and time.

2. What is simple harmonic motion (SHM) of a charge ring?

SHM refers to a type of periodic motion in which an object moves back and forth along a straight line. In the case of a charge ring, the motion is caused by the interaction between the charge on the ring and an external electric field.

3. How is the equation of SHM derived for a charge ring?

The equation for SHM of a charge ring is derived using the principles of electromagnetism and Newton's laws of motion. It involves calculating the forces acting on the ring and setting them equal to the mass times the acceleration.

4. What is an electric quadrupole and how does it affect the motion of a charged particle?

An electric quadrupole refers to a system of two equal and opposite charges separated by a certain distance. It affects the motion of a charged particle by creating an electric field that can either attract or repel the particle, depending on its charge and position relative to the quadrupole.

5. How can problems involving projectile motion and SHM of a charge ring be solved using numerical methods?

Numerical methods, such as Euler's method or the Runge-Kutta method, can be used to solve problems involving projectile motion and SHM of a charge ring. These methods involve breaking down the problem into smaller steps and using numerical calculations to approximate the solution.

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