A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $\mathrm{x}$-axis are shown in the figure. These lines suggest that $Image$

$(A)$ $\left|Q_1\right|>\left|Q_2\right|$

$(B)$ $\left|Q_1\right|<\left|Q_2\right|$

$(C)$ at a finite distance to the left of $\mathrm{Q}_1$ the electric field is zero

$(D)$ at a finite distance to the right of $\mathrm{Q}_2$ the electric field is zero

A charged shell of radius $R$ carries a total charge $Q$. Given $\Phi$ as the flux of electric field through a closed cylindrical surface of height $h$, radius $r$ and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct ? $\epsilon_0$ is the permittivity of free space]

$(1)$ If $h >2 R$ and $r > R$ then $\Phi=\frac{ Q }{\epsilon_0}$

$(2)$ If $h <\frac{8 R }{5}$ and $r =\frac{3 R }{5}$ then $\Phi=0$

$(3)$ If $h >2 R$ and $r =\frac{4 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$

$(4)$ If $h >2 R$ and $r =\frac{3 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3} \;Nm ^{2} / C .$

$(a)$ What is the net charge inside the box?

$(b)$ If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Figure shows electric field lines due to a charge configuration, from this we conclude that

Gauss’s law is true only if force due to a charge varies as

What is called Gaussian surface ?