Geometry of the field lines

Suppose that the left-hand charge is at the point $(-a,0)$ , and that the right-hand charge is at the point $(a,0)$ . Then the electrostatic potential $\phi$ at the point $P$ of the pair of charges is $\phi(P) \; = \; \frac{q}{4\pi\varepsilon_0}\Big(\frac{1}{r_1} + \frac{1}{r_2}\Big)$ where $r_1$ is distance of $P$ from the left-hand particle, and $r_2$ is the distance of $P$ from the right-hand particle. This potential is axially symmetric, and so we can restrict ourselves to considering a two-dimensional problem. Thus $\mathbf{E}\; = \; -\nabla \phi \; = \; \frac{q}{4\pi\varepsilon_0}\Big[r_1^{-3}{x+a \choose y} + r_2^{-3}{x-a \choose y} \Big]$ and hence the electric field lines are solutions of the differential equation $\frac{dx}{r_1^{-3}(x+a) + r_2^{-3}(x-a)} \; = \; \frac{dy}{r_1^{-3}y + r_2^{-3}y}$ or $\big[r_1^{-3}y\,dx - r_1^{-3}(x+a)\,dy\big] + \big[r_2^{-3}y\,dx - r_2^{-3}(x-a)\,dy\big] \; = \; 0$ If we define $\theta_1$ and $\theta_2$ to be the angles that the lines from $P$ to the left-hand and right-hand charges make with the positive $x$ -axis, then $\cos\theta_1 \; = \; \frac{x+a}{r_1} \qquad \cos\theta_2 \; = \; \frac{x-a}{r_2}$ and hence it can be shown that $\begin{array}{rcl} -\sin\theta_1\,d\theta_1 & = & y[y\,dx - (x+a)\,dy]r_1^{-3} \\ -\sin\theta_2\,d\theta_2 & = & y\big[y\,dx - (x-a)\,dy]r_2^{-3} \end{array}$ and hence the general equation of an electric field line is $\cos\theta_1 + \cos\theta_2 \; = \; c$ For line $A$ , we have $\theta_1 \to \alpha$ and $\theta_2 \to \pi$ as $P$ approaches the left-hand charge, where $\alpha=\tfrac14\pi$ . Similarly, for the line $B$ we have $\theta_1 \to 0$ and $\theta_2 \to \pi-\alpha$ as $P$ approaches the right-hand charge. Thus the equations of the two lines are $\cos\theta_1 + \cos\theta_2 \; = \; \cos\alpha-1 \qquad \qquad \cos\theta_1 + \cos\theta_2 \; = \; 1 - \cos\alpha$ It is clear that lines $A$ and $B$ are mirror images of each other in the central line $x=0$ . Thus the smallest distance $d$ between them is equal to $2x_\mathrm{min}$ , where $x_{\mathrm{min}}$ is the least positive value of $x$ attained on the line $B$ .

At the point $x=x_\mathrm{min}$ , we have $\tfrac{dx}{dy} = 0$ , and hence $(x+a)r_1^{-3} + (x-a)r_2^{-3} = 0$ . Thus we need to solve the simultaneous equations $\cos\theta_1 + \cos\theta_2 \; = \; 1 - \cos\alpha \qquad \qquad r_1^{-2}\cos\theta_1 + r_2^{-2}\cos\theta_2 \; = \; 0$ Multiplying the second by $y^2\,=\,r_1^2\sin^2\theta_1\,=\,r_2^2\sin^2\theta_2$ , we deduce that $\begin{array}{rcl} 0 & = & \sin^2\theta_1\cos\theta_1 + \sin^2\theta_2\cos\theta_2\\ & = & \cos\theta_1+\cos\theta_2 - \cos^3\theta_1-\cos^3\theta_2 \\ & = & (\cos\theta_1 + \cos\theta_2)(1 - \cos^2\theta_1 + \cos\theta_1\cos\theta_2 - \cos^2\theta_2) \\ & = & (\cos\theta_1 + \cos\theta_2)\big[1 - (\cos\theta_1+\cos\theta_2)^2 + 3\cos\theta_1\cos\theta_2\big] \end{array}$ Hence $3\cos\theta_1\cos\theta_2 \,=\, (1-\cos\alpha)^2-1 \,=\,4\sin^4\tfrac12\alpha - 1$ . Thus $\cos\theta_1) and $\cos\theta_2$ are the roots of the quadratic $3X^2 - 6\sin^2\tfrac12\alpha \,X + 4\sin^4\tfrac12\alpha - 1 \; = \; 0$ with $\cos\theta_1 > 0 > \cos\theta_2$ . For these values of $\theta_1$ and $\theta_2$ we have $r_1\cos\theta_1 - r_2\cos\theta_2 \,=\,2a\qquad \qquad r_1\sin\theta_1 \,=\,r_2\sin\theta_2$ and hence $r_1 \; = \; \frac{2a \sin\theta_2}{\sin(\theta_2-\theta_1)}$ and thus $x_{\mathrm{min}}\; = \; r_1\cos\theta_1 - a \; = \; \frac{2a \cos\theta_1\sin\theta_2}{\sin(\theta_2-\theta_1)} - a$ With $\alpha = \tfrac14\pi$ and $a=5$ , we obtain $d \; = \; 2x_\mathrm{min} \; = \; 3.743$

$d$ will be min when electric field line is perpendicular to x axis.

Let the distance along axis of the point from the left charge where this happens be x and the y - coordinate be y.

Now let us consider a circle tangent to this field line passing through this point and another such point in downward direction having its plane along Y-Z plane. Let us consider another such circle at the beginning of the field line tangent to it and very close to $x = -5$ and in Y-Z plane.

Flux through both the circles must be equal.

$\Rightarrow \frac{q}{2 \in_{0}}[ 1 - \frac{1}{\sqrt{2}}] = \frac{q}{2 \in_{0}}[ [1 - \frac{x}{\sqrt{x^2 + y^2}} ] - [ 1- \frac{(d - x)}{\sqrt{(d - x)^2 + y^2}}]]$

$\Rightarrow \frac{(d - x)}{\sqrt{(d - x)^2 + y^2}} - \frac{x}{\sqrt{x^2 + y^2}} = 1 - \frac{1}{\sqrt{2}} ...... (i)$

Now , $E_{x}$ = 0 at this point.Hence $E_{x}$ of both charges will cancel.

$\Rightarrow \frac{kq}{x^2 + y^2}\frac{x}{\sqrt{x^2 + y^2}} = \frac{kq}{(10 - x)^2 + y^2}\frac{(10 - x)}{\sqrt{(10 - )x^2 + y^2}} ...... (ii)$

Solving $(i)$ and $(ii)$ , we get

$x = \fbox{3.13}$ and $y= \fbox{6.67}$

Hence , $d_{min} = 10 - 2*(3.13) = \fbox{3.74}$ .