Electric field and force

Can we solve this problem Anayltically
I will share my attempt within some hours.

Note by Talulah Riley
2 months, 2 weeks ago

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1 vote

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Is the answer:

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

Karan Chatrath - 2 months, 2 weeks ago

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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.

Talulah Riley - 2 months, 2 weeks ago

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Maybe I am missing something. Cannot spot my mistake.

Karan Chatrath - 2 months, 2 weeks ago

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@Karan Chatrath @Karan Chatrath if you don't mind me, your answer is also incorrect with dimension.
By the way, do you like to see answer?

Talulah Riley - 2 months, 2 weeks ago

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@Talulah Riley I think my answer is dimensionally correct. Please recheck this. Also, you may share the correct answer.

Karan Chatrath - 2 months, 2 weeks ago

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

Talulah Riley - 2 months, 2 weeks ago

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@Talulah Riley 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 - 2 months, 2 weeks ago

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@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.

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import math

dt = 10.0**(-5.0)

q = 2.0
m = 3.0
M = 5.0
E = 7.0
L = 11.0
e0 = 1.0

k = 1.0/(4.0*math.pi*e0)

Dref = q*(M+m)/(4.0*math.pi*e0*E*L*(M-m))

#################################

t = 0.0
count = 0

xm = 0.0
xM = L

xmd = 0.0
xMd = 0.0

xmdd = 0.0
xMdd = 0.0

while t <= 200.0:

    xm = xm + xmd*dt
    xM = xM + xMd*dt

    xmd = xmd + xmdd*dt
    xMd = xMd + xMdd*dt

    D = xM - xm

    F = k*q*q/(D**2.0)

    Fm = q*E - F
    FM = q*E + F

    xmdd = Fm/m
    xMdd = FM/M

    t = t + dt
    count = count + 1

    if count % 1000 == 0:
        print t,D

Steven Chase - 2 months, 2 weeks ago

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@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 - 2 months, 2 weeks ago

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@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 L and zero in perpetuity.

Steven Chase - 2 months, 2 weeks ago

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@Steven Chase I think it oscillates between 0.0830.083 and LL. 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.

Karan Chatrath - 2 months, 2 weeks ago

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@Karan Chatrath So the max separation is L L for those numbers. But the expression doesn't reduce to L L , does it?

Steven Chase - 2 months, 2 weeks ago

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@Steven Chase Indeed it does not. For the numbers you have chosen, the expression yields the minimum separation

Karan Chatrath - 2 months, 2 weeks ago

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@Karan Chatrath I'm getting the impression that the qualitative behavior depends on the particular constants chosen.

Steven Chase - 2 months, 2 weeks ago

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@Steven Chase Yes, initially, I just solved for the fixed points of x˙=f(x)\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.

Karan Chatrath - 2 months, 2 weeks ago

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@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.

q=2q = \sqrt{2} E=5E = \sqrt{5} L=π20L = \frac{\pi}{20} m=em = \mathrm{e} M=e2M = \mathrm{e}^2 ϵo=1\epsilon_o = 1

Karan Chatrath - 2 months, 2 weeks ago

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@Talulah Riley I look forward to seeing your attempt, after which I will share mine. Later, though.

Karan Chatrath - 2 months, 2 weeks ago

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@Steven Chase My attempt:

Coordinate of MM is x1x_1 and that of mm is x2x_2.

Let:

x=x1x2x = x_1 - x_2

Applying Newton's second law gives:

Mx¨1=qE+Kq2x2M\ddot{x}_1 = qE + \frac{Kq^2}{x^2} mx¨2=qEKq2x2m\ddot{x}_2 = qE - \frac{Kq^2}{x^2}

Manipulating and rearranging both equations gives:

x¨1x¨2=Kq2x2(1M+1m)+qE(1M1m)\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)

    x¨=A+Bx2\implies \ddot{x} = A + \frac{B}{x^2}

A=qE(1M1m)A = qE\left(\frac{1}{M} - \frac{1}{m}\right) B=Kq2(1M+1m)B =Kq^2\left(\frac{1}{M} + \frac{1}{m}\right)

x(0)=Lx(0) = L x˙(0)=0\dot{x}(0) = 0

x¨=A+Bx2\ddot{x} = A + \frac{B}{x^2}     x˙d(x˙)=(A+Bx2) dx\implies \dot{x} d(\dot{x}) = \left(A + \frac{B}{x^2}\right) \ dx

Integrating gives:

x˙22=A+Bx2+C\frac{\dot{x}^2}{2} = A + \frac{B}{x^2} + C

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

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

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

Karan Chatrath - 2 months, 2 weeks ago

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Does this method distinguish between local and global maxima?

Steven Chase - 2 months, 2 weeks ago

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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.

Karan Chatrath - 2 months, 2 weeks ago

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

Talulah Riley - 2 months, 2 weeks ago

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Nice catch. Corrected now.

Karan Chatrath - 2 months, 2 weeks ago

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@Lil Doug Bro your problems are really hard. Lol.

Krishna Karthik - 2 months, 2 weeks ago

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