Story of e x e^{x} and Flux

A particle with some charge is position at N ( x , y , z ) = ( 0 , 5 , 1 ) N (x, y, z)=(0,5,1) at that green dot as shown in figure. If the flux passing between the equations e x e^{x} and y = 0 y=0 for x < 0 x<0 is ϕ 1 \phi_{1} and the flux passing between the equations e x e^{x} and y = 0 y=0 for x > 0 x>0 is ϕ 2 \phi_{2} .Determine the ratio: ϕ 2 ϕ 1 \frac{\phi_{2}}{\phi_{1}} .


The answer is 110.534.

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1 solution

Steven Chase
Mar 15, 2020

My standard numerical integration approach doesn't work as well for this one, since the integration regions are infinite. So I outsourced the computations to Wolfram. I left out some of the scalar multipliers since they cancel out in the ratio (hence the "prime" superscripts).

ϕ 1 = 0 e x 0 d x d y ( x 2 + ( y 5 ) 2 + 1 ) 3 / 2 0.00818958 ϕ 2 = 0 e x 0 d x d y ( x 2 + ( y 5 ) 2 + 1 ) 3 / 2 0.90523 \phi'_1 = \int_0^{e^x} \int_{-\infty}^0 \frac{dx \, dy}{\Big(x^2 + (y-5)^2 + 1 \Big)^{3/2}} \approx 0.00818958 \\ \phi'_2 = \int_0^{e^x} \int_0^{\infty} \frac{dx \, dy}{\Big(x^2 + (y-5)^2 + 1 \Big)^{3/2}} \approx 0.90523

Did it the same way. I like this problem, especially because the moment I saw it, I thought that the resulting integrals are likely to diverge. But this is not so, since the electric field magnitude also tends to zero as x x approaches infinity on either side of the number line, resulting in convergence. Interesting how the math works out here.

Karan Chatrath - 1 year, 2 months ago

I suppose the integrals can't diverge, because the total flux is a finite number (whatever would go through a sphere surrounding the charge). And of that, exactly half will go through the x y xy plane. And then the fluxes under consideration are fractions of the flux going through the x y xy plane.

Steven Chase - 1 year, 2 months ago

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Oh, yes, that is a good argument. Had not thought of that.

Karan Chatrath - 1 year, 2 months ago

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