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@Karan Chatrath
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@Karan Chatrath OMG there is a coincidence if you watch answer basically the $L$ in your answer is managing the need of square root.
That's really sexy.

@Lil Doug
–
I think that the book is wrong here. I have done a numerical simulation and checked the expression that I got and the one in the book, and the one I have obtained seems correct. I even checked various possible cases by varying the initial separation parameter.

@Steven Chase could you validate this observation when possible?

@Karan Chatrath
–
I'm not even convinced that there is a maximum distance. I got the following results with the numerical values given in the code. It looks like the local maxima might just keep increasing.

@Steven Chase
–
I ran a quick simulation with your code and I see that the progressively increasing separation is a numerical issue that you seem to be having. If I use the explicit Euler solver, I get a result similar to yours. I am using a higher-order accurate integrator.

@Karan Chatrath
–
Yes, I neglected to check with multiple time steps and compare. When I run with a microsecond time step, the drift upwards is much smaller, but still there. It almost seems as if with a very small time step, the particle separation will oscillate between $L$ and zero in perpetuity.

@Steven Chase
–
I think it oscillates between $0.083$ and $L$. This is the value I get with the expression and also validated using simulations. From the plots, it looks like zero as 0.083 is a relatively small magnitude.

@Steven Chase
–
Yes, initially, I just solved for the fixed points of $\dot{x}=f(x)$ and thought that my job is done. But when @Lil Doug shared the answer with me, showing the various cases, I realised that I had not thought it through. Having said that, I do think that the expression that the book shows is not correct.

@Steven Chase
–
I ran the simulation using the following parameters and got the result with matches with the expression that I derived. While testing with closed-form expressions, I like to use irrational numbers as parameters.

Apply initial conditions and solve for integration constant $C$. Finally, the separation is max or min when $\dot{x} = 0$. Equating $\dot{x}=0$ gives a quadratic equations, one root of which is obviously $L$. The other is:

$d = \frac{q(M+m)}{4 \pi \epsilon_o EL(M-m)}$

Whether $d$ is a max or min can be confirmed by a second derivative test, which gives rise to cases.

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## Comments

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$d_{max} = \frac{q(M+m)}{4 \pi \epsilon_o EL(M-m)}$

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@Karan Chatrath No,but very very near to correct answer. I am not telling you answer because you always say, don't tell answer.

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Maybe I am missing something. Cannot spot my mistake.

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@Karan Chatrath if you don't mind me, your answer is also incorrect with dimension.

By the way, do you like to see answer?

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$L$ in your answer is managing the need of square root.

@Karan Chatrath OMG there is a coincidence if you watch answer basically theThat's really sexy.

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@Steven Chase could you validate this observation when possible?

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$L$ and zero in perpetuity.

Yes, I neglected to check with multiple time steps and compare. When I run with a microsecond time step, the drift upwards is much smaller, but still there. It almost seems as if with a very small time step, the particle separation will oscillate betweenLog in to reply

$0.083$ and $L$. This is the value I get with the expression and also validated using simulations. From the plots, it looks like zero as 0.083 is a relatively small magnitude.

I think it oscillates betweenLog in to reply

$L$ for those numbers. But the expression doesn't reduce to $L$, does it?

So the max separation isLog in to reply

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$\dot{x}=f(x)$ and thought that my job is done. But when @Lil Doug shared the answer with me, showing the various cases, I realised that I had not thought it through. Having said that, I do think that the expression that the book shows is not correct.

Yes, initially, I just solved for the fixed points ofLog in to reply

$q = \sqrt{2}$ $E = \sqrt{5}$ $L = \frac{\pi}{20}$ $m = \mathrm{e}$ $M = \mathrm{e}^2$ $\epsilon_o = 1$

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@Steven Chase My attempt:

Coordinate of $M$ is $x_1$ and that of $m$ is $x_2$.

Let:

$x = x_1 - x_2$

Applying Newton's second law gives:

$M\ddot{x}_1 = qE + \frac{Kq^2}{x^2}$ $m\ddot{x}_2 = qE - \frac{Kq^2}{x^2}$

Manipulating and rearranging both equations gives:

$\ddot{x}_1 - \ddot{x}_2 = \frac{Kq^2}{x^2}\left(\frac{1}{M} + \frac{1}{m}\right) + qE\left(\frac{1}{M} - \frac{1}{m}\right)$

$\implies \ddot{x} = A + \frac{B}{x^2}$

$A = qE\left(\frac{1}{M} - \frac{1}{m}\right)$ $B =Kq^2\left(\frac{1}{M} + \frac{1}{m}\right)$

$x(0) = L$ $\dot{x}(0) = 0$

$\ddot{x} = A + \frac{B}{x^2}$ $\implies \dot{x} d(\dot{x}) = \left(A + \frac{B}{x^2}\right) \ dx$

Integrating gives:

$\frac{\dot{x}^2}{2} = A + \frac{B}{x^2} + C$

Apply initial conditions and solve for integration constant $C$. Finally, the separation is max or min when $\dot{x} = 0$. Equating $\dot{x}=0$ gives a quadratic equations, one root of which is obviously $L$. The other is:

$d = \frac{q(M+m)}{4 \pi \epsilon_o EL(M-m)}$

Whether $d$ is a max or min can be confirmed by a second derivative test, which gives rise to cases.

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Does this method distinguish between local and global maxima?

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Since we are solving a quadratic equation which has only two solutions, the local maxima and minima are actually the global maxima and minima.

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@Karan Chatrath in the 3rd line of your solution it will be $m$

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Nice catch. Corrected now.

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@Lil Doug Bro your problems are

really hard. Lol.Log in to reply