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Path of a deflected charge? Solvable ODE's?

  • #1

Homework Statement


Suppose a particle of mass m and charge q is moving with velocity v(t) to the right in the second quadrant of a coordinate system C and whose position is described by the vector r(t) pointing from the origin to the particle's location in C. A charge with magnitude Q is fixed at the origin of this coordinate system. What would be the path of the moving charge as it is deflected by the fixed, stationary charge Q?

Homework Equations


$$\vec F_e=\frac {qQ}{4\pi\epsilon_0[r(t)]^2} \hat r$$ $$\vec F =m\vec a $$
$$\hat r = \frac {\vec r}{r(t)}$$
$$\vec r = x(t) \hat i + y(t) \hat j$$
$$\vec a = \frac {d^2x}{dt^2} \hat i + \frac {d^2y}{dt^2} \hat j$$ $$r(t) = \sqrt{x^2+y^2}$$

The Attempt at a Solution


The electric force exherted on q by the fixed charge will be equal to Sir Isaac Newton's Second Law of Motion (since Q will impart an acceleration to the components of q). Thus, $$\frac {qQ}{4\pi\epsilon_0[r(t)]^2} \hat r = m\vec a $$ and after some substituting, $$\frac {qQ}{4\pi\epsilon_0[r(t)]^3} [x(t) \hat i + y(t) \hat j] - m[\frac {d^2x}{dt^2} \hat i + \frac {d^2y}{dt^2} \hat j]= 0$$ Let $$ k=\frac {qQ}{4\pi\epsilon_0 m} $$ and then, to satisfy the preceeding equation, it must be that $$\frac {d^2x}{dt^2}=\frac {kx}{\sqrt{(x^2+y^2)^3}} $$ $$\frac {d^2y}{dt^2}=\frac {ky}{\sqrt{(x^2+y^2)^3}} $$ My thinking is that, if the parametrized coordinates x and y can be solved for, then we can trace out the path given by the vector r. If we let $$x=sin t$$ and $$y=cos t$$ the system of equations is solved, but this yields a circle of radius one for all values of t. Is my approach correct? Does this system of differential equations have an analytical solution? Should I be using also the Lorentz Force Law since the moving particle will emit radiation? I would appreciate any input on any information anybody may have and put me on the right track if I am making any mistakes or have any misconceptions. Thanks! :)
 

Answers and Replies

  • #2
BvU
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Suppose a particle of mass m and charge q is moving with velocity v(t) to the right in the second quadrant of a coordinate system C and whose position is described by the vector r(t) pointing from the origin to the particle's location in C
It seems to me this isn't a good problem statement. Usually there are initial conditions, not trajectory constraints, for second-order differential equations. So I'd expect something like

is moving with velocity ##\vec v(0) = (v_0, 0) ## to the right in the second quadrant of a coordinate system C and whose position is described by the vector ##\vec r(0) = (x_0, y_0) ## ...

Your solution has no room for initial conditions
 
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  • #3
George Jones
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$$\frac {d^2x}{dt^2}=\frac {kx}{\sqrt{(x^2+y^2)^3}} $$ $$\frac {d^2y}{dt^2}=\frac {ky}{\sqrt{(x^2+y^2)^3}} $$ My thinking is that, if the parametrized coordinates x and y can be solved for, then we can trace out the path given by the vector r. If we let $$x=sin t$$ and $$y=cos t$$ the system of equations is solved, but this yields a circle of radius one for all values of t. Is my approach correct?
Your trial solution does not satisfy the system of DEs, but it is close to satisfying the system. Plug your solution into the system of DEs, and see what happens!

Once you fix this up, you will have a particular solution to a particular case.

There are two very different cases to consider: 1) the charges have the same sign; 2) the charges have differing signs.

Your are close to having a particular solution to case 2). There is a different physical situation (not involving charges) that results in the same equations as case 2). This different physical situation, with which you are very familiar, will you tell you the types of trajectories possible for your problem.
 
  • #4
It seems to me this isn't a good problem statement. Usually there are initial conditions, not trajectory constraints, for second-order differential equations. So I'd expect something like

is moving with velocity ##\vec v(0) = (v_0, 0) ## to the right in the second quadrant of a coordinate system C and whose position is described by the vector ##\vec r(0) = (x_0, y_0) ## ...

