$A$ charge is uniformly distributed on the surface of a spherical rubber balloon. As it is blown up,the total electric flux coming out of the surface

The electric field in the region is $\vec{E} = a\hat{i} + b\hat{j}$ where $a$ and $b$ are constants. The net electric flux passing through a square area of side $l$ parallel to the $Y-Z$ plane is

$A$ square loop of sides $a=1 \ m$ is held normally in front of a point charge $q=1 \ C$. The charge is placed at a distance of $a/2$ from the center of the square. The flux of the electric field through the shaded region is $\frac{5}{p} \times \frac{1}{\varepsilon_0} \frac{N m^2}{C}$,where the value of $p$ is . . . . . . .

$(a)$ An electrostatic field line is a continuous curve. That is,a field line cannot have sudden breaks. Why not?
$(b)$ Explain why two field lines never cross each other at any point?

For a given surface,the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this,we can conclude that:

$A$ hollow cylinder has a charge '$q$' $C$ within it. If '$\phi$' is the electric flux associated with the curved surface $B$,the flux linked with the plane surface $A$ will be