What will be the total flux through the faces of the cube as in figure with side of length $a$ if a charge $q$ is placed at ?

$(a)$ $A$ $:$ a corner of the cube.

$(b)$ $B$ $:$ midpoint of an edge of the cube.

The electric field in a region is given $\overrightarrow{ E }=\left(\frac{3}{5} E _{0} \hat{ i }+\frac{4}{5} E _{0} \hat{ j }\right) \frac{ N }{ C } .$ The ratio of flux of reported field through the rectangular surface of area $0.2\, m ^{2}$ (parallel to $y – z$ plane) to that of the surface of area $0.3\, m ^{2}$ (parallel to $x – z$ plane $)$ is $a : b ,$ where $a =$ ………….

[Here $\hat{ i }, \hat{ j }$ and $\hat{ k }$ are unit vectors along $x , y$ and $z-$axes respectively]

In figure $+Q$ charge is located at one of the edge of the cube, then electric flux through cube due to $+Q$ charge is

The electric field components in Figure are $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0,$ in which $\alpha=800 \;N / C\, m ^{1 / 2} .$ Calculate

$(a)$ the flux through the cube, and

$(b)$ the charge within the cube. Assume that $a=0.1 \;m$

Three positive charges of equal value $q$ are placed at vertices of an equilateral triangle. The resulting lines of force should be sketched as in