A short electric dipole with dipole moment $\overrightarrow{p}$  is placed coaxially with a ring of radius R at a distance $r_0$  from the center of the ring as shown in figure. The flux of the electric field due to dipole through the area bounded by the ring is $\varphi =\frac{\left|\overrightarrow{p}\right|R^{\alpha }}{4{\epsilon }_0{[R^2+r_0^2]}^{\frac{\beta }{2}}}.$ The product  $\alpha \beta$  is _____.

$E\left(n\right)=\frac{\overline{P}}{4\pi {\epsilon }_0{z}^3}\sqrt{3{\cos }^2\theta +1}$

and  $\tan \theta =\frac{\tan \theta }{2}$

$$\begin{gathered} \therefore d\varphi =\frac{2\pi \times dx.\overline{P}}{4\pi {\epsilon }_0{z}^3}.\sqrt{3{\cos }^2\theta +1} \\ \left[\text{cosαcosθ-sinαsinθ}\right] \\ \Rightarrow d\varphi =\frac{2\pi }{4\pi {\epsilon }_0}\frac{\times dx}{z^3}\sqrt{3{\cos }^2\theta +1} \\ \left[\frac{{\text{cos}}^{\text{2}}\text{θ.2}}{\sqrt{{\text{3cos}}^{\text{2}}\text{θ+1}}}\text{-}\frac{{\text{sin}}^{\text{2}}\text{x}}{\sqrt{{\text{3cos}}^{\text{2}}\text{θ+1}}}\right] \\ \Rightarrow d\varphi \frac{2\pi \left|\overline{P}\right|xdx}{4\pi {\epsilon }_0{z}^3}[3{\cos }^2\theta -1] \\ \Rightarrow d\varphi \frac{\overline{P}}{4{\epsilon }_0{[x^2+r_0^2]}^{\frac{3}{2}}}[\frac{3r_0^2}{x^2+r_0^2}-1]xdx \\ \Rightarrow \varphi =\underset{0}{\overset{R}{\int }}\frac{\overline{P}}{4{\epsilon }_0}[\frac{3r_0^2}{{(x^2+r_0^2)}^{\frac{5}{2}}}-\frac{1}{{(x^2+r_0^2)}^{\frac{3}{2}}}]2xdx \\ \Rightarrow \varphi =\frac{P}{4{\epsilon }_0}{[-\frac{3r_0^2\times 2}{3{[x^2+r_0^2]}^{\frac{3}{2}}}+\frac{2}{{(x^2+r_0^2)}^{\frac{1}{2}}}]}_0^R \\ \Rightarrow \varphi =\frac{\left|\overline{P}\right|}{4{\epsilon }_0}[1-\frac{r_0^2}{(x^2+r_0^2)}].\frac{2}{{(x^2+r_0^2)}^{\frac{1}{2}}} \\ {\Rightarrow \varphi =\frac{\left|\overline{P}\right|}{4{\epsilon }_0}\frac{2x^2}{[x^2+r_0^2]}|}_0^R \\ \therefore \varphi =\frac{\left|\overline{P}\right|.R^2}{4{\epsilon }_0{[R^2+r_0^2]}^{\frac{3}{2}}} \end{gathered}$$