Your solution has no room for initial conditions
I absolutely understand and the initial conditions to be supposed are also something I am a little unsure of. I am assuming that the charge q approaches from negative infinity, so would it suffice to reckon' that $$ \vec r(0) =x_0 \hat i + y_0 \hat j $$ where $$x(0)=x_0$$ and $$ y(0)=y_0$$ for instance? I'm not sure if that would help my cause (If I'm thinking about this correctly) but I realize I have not thought out the initial conditions and that more focus on that topic may be required to sufficiently propose the problem in a context that may provide a suitable solution.

EDIT: I realize that what I have stated in terms of initial conditions regarding the vector r is equivalent to what you have stated. The other condition that $$\vec v(0) = v_{x0} \hat i +0 \hat j $$ is suitable for my considerations as well. I can say that I completely understand what you have explained. Thank you for your help, it is very valuable and very much appreciated.
 
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  • #5
Your trial solution does not satisfy the system of DEs, but it is close to satisfying the system. Plug your solution into the system of DEs, and see what happens!

Once you fix this up, you will have a particular solution to a particular case.

There are two very different cases to consider: 1) the charges have the same sign; 2) the charges have differing signs.

Your are close to having a particular solution to case 2). There is a different physical situation (not involving charges) that results in the same equations as case 2). This different physical situation, with which you are very familiar, will you tell you the types of trajectories possible for your problem.
Whoops! Yes, so the solution would be $$ x(t)=sin\sqrt{k}t $$ and $$ y(t) = cos\sqrt{k}t $$ My problem is that these solutions represent a circle with a radius of 1. Varying the value of k just seems to vary the range of t values.. do these ranges of t correspond to different ranges of the values of the vector velocity v(t) and hence correspond to circles with different curvatures depending on v? So that the path taken by the particle will be a circular curve depending upon the velocity? Also, I mistakenly omitted the fact that I am considering the particle to be moving with a constant speed from negative infinity and I deeply apologize for my sloppiness.
 
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  • #6
George Jones
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Whoops! Yes, so the solution would be $$ x(t)=sin\sqrt{k}t $$ and $$ y(t) = cos\sqrt{k}t $$ My problem is that these solutions represent a circle with a radius of 1.
What about
$$x\left(t\right) = x_0 \sin\sqrt{k}t$$
$$y\left(t\right) = y_0 \cos\sqrt{k}t?$$

Is this also a solution to your system of DEs?

Also, I mistakenly omitted the fact that I am considering the particle to be moving with a constant speed from negative infinity
I don't see how this is possible. If the separation distance between the charges changes, then potential energy changes. Since total mechanical energy is conserved, this means that kinetic energy (and thus speed) changes.
 
  • #7
What about
$$x\left(t\right) = x_0 \sin\sqrt{k}t$$
$$y\left(t\right) = y_0 \cos\sqrt{k}t?$$

Is this also a solution to your system of DEs?



I don't see how this is possible. If the separation distance between the charges changes, then potential energy changes. Since total mechanical energy is conserved, this means that kinetic energy (and thus speed) changes.
Hmm, okay so then, $$ -x_0ksin\sqrt{k}t = \frac {x_0ksin\sqrt{k}t}{\sqrt{(x_0^2sin^2\sqrt{k}t+y_0^2cos^2\sqrt{k}t)^3}}$$ and if we solve for the square root in the denominator, $$\sqrt{(x_0^2sin^2\sqrt{k}t+y_0^2cos^2\sqrt{k}t)^3} = -1 $$ so hence we have a solution to the system (the other DE works out the same). So then q would take the path of an elliptic curve, for the case of a negative charge.
 
  • #8
vela
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It would be easier to solve this problem using polar coordinates. Any reason you're sticking with Cartesian coordinates?
 
  • #9
It would be easier to solve this problem using polar coordinates. Any reason you're sticking with Cartesian coordinates?
Honestly, my first approach was with polar coordinates and I found that to be much more cumbersome; my approach in regards to that aspect may have been off but that was what originally lead me to the cartesian analysis.
 
  • #10
What about
$$x\left(t\right) = x_0 \sin\sqrt{k}t$$
$$y\left(t\right) = y_0 \cos\sqrt{k}t?$$

Is this also a solution to your system of DEs?
So upon investigating the units, it has turned out for me that, for the x-component of the speed, which would according to the solution be $$ v_x(t) = \sqrt{k}x_0 cos \sqrt{k}t, $$ the units would work out as $$ L\sqrt{\frac{C^2KgL^3}{KgT^2C^2}}= \sqrt{\frac{L^5}{T^2}}=\frac{L^2\sqrt{L}}{T}$$ where L is the unit of length, T is the unit of time, C is the unit of electric charge, N is the unit of force and Kg is the unit of mass. The preceeding unit analysis does not yield a quantity of speed (assuming I haven't made any errors in my calculations). In order for the argument of the cosine function to be dimensionally consistent, then $$\sqrt{k} =T^{-1}$$ Can this be resolved or is this already true and I'm just missing something? Also, at t=0, the position vector r takes the form $$\vec r(0) = x_0sin(0)\hat i + y_0cos(0)\hat j =0\hat i +y_0 \hat j $$ which implies that the particle starts initially directly above the charge Q. I suppose imposing the desired initial conditions into the solution of the system of DE's may clear up this issue (assuming that the solutions provided are in fact solutions to the system)?



I don't see how this is possible. If the separation distance between the charges changes, then potential energy changes. Since total mechanical energy is conserved, this means that kinetic energy (and thus speed) changes.
I see, that is correct. Absolutely understood.
 
  • #11
George Jones
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What about
$$x\left(t\right) = x_0 \sin\sqrt{k}t$$
$$y\left(t\right) = y_0 \cos\sqrt{k}t?$$
Oops, should be
$$x\left(t\right) = R \sin\sqrt{\frac{K}{R^3}}t\\
y\left(t\right) = R \cos\sqrt{\frac{K}{R^3}}t$$
with
$$K = -k = - \frac {qQ}{4\pi\epsilon_0 m}.$$
 
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  • #12
George Jones
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Oops, should be
$$x\left(t\right) = R \sin\sqrt{\frac{K}{R^3}}t\\
y\left(t\right) = R \cos\sqrt{\frac{K}{R^3}}t$$
with
$$K = -k = - \frac {qQ}{4\pi\epsilon_0 m}.$$
Then.
$$\begin{align} \frac{dx}{dt} &= \sqrt{\frac{K}{R^3}} R \cos\sqrt{\frac{K}{R^3}}\\
\frac{d^2x}{dt^2} &= - \frac{K}{R^3} R \sin\sqrt{\frac{K}{R^3}} = - \frac{K}{R^3} x = \frac{k}{R^3} x \end{align}$$
 
  • #13
vela
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Honestly, my first approach was with polar coordinates and I found that to be much more cumbersome; my approach in regards to that aspect may have been off but that was what originally lead me to the cartesian analysis.
The solutions to this problem are well known to be the conic sections (if you neglect radiation). You're zeroing in on the circular-orbit solution, but the other ones are probably not as straightforward to solve for using cartesian coordinates.
 
  • #14
The solutions to this problem are well known to be the conic sections (if you neglect radiation). You're zeroing in on the circular-orbit solution, but the other ones are probably not as straightforward to solve for using cartesian coordinates.
Thank you for your input.
 
  • #15
Ray Vickson
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Homework Statement


Suppose a particle of mass m and charge q is moving with velocity v(t) to the right in the second quadrant of a coordinate system C and whose position is described by the vector r(t) pointing from the origin to the particle's location in C. A charge with magnitude Q is fixed at the origin of this coordinate system. What would be the path of the moving charge as it is deflected by the fixed, stationary charge Q?
This is a problem tackled by Rutherford in the early 20th century. Google "Rutherford scattering" to find some articles containing solutions. The Wikipedia entry is quite helpful.
 
  • #16
This is a problem tackled by Rutherford in the early 20th century. Google "Rutherford scattering" to find some articles containing solutions. The Wikipedia entry is quite helpful.
Awesome, thank you for your useful information.
 

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