Asymptotic Field Lines

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NOTE: I didn't have a punctilious method to solve this problem, as it proved to be too messy. So, I came up with this kind of logical solution. I don't think this is the proper way to solve it, but anyway.

Above, I have shown 2 vector addition diagrams, in which, one of the vectors is varied in magnitude. Keeping this picture in mind, let us go through a thought process.

If $\displaystyle E_1,E_2$ represent the Electric Field due to $\displaystyle q_1,q_2$ at a general point which lies on the path traced by the field in the problem.

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$\bullet$ If $\displaystyle q_1$ grows in magnitude, $\displaystyle E_1$ will grow proportionally, and the resultant Electric field will become more steeper (taking reference from the first image).
Thus, if the line becomes more steeper, note that the length $\displaystyle BC$ will also grow proportionally .

$\bullet$ If $\displaystyle q_2$ grows in magnitude, $\displaystyle E_2$ will grow proportionally, and the resultant Electric field will become less steeper (taking reference from the first image).
Thus, if the line becomes less steeper, note that the length $\displaystyle AC$ will also grow proportionally .

Thus, if we let $\displaystyle AC = \ell$ , then clearly $\displaystyle BC = 10-\ell$ , then,
the mathematical translations of the two points given above are,

$\displaystyle q_1 \propto (10-\ell)$ and
$\displaystyle q_2 \propto \ell$

Dividing these two relations,
$\displaystyle \frac{q_2}{q_1} = \frac{\ell}{10-\ell}$

Substituting the values of $\displaystyle q_1$ and $\displaystyle q_2$ , we get,
$\displaystyle \ell = 2.5 cm$

Hence,
$\displaystyle\boxed{AC=2.5cm}$

I even don't know if this is the correct way to arrive at the solution. I just thought that there must be some simpler way to solve this problem. So, I thought about this, I applied it, and it worked. I worked hard on the images and the typing, so please be nice in the comments even if you don't feel that this solution is correct.

Considering At infinity we can assume that total charge were located at their common centre of charge and thus asymptote showing the direction of field will pass though the centre of charge 2.5 cm away from A

Here is a fully analytic solution. Let $\theta$ and $\phi$ be the angles that lines from the point $(x,y)$ to the points $A$ and $B$ make with the $x$ -axis. We have chosen $A$ and $B$ to lie on the $x$ -axis, and have chosen the origin $O$ to be the midpoint of $AB$ , so that $A$ and $B$ have coordinates $(-5,0)$ and $(5,0)$ respectively. A standard result is that the equation of a field line for this problem is $3q \cos\theta + q\cos\phi \; = \; qc$ for some constant $c$ , where $3q$ and $q$ are the charges at $A$ and $B$ . For this particular problem (which has $\theta = \tfrac14\pi$ and $\phi = \pi$ as initial values), we have $c = \tfrac32\sqrt{2} - 1$ , but this is unimportant. Regarding $\phi$ as a function of $\theta$ , we deduce that $3\sin\theta + \sin\phi\,\phi' \; = \; 0$ Using the Sine Rule we have $r \; = \; \frac{10 \sin\phi}{\sin(\phi-\theta)}$ and hence $x(\theta) \; = \; \frac{10 \cos\theta\sin\phi}{\sin(\phi-\theta)} - 5 \qquad \qquad y(\theta) \; = \; \frac{10 \sin\theta\sin\phi}{\sin(\phi-\theta)}$ But then $\begin{array}{rcl} x'(\theta) & = & \displaystyle \frac{10}{\sin^2(\phi-\theta)}\big[-\sin\theta\cos\theta \phi' + \sin\phi \cos\phi\big] \; = \; \frac{10}{\sin\phi\sin^2(\phi-\theta)}\big[3\sin^2\theta\cos\theta + \sin^2\phi\cos\phi\big] \\ y'(\theta) & = & \displaystyle \frac{10}{\sin^2(\phi-\theta)}\big[-\sin^2\theta \phi' + \sin^2\phi\big] \; = \; \frac{10}{\sin\phi\sin^2(\phi-\theta)}\big[3\sin^3\theta + \sin^3\phi\big] \end{array}$ and hence $(5+x(\theta))y'(\theta) - y(\theta)x'(\theta) \; = \; \frac{100\sin^2\phi}{\sin^2(\phi-\theta)}$ The tangent to the fieldline at the point $x(\theta),y(\theta))$ has equation $\big(x - x(\theta)\big)y'(\theta) \; = \; \big(y - y(\theta)\big)x'(\theta)$ and hence meets the $x$ -axis at the point $(X(\theta),0)$ , where $X(\theta) \; = \; \frac{x(\theta)y'(\theta) - y(\theta)x'(\theta)}{y'(\theta)} \; = \; \frac{10\sin^3\phi}{3\sin^3\theta + \sin^3\phi} -5$ In the asymptotic limit, the fieldline tends towards the situation with $\theta = \phi = \alpha$ , where $4\cos\alpha = c$ . It is clear that the limit of $X(\theta)$ is $\tfrac{10}{4} - 5 = -2.5$ , and hence $AC = 2.5$ cm.

It doesnt matter what the angle of emmision is,,

the simple concept is to find an analogy which in this case is centre of mass, you see at large distances the system will behave like a single charge with charge 4uC

and its position is the same as the "centre of Charge" of this system which is at

10/(1+3) = 2.5 cm from the charge q1 hence the asymptote s of all field lines will amanate from this point,, hence the distance AC is 2.5 cm

such a simple and beautiful question