Angles formed from power lines

I feel like banging my head on the door… I don’t understand the problem well… I am assuming power line means line of force… or maybe electric field line…

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I don’t even know why the electric field line should intersect the x-axis. Maybe we need to find \beta as denoted in my drawing?

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Yeah exactly… and I dont even know why or how it will intersect at B… maybe something to do with solid angle and flux etc… one of my friends was saying.

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Where do you think C is present? I feel like your diagram is incorrect…

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So what exactly is final tangent?

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Bruh that is what I am asking… :sob::sob:

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So confusing

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@Ersultan yo check this out

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We can write the conservation of flux along the power tube created by power lines emanating from q that form the same angle \alpha with AB.

If the power line ends on the charge q', we will have 2\pi q(1-\cos(\alpha)) = 2\pi q'(1-\cos(\beta) because near each point charge flux of the other charge one can be neglected. Then we will have \displaystyle \sin \left(\frac{\beta}{2}\right)^2 = \frac{q}{q'} \sin \left(\frac{\alpha}{2}\right)^2 , and power line will intersect x in the point B, where q' is located. (2\pi (1-\cos(\alpha) came from solid angle)

However, if \displaystyle \frac{q}{q'} \sin \left(\frac{\alpha}{2}\right)^2>1 the equation for \beta will not work, and we will need to consider more general equation for the power line. By using the same method of flux conservation we can get the r(x) as an implicit function.

Calculating the flux through the circle of radius r(x) by each charge, and requiring the sum of them to be constant we will get \displaystyle q(1- \frac{x}{\sqrt{r^2+x^2}}) - q' (1 - \frac{x-l}{\sqrt{r^2+(x-l)^2}})=C.

After using the fact that C = q(1-\cos(\alpha)) (the initial flux), we will get \displaystyle q(\cos(\alpha) - \frac{x}{\sqrt{r^2+x^2}}) - q' (1 - \frac{x-l}{\sqrt{r^2+(x-l)^2}})=0 .

It is obvious that in this case power line must go to infinity, and therefore we can find an asymptote of this equation by taking the limit r \rightarrow \infin, which will give us the straight line equation (q \cos(\alpha) - q')r = (q - q') x + q'l. The gradient of this line will be the tan(\beta), and x for r=0 will be the location of intersection of the final tangent.

The intersection of this final tangent will be \displaystyle x = - \frac{q'}{q - q'}l, and the angle will be \tan(\beta) = \frac{q-q'}{q \cos(\alpha) - q'}

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oh well, I could never have thought of that myself… Thanks a lot! You’re awesome…

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