Two charges +q and –q are fixed closely on x-axis as shown. Consider a region in y-z plane ${\mathrm{a}}^2\le {\mathrm{y}}^2+{\mathrm{z}}^2\le {\mathrm{b}}^2$ ,   (a>>>d).

Choose the correct statement(s).

For the considered region given charge distribution will act as a dipole system. Therefore electric field anywhere in the region is directed towards +ve x-axis. 

Due to symmetry we can conclude that electric potential anywhere in y-z plane will be zero. So the work done from bringing a positive test change from $\left(0,\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}\right)$ to $\left(a,\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}\right)$ is zero.

Electric flux crossing the region:

Flux due to $+q$: ${\mathrm{\varphi }}_{+\mathrm{q}}=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}\left(1-\mathrm{cos\beta }\right)-\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}\left(1-\mathrm{cos\alpha }\right)=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}\left(\mathrm{cos\alpha }-\mathrm{cos\beta }\right)$

similarly, Flux due to $-q :$ ${\mathrm{\varphi }}_{-\mathrm{q}}=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}\left(1-\mathrm{cos\beta }\right)-\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}\left(1-\mathrm{cos\alpha }\right)=\frac{\mathrm{q}}{2{\mathrm{\epsilon }}_0}\left(\mathrm{cos\alpha }-\mathrm{cos\beta }\right)$

$\because \mathrm{d}<<<\mathrm{a} \Rightarrow \mathrm{cos\alpha }=\frac{\mathrm{d}}{\sqrt{{\mathrm{a}}^2+{\mathrm{d}}^2}}\approx \frac{\mathrm{d}}{\mathrm{a}},\mathrm{cos\beta }=\frac{\mathrm{d}}{\sqrt{{\mathrm{b}}^2+{\mathrm{d}}^2}}\approx \frac{\mathrm{d}}{\mathrm{b}}$

Net flux through the region: ${\mathrm{\varphi }}_{\mathrm{T}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\left(\mathrm{cos\alpha }-\mathrm{cos\beta }\right)=\frac{\mathrm{dq}}{{ϵ}_0}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right)$