Warped Gravity

The repulsive force between the two rods is perpendicular to the rods, and of magnitude $\begin{aligned} F_1 & = \; \frac{\lambda^2}{4\pi\varepsilon_0}\int_0^L\int_0^L \frac{dx\,dy}{\Delta^2 + (x-y)^2} \times \frac{\Delta}{\sqrt{\Delta^2 + (x-y)^2}} \; = \; \frac{\lambda^2D}{4\pi\varepsilon_0}\int_0^1\int_0^1 \frac{dX\,dY}{(D^2 + (X-Y)^2)^{\frac32}} \end{aligned}$ where $\Delta = DL$ , using the substitution $x = LX$ , $y = LY$ . With the double substitution $U = X+Y$ , $V + X-Y$ , we have $\begin{aligned} \int_0^1\int_0^1 \frac{dX\,dY}{(D^2 + (X-Y)^2)^{\frac32}} & = \; \tfrac12\int_{-1}^1\,dV \int_{|V|}^{2-|V|} \frac{dU}{(D^2 + V^2)^{\frac32}} \; = \; \int_{-1}^1 \frac{1-|V|}{(D^2 + V^2)^{\frac32}}\,dV \\ & = \; 2\int_0^1 \frac{1-V}{(D^2 + V^2)^{\frac32}} \; = \; \frac{2(\sqrt{D^2+1} - D)}{D^2} \end{aligned}$ where the last integral is performed by standard tricks, and so $F_1 \;= \; \frac{2\lambda^2}{4\pi\varepsilon_0}\frac{\sqrt{D^2+1} - D}{D}$ Smiilarly, the attractive gravitational force between the rods, and of magnitude $F_2 \; = \; G\beta^2\int_0^L\int_0^L (\Delta^2 + (x-y)^2) \frac{\Delta}{\sqrt{\Delta^2 + (x-y)^2}}\,dx\,dy \; = \; G\beta^2L^4D\int_0^1 \int_0^1\sqrt{D^2 + (X-Y)^2}\,dX\,dY$ and the same substitution and similar standard integration tricks give $F_2 \; = \; \tfrac13G\beta^2L^4D\left\{\sqrt{D^2+1} + 2D^3 - 2D^2\sqrt{D^2+1} + 3D^2\ln\left(\tfrac{\sqrt{D^2+1}+1}{D}\right)\right\}$ We want to find the equilibrium position, namely the value of $D$ for which $F_1= F_2$ . This simplifies down to the equation $\frac{6\lambda^2}{4\pi\varepsilon_0 G\beta^2 L^4} \; = \; D^2\left[1 - D^2 + D\sqrt{D^2+1} + 3D^2(\sqrt{D^2+1} + D)\ln\left(\tfrac{\sqrt{D^2+1} + 1}{D}\right)\right]$ With the given values of the parameters, we can solve this equation numerically to obtain $D = 0.29166666667$ , and hence $\Delta = \boxed{7}$ .

It is a little weird that $G$ is given to such precision, and yet $\tfrac{1}{4\pi\varepsilon_0}$ is given to just one significant figure